Book Name | : Ganit Prakash |
Subject | : Mathematics (Maths) |
Class | : 9 (Madhyamik/WB) |
Publisher | : Prof. Nabanita Chatterjee |
Chapter Name | : Area And Perimeter Of Triangle and Quadrilateral (15th Chapter) |
Let us Do – 15.1
I see the following figure and find out the perimeter.
Perimeter = 12 cm +13cm + 14.8cm + 16.2cm + 10cm = 66cm (Ans.)
Perimeter = 7cm+19.4cm+21cm+ 10cm = 57.4 cm (Ans.)
Perimeter = 3cm+5.6cm+19cm+ 12cm = 39.6 cm (Ans.)
Perimeter = 8cm+19cm+6cm+15cm +9cm+6cm = 63cm (Ans.)
Perimeter = 9cm+16cm+26cm+ 12cm= 63cm (Ans.)
Let us Do – 15.2
1. If in a square land the length of diagonal is 20 \sqrt{2} meter, let us write by calculating that how much length in meter is required for fencing a wall surrounding of it.
Solution :
Let ‘ a ‘ e the side of a square
\therefore Diagonal = a \sqrt{2}
But, \quad a \sqrt{2}=20 \sqrt{2}
\therefore a = 20
\therefore \quad Length for fencing a wall surrounding of square
=4 \times \text { side } \\
=4 \times 20 \mathrm{~m}=80 \mathrm{~m} \text { (Ans.) }
2. The rectangular land of Pritma has 5 meter wide path all around it on the out side. The length and width of the rectangular land 2.5Dm and 1.7Dm respectively. Let us write by calulating the how much cost will be required for fencing around the other side of path at the rate of Rs. 18 per in meter.
Solution :
The rectangular land of Pritma has 5 meter wide path all around it on the out side.
\therefore Length of rectangular land = 2.5Dm = 25m.
Width \text{ " " " " } = 1.7Dm = 17m.
Length of the rectangular field with path
=(27+2 \times 5) \mathrm{cm}. \\
=(25+10) \mathrm{cm} \quad=35 \mathrm{~cm}.
Breadth of the rectangular field with path
=(17+2 \times 5) \\
=(17+10) \mathrm{cm}=27 \mathrm{~cm}.
\therefore Length for fencing all around the rectangular field
=2 \text { (Length }+ \text { Breadth) } \\
=2(35+27) \mathrm{cm}. =2 \times 62 \mathrm{~cm}=124 \mathrm{~cm}
Cost for fencing = Rs. 18 \times 124
= Rs. 2232 (Ans.)
3. Let us see the card below, find perimeter and let us write by calculating what will be the length of one side of equilateral triangle with same perimeter.
Solution:
1.
Perimeter = 2(18+12) \mathrm{cm} \\
=2 \times 30 \mathrm{~cm}=60 \mathrm{~cm} \text {. (Ans) }
2.
Perimeter = 4 \times 9 \\
= 36 \mathrm{~cm} \text {. (Ans) }
3.
Perimeter = 8 \mathrm{~cm}+9 \mathrm{~cm}+15 \mathrm{~cm}+7 \mathrm{~cm} \\
=39 \mathrm{~cm} \text {. (Ans) }
4.
Perimeter = 12 \mathrm{~cm}+12 \mathrm{~cm}+21 \mathrm{~cm}+21 \mathrm{~cm}
=66 \mathrm{~cm}. (Ans)
5.
Perimeter = 5 \mathrm{~cm}+13 \mathrm{~cm}+12 \mathrm{~cm} \\
=30 \mathrm{~cm} \text {. (Ans) }
6.
Perimeter = 14 \mathrm{~cm}+14 \mathrm{~cm}+17 \mathrm{~cm} \\
=45 \mathrm{~cm} \text {. (Ans) }
Let us Do – 15.3
1. Look at the figures below and let us write by calculation, the area.
(i)
(ii)
(iii)
(iv)
Solution:
(i) We have,
AC = 13cm, BC = 5cm
For right angled triangle ABC, we get,
A B^2=A C^2-B C^2
or, A B^2=(13)^2-(5)^2
or, A B^2=169-5
or, \mathrm{AB}^2=144
or, \cdot \mathrm{AB}=\sqrt{144}
or, \mathrm{AB}=12
\therefore Area of \triangle \mathrm{ABC} =\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB} \\
=\frac{1}{2} \times 5 \times 12 \mathrm{Sq} . \mathrm{m} \\
=30 \text { Sq.m (Ans.) }
(ii) Area of \triangle \mathrm{ABC} =\frac{\sqrt{3}}{4} \times(6)^2 \\
=\frac{\sqrt{3}}{4} \times{ }^9 36 \\
=9 \sqrt{3} \text { Sq.cm (Ans.) }
(iii) Area of \triangle \mathrm{ABC}</p>
<p dir="auto">=\frac{1}{2} \times 8 \times \sqrt{(6)^2-(\frac{8}{2})^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{\cancel2} \times \cancel8 \times \sqrt{(36)-16} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \sqrt{20} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \sqrt{2 \times 2 \times 5} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times \sqrt{5} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=8 \sqrt{5} \text { Sq.cm (Ans.) }</span>
<p dir="auto"><img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23565" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3290-300x238.jpg" alt="IMG 3290" width="300" height="238" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 18" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3290-300x238.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3290-1024x812.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3290-768x609.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3290-24x19.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3290-36x29.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3290-48x38.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3290.jpg 1526w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto">(iv) Let a = 16cm, b = 14cm, c = 10cm.</p>
<p dir="auto">s <span class="katex-eq" data-katex-display="false"> =\frac{a+b+c}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{16+1+\therefore}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{40}{2} \mathrm{~cm}=20 \mathrm{~cm} .</span>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\triangle B C D =\sqrt{s(s-a)(s-b)(s-c)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{20(20-16)(20-14)(20-10)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{20 \times 4 \times 6 \times 10} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{10 \times 2 \times 2 \times 2 \times 2 \times 3 \times 10} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 2 \times 10 \sqrt{3} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=40 \sqrt{3} \mathrm{Sq} . \mathrm{cm}(\text { Ans. })</span>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\angle \mathrm{ABCD}=2 \times \text { area of } \triangle \mathrm{BCD} =2 \times 40 \sqrt{3} \text { Sq.cm. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=80 \sqrt{3} \quad \text { (Ans.) }</span>
<p dir="auto"><img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23566" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3291-300x191.jpg" alt="IMG 3291" width="300" height="191" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 19" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3291-300x191.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3291-1024x653.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3291-768x490.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3291-24x15.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3291-36x23.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3291-48x31.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3291.jpg 1043w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto"><strong>2. In a lake of Botanical Garden the tip of lotus was seen 2cm. above the surface of water. Being forced by the wind, it gradually advanced and submerged at a distance of 15cm. from the previous position. Let us write by calculating the depth of water</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto">Let the depth of water be x.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad A C=x, A B=15 \mathrm{~cm} \text {. }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad B C=(x+2) \mathrm{cm} \text {. }</span>
<p dir="auto">We have,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BC}^2=\mathrm{AB}^2+\mathrm{AC}^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(x+2)^2=(15)^2+(x)^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(x)^2+2 \cdot x \cdot 2+(2)^2=225+x^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \cancel{x^2}+4 x+4=225+\cancel{x^2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">4 x=225-4</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 4 \mathrm{x}=221</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x=\frac{221}{4}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{x}=55.25</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span>Depth of water =55.25cm (Ans.)</p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23517" src="https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C-300x244.jpeg" alt="F561C260 2948 4FCC 8DB9 C55C2CDCC67C" width="470" height="382" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 20" srcset="https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C-300x244.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C-1024x831.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C-768x623.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C-1536x1247.jpeg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C-24x19.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C-36x29.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C-48x39.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/F561C260-2948-4FCC-8DB9-C55C2CDCC67C.jpeg 1626w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><strong>3. The length of hypotenuse of an isosceles right-angle triangle is <span class="katex-eq" data-katex-display="false">12 \sqrt{2} \mathrm{~cm},</span> Let us write by calculating what wi be the area of that fied.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\because \triangle A B C</span> is an lsosceles right angled triangle.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{AB}=\mathrm{BC}</span>
<p dir="auto">Let AB = BC = x</p>
<p dir="auto">We know that,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(x)^2+(x)^2=(12 \sqrt{2})^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x^2+x^2=144 \times 2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\cancel2 x^2=144 \times \cancel2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x^2=144</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } x=\sqrt{144} \quad \therefore \quad x=12</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{AB}=\mathrm{BC}=12 \mathrm{~cm}.</span>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\triangle A B C=\frac{1}{2} \times B C \times A B</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 12 \times 12 \mathrm{sq} \cdot \mathrm{cm}=72 \mathrm{Sq} \cdot \mathrm{cm}</span>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23516" src="https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F-254x300.jpeg" alt="076A9D59 2939 42B6 8174 F763E226394F" width="470" height="555" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 21" srcset="https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F-254x300.jpeg 254w, https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F-867x1024.jpeg 867w, https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F-768x907.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F-1301x1536.jpeg 1301w, https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F-20x24.jpeg 20w, https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F-30x36.jpeg 30w, https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F-41x48.jpeg 41w, https://flasheducation.online/wp-content/uploads/2023/06/076A9D59-2939-42B6-8174-F763E226394F.jpeg 1725w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><strong>4. The lengths of three sides of our trianglular park are 65m, 70m and 75m. Let us write by calculating the length of perpendicular drawn from opposite vertex on the long side.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let } a=65 m, b=70 m, c=75 m \\</span>
<p dir="auto">s<span class="katex-eq" data-katex-display="false">=\frac{a+b+c}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{65+70+75}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{210}{2}=105 \mathrm{~m}</span>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\triangle A B C =\sqrt{s(s-a)(s-b)(s-c)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{105(105-65)(105-70)(105-75)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{105 \times 40 \times 35 \times 30} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{35 \times 3 \times 10 \times 2 \times 2 \times 35 \times 3 \times 10} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=35 \times 3 \times 2 \times 10 \mathrm{Sq} . \mathrm{m} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2100 \mathrm{Sq} . \mathrm{m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> We also know that,</p>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{CD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 2100=\frac{1}{2} \times 75 \times C D \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 75 \mathrm{CD}=4200 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } C D=\frac{4200}{75} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad C D=56 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Length of perpendicular }=56 \mathrm{~m} \text { (Ans.) } \\</span>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23515" src="https://flasheducation.online/wp-content/uploads/2023/06/733D086B-AEC4-44A3-972C-1AA9420A7E6C-300x211.jpeg" alt="733D086B AEC4 44A3 972C 1AA9420A7E6C" width="470" height="331" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 22" srcset="https://flasheducation.online/wp-content/uploads/2023/06/733D086B-AEC4-44A3-972C-1AA9420A7E6C-300x211.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/733D086B-AEC4-44A3-972C-1AA9420A7E6C-1024x720.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/733D086B-AEC4-44A3-972C-1AA9420A7E6C-768x540.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/733D086B-AEC4-44A3-972C-1AA9420A7E6C-24x17.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/733D086B-AEC4-44A3-972C-1AA9420A7E6C-36x25.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/733D086B-AEC4-44A3-972C-1AA9420A7E6C-48x34.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/733D086B-AEC4-44A3-972C-1AA9420A7E6C.jpeg 1203w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><strong>5. The ratio of height of the two triangles which are drawn by Suja and I is 3 : 4 and the ratio of their area is 4 : 3. Let us write by calculating what will be the ratio of two bases.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> We have.</p>
<p dir="auto">Ratio of the twa triangles = 3 : 4</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Ratio of area of the two triangles = 4 : 3</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \frac{\text { Area of } 1 \text { st. triangle }}{\text { Area of } 2 \text { nd. triangle }}=\frac{4}{3}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \frac{\text { Base of } 1 \text { st. triangle }}{\text { Base of } 2 \text { nd. triangle }} \times \frac{\text { Height } 1 \text { st triangle }}{\text { Height } 2 \text { nd triangle }}=\frac{4}{3}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{\text { Base of } 1 \text { st. triangle }}{\text { Base of } 2 \text { nd. triangle }} \times \frac{3}{4}=\frac{4}{3}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \frac{\text { Base of } 1 \text { st. triangle }}{\text { Base of } 2 \text { nd triangle }} \times \frac{16}{9}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Ratio of two bases = 16 : 9 (Ans.)</p>
<hr />
<h2 dir="auto" style="text-align: center;"><span class="ez-toc-section" id="Let_us_work_out_-_151"></span><em><span style="text-decoration: underline;">Let us work out - 15.1</span></em><span class="ez-toc-section-end"></span></h2>
<p dir="auto"><strong>1. I see the house of Kamal and let us find the answers.</strong></p>
<p dir="auto"><strong>(i) Let us write by calculating the area of Kamal's garden.</strong></p>
<p dir="auto"><strong>(ii) Let us write by calculating how much cost is required to repair the foor of Kamal's varandah at the rate of Rs. 30/m.</strong></p>
<p dir="auto"><strong>(iii) Kamal wants to cover the floor of his reading room with tiles. Let us write by calculating how many tiles will be required to cover his floor of reading room with size of tiles 25cm. × 25cm.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto">(i) Area of Kamal's garden <span class="katex-eq" data-katex-display="false">=20 \mathrm{~m} \times 20 \mathrm{~m}=400 \mathrm{Sq} . \mathrm{m}. (Ans)</span>
<p dir="auto">(ii) Area of Kamal's varandah <span class="katex-eq" data-katex-display="false">=10 \mathrm{~m} \times 5 \mathrm{~m}=50 \mathrm{sq} . \mathrm{m}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Cost of repair the floor of Kamal's varandah <span class="katex-eq" data-katex-display="false">= Rs. 30 \times 50</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\mathrm{Rs} .1500(\mathrm{Ans} \text {.) }</span>
<p dir="auto">(iii) Length of reading room <span class="katex-eq" data-katex-display="false">=6 \mathrm{~m}=600 \mathrm{~cm}.</span>
<p dir="auto">Breadth <span class="katex-eq" data-katex-display="false">\text{ " " " " " } =5 \mathrm{~m}=500 \mathrm{~cm}.</span>
<p dir="auto">Size of one tile <span class="katex-eq" data-katex-display="false">=25 \mathrm{~cm} . \times 25 \mathrm{~cm}.</span>
<p dir="auto">No. of tiles <span class="katex-eq" data-katex-display="false">=\frac{600 \times 500}{25 \times 25}=480 (Ans.)</span>
<p dir="auto"><strong>2. Let us see the following pictures and calculate the area of its colored part.</strong></p>
<p dir="auto">(i) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23518" src="https://flasheducation.online/wp-content/uploads/2023/06/2E7DDD6C-C6B8-4AAD-9F52-9B989DC8C8C3-300x235.jpeg" alt="2E7DDD6C C6B8 4AAD 9F52 9B989DC8C8C3" width="300" height="235" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 23" srcset="https://flasheducation.online/wp-content/uploads/2023/06/2E7DDD6C-C6B8-4AAD-9F52-9B989DC8C8C3-300x235.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/2E7DDD6C-C6B8-4AAD-9F52-9B989DC8C8C3-1024x801.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/2E7DDD6C-C6B8-4AAD-9F52-9B989DC8C8C3-768x601.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/2E7DDD6C-C6B8-4AAD-9F52-9B989DC8C8C3-24x19.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/2E7DDD6C-C6B8-4AAD-9F52-9B989DC8C8C3-36x28.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/2E7DDD6C-C6B8-4AAD-9F52-9B989DC8C8C3-48x38.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/2E7DDD6C-C6B8-4AAD-9F52-9B989DC8C8C3.jpeg 1173w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto">(ii) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23520" src="https://flasheducation.online/wp-content/uploads/2023/06/6D4BD9B8-2722-40AE-9C1E-3DCE548CFC73-300x244.jpeg" alt="6D4BD9B8 2722 40AE 9C1E 3DCE548CFC73" width="300" height="244" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 24" srcset="https://flasheducation.online/wp-content/uploads/2023/06/6D4BD9B8-2722-40AE-9C1E-3DCE548CFC73-300x244.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/6D4BD9B8-2722-40AE-9C1E-3DCE548CFC73-1024x833.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/6D4BD9B8-2722-40AE-9C1E-3DCE548CFC73-768x625.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/6D4BD9B8-2722-40AE-9C1E-3DCE548CFC73-24x20.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/6D4BD9B8-2722-40AE-9C1E-3DCE548CFC73-36x29.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/6D4BD9B8-2722-40AE-9C1E-3DCE548CFC73-48x39.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/6D4BD9B8-2722-40AE-9C1E-3DCE548CFC73.jpeg 1092w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto">(iii) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23522" src="https://flasheducation.online/wp-content/uploads/2023/06/D2C79C6E-E03D-45CB-812C-5E7949F0813B-300x212.jpeg" alt="D2C79C6E E03D 45CB 812C 5E7949F0813B" width="300" height="212" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 25" srcset="https://flasheducation.online/wp-content/uploads/2023/06/D2C79C6E-E03D-45CB-812C-5E7949F0813B-300x212.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/D2C79C6E-E03D-45CB-812C-5E7949F0813B-1024x722.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/D2C79C6E-E03D-45CB-812C-5E7949F0813B-768x542.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/D2C79C6E-E03D-45CB-812C-5E7949F0813B-24x17.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/D2C79C6E-E03D-45CB-812C-5E7949F0813B-36x25.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/D2C79C6E-E03D-45CB-812C-5E7949F0813B-48x34.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/D2C79C6E-E03D-45CB-812C-5E7949F0813B.jpeg 1038w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto">(iv) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23521" src="https://flasheducation.online/wp-content/uploads/2023/06/46766ED3-C6E6-4852-A0D7-E4890DCFC984-300x199.jpeg" alt="46766ED3 C6E6 4852 A0D7 E4890DCFC984" width="300" height="199" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 26" srcset="https://flasheducation.online/wp-content/uploads/2023/06/46766ED3-C6E6-4852-A0D7-E4890DCFC984-300x199.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/46766ED3-C6E6-4852-A0D7-E4890DCFC984-1024x678.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/46766ED3-C6E6-4852-A0D7-E4890DCFC984-768x508.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/46766ED3-C6E6-4852-A0D7-E4890DCFC984-24x16.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/46766ED3-C6E6-4852-A0D7-E4890DCFC984-36x24.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/46766ED3-C6E6-4852-A0D7-E4890DCFC984-48x32.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/46766ED3-C6E6-4852-A0D7-E4890DCFC984.jpeg 1278w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto">(v) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23519" src="https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084-300x142.jpeg" alt="3A4353DD B277 4B1C 9BF6 9EB2E6D21084" width="300" height="142" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 27" srcset="https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084-300x142.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084-1024x484.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084-768x363.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084-1536x726.jpeg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084-24x11.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084-36x17.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084-48x23.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/3A4353DD-B277-4B1C-9BF6-9EB2E6D21084.jpeg 1974w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto">(i) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23567" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293-300x222.jpg" alt="IMG 3293" width="300" height="222" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 28" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293-300x222.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293-1024x757.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293-768x568.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293-1536x1136.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293-24x18.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293-36x27.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293-48x36.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3293.jpg 1648w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto">Area of ABCD = AB × AD</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=12 \mathrm{~m} \times 8 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=96 \text { sq. } \mathrm{m} \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{AE}=12 \mathrm{~m}-3 \mathrm{~m}=9 \mathrm{~m} \text {. } \\</span>
<p dir="auto">AG = 8m - 3m = 5m</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \mathrm{AEFG}=\mathrm{AE} \times \mathrm{AG} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=9 \mathrm{~m} \times 5 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=45 \text { sq. } \mathrm{m} \text {. } \\</span>
<p dir="auto">Area of coloured part <span class="katex-eq" data-katex-display="false">=(96-45) Sq.m.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=51 \mathrm{sq} \cdot \mathrm{m} \text {. (Ans.) }</span>
<p dir="auto">(ii) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23568" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3294-300x219.jpg" alt="IMG 3294" width="300" height="219" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 29" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3294-300x219.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3294-1024x749.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3294-768x562.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3294-24x18.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3294-36x26.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3294-48x35.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3294.jpg 1422w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AB}=14 \mathrm{~m}, \quad \mathrm{BC}=26 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \mathrm{ABCD}=\mathrm{AB} \times \mathrm{BC} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=14 \mathrm{~m} \times 26 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=364 \mathrm{sq} . \mathrm{m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{DE}=\frac{26-3}{2}=\frac{23}{2} \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{DG}=\frac{14-3}{2} \mathrm{~m}=\frac{11}{2} \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of DEFG }=\mathrm{DE} \times \mathrm{DG} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{23}{2} \mathrm{~m} \times \frac{11}{2} \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{253}{4} \mathrm{Sq} . \mathrm{m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } 4 \text { small square land }=4 \times \frac{253}{4} \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=253 \mathrm{Sq} \cdot \mathrm{m} \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of coloured part }=(364-253) \text { Sq.m } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=111 \text { Sq.m. (Ans.) } \\</span>
<p dir="auto">(iii) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23569" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3295-300x242.jpg" alt="IMG 3295" width="300" height="242" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 30" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3295-300x242.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3295-1024x825.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3295-768x619.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3295-24x19.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3295-36x29.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3295-48x39.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3295.jpg 1507w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\because \quad A B=16 \mathrm{~m}, \quad A D=9 \mathrm{~m} \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{EF}=16 \mathrm{~m}+2 \times 4 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=16 m+8 m=24 m \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{HE}=9 \mathrm{~m}+2 \times 4 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=9 m+8 m=17 m \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \mathrm{ABCD}=\mathrm{AB} \times \mathrm{AD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=16 \mathrm{~m} \times 9 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=144 \mathrm{Sq} . \mathrm{m} \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \mathrm{EFGH}=\mathrm{EF} \times \mathrm{HE} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=24 \mathrm{~m} \times 17 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=408 \mathrm{Sq} . \mathrm{m} \text {. } \\</span>
<p dir="auto">Area of coloured part <span class="katex-eq" data-katex-display="false">=(408-144) \mathrm{Sqm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=264 \text { Sq.m. (Ans.) }</span>
<p dir="auto">(iv) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23570" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296-300x186.jpg" alt="IMG 3296" width="300" height="186" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 31" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296-300x186.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296-1024x635.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296-768x476.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296-1536x953.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296-24x15.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296-36x22.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296-48x30.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3296.jpg 1614w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\because \quad A B =28 \mathrm{~m}, \quad B C=20 \mathrm{~m} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{EF} =(28-2 \times 3) \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(28-6) \mathrm{m}=22 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{FG} =(20-2 \times 3) \mathrm{m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(20-6) \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=14 \mathrm{~m} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \mathrm{ABCD} =\mathrm{AB} \times \mathrm{BC} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=28 \mathrm{~m} \times 20 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=560 \mathrm{Sq} \cdot \mathrm{m} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of EFGH } =\mathrm{EF} \times \mathrm{FG} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=22 \mathrm{~m} \times 14 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=308 \mathrm{Sq} . \mathrm{m} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of coloured part } =(560-308) \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=252 \mathrm{Sq} \cdot \mathrm{m} . \text { (Ans.) }</span>
<p dir="auto">(v) <img loading="lazy" decoding="async" class="alignnone size-medium wp-image-23571" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-300x132.jpg" alt="IMG 3297" width="300" height="132" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 32" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-300x132.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-1024x451.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-768x338.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-1536x676.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-2048x902.jpg 2048w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-24x11.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-36x16.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3297-48x21.jpg 48w" sizes="(max-width: 300px) 100vw, 300px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\because \quad \mathrm{BC}=120 \mathrm{~cm}, \quad \mathrm{CD}=90 \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \mathrm{ABCD}=\mathrm{BC} \times \mathrm{CD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=120 \mathrm{~cm} \times 90 \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=10800 \mathrm{Sq} . \mathrm{cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{II}=\mathrm{BC}=120 \mathrm{~cm}, \mathrm{IJ}=3 \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of IJKL }=\mathrm{BC} \times \mathrm{IJ} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=120 \mathrm{~cm} \times 3 \mathrm{~cm}=360 \mathrm{Sq} . \mathrm{cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{EH}=3 \mathrm{~cm}, \quad \mathrm{FE}=\mathrm{BJ}=\frac{90-3}{2}=\frac{87}{2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{IL}=\mathrm{BC}=120 \mathrm{~cm}, \mathrm{IJ}=3 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{EH}=3 \mathrm{~cm}, \quad \mathrm{FE}=\mathrm{BJ}=\frac{90-3}{2}=\frac{87}{2}</span>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\mathrm{EFGH}=\mathrm{EH} \times \mathrm{FE}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=3 \mathrm{~cm} \times \frac{87}{2} \mathrm{~cm}=\frac{261}{2} \text { Sq.cm }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } 4 \text { small rectangle } =4 \times \frac{261}{2} \text { Sq.cm } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=522 \mathrm{Sq} . \mathrm{cm} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of coloured part } =(360+522) \text { Sq.cm. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=882 \text { Sq.cm.(Ans.) }</span>
<p dir="auto"><strong>3. The length and breadth of rectangular tield of Birati Mahajati Sangha are in the ratio 4 : 3. The path of 336 meter is covered by walking once round the field. Let us write by calculating the area of the field.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23535" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3273-300x170.jpg" alt="IMG 3273" width="470" height="266" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 33" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3273-300x170.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3273-1024x579.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3273-768x434.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3273-24x14.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3273-36x20.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3273-48x27.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3273.jpg 1038w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let length } =4 x, \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { breadth } =3 x,</span>
<p dir="auto">x, being the common ration.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Perimeter } =2(4 x+3 x) \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 7 x \\</span>
<p dir="auto">= 14x</p>
<p dir="auto">According to the condition of the problem,</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">14 \mathrm{x}=336</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } x=\frac{336}{14} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{x}=24 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Length }=4 x=4 \times 24=96 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Breadth }=3 \mathrm{x}=3 \times 24=72 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Area of the rectangular field }=\text { Length } \times \text { Breadth } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=96 \mathrm{~m} \times 72 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=6912 \text { Sq.m (Ans.) } \\</span>
<p dir="auto"><strong>4. The cost of farming a square land of Samar at the rate of Rs.3.50 per sq. meter is Rs.1400. Let us calculate how much cost will be for fencing around its four sides with same height of Samar's land at the rate of Rs. 8.5 per meter.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> The cost of farming a square land of Samar at the rate of Rs. 3.50 per sq.meter is Rs. 1400 .</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Area of the square land <span class="katex-eq" data-katex-display="false">=(1400 \div 3.50) sq.m.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\cancel{1400} \times \frac{100}{\cancel{350}} \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= 400 \text { Sq.m. }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of side of a square land } =\sqrt{\text { Area }} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{400} \quad=20 \mathrm{~m} .</span>
<p dir="auto">Length of fencing around its four sides</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \dot{\times} \text { side } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times 20 \mathrm{~m}=80 \mathrm{~m} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Total cost of fencing }=\text { Rs. } 8.50 \times 80 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{850}{100} \times 80 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Rs. } 680 \text { (Ans.) }</span>
<p dir="auto"><strong>5. The area of rectangular land of Suhas's is 500sq. meters. If length of land is decreased by 3 meter and breadth is increased by 2 meter, then the land formed a square. Let us write by calculating the length and breadth of land of Suhas's.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> Let the length of the rectangular land be x</p>
<p dir="auto">and " Breadth " " " " " " " " " y</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Area of rectangular land = 500 sq.m</p>
<p dir="auto">or, Length <span class="katex-eq" data-katex-display="false">\times Breadth =500 \mathrm{sq} . \mathrm{m}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{xy}=500</span>............. (i)</p>
<p dir="auto">If length of land is decreased by 3m and breadth is increased by 2m, then the land formed a square.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad x-3=y+2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x=y+2+3</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x=y+5</span>................ (ii)</p>
<p dir="auto">Putting the value of x in equation</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">(y+5) y=500</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">y^2+5 y-500=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">y^2+25 y-20 y-500=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">y(y+25)-20(y+25)=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(y+25)(y-20)=0</span>
<p dir="auto">either,</p>
<p dir="auto">y + 25 = 0</p>
<p dir="auto">y - 20 = 0</p>
<p dir="auto">y = -25</p>
<p dir="auto">y = 20</p>
<p dir="auto">Since distance can not be negative.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> y = 20</p>
<p dir="auto">From (ii),</p>
<p dir="auto">x = y + 5</p>
<p dir="auto">= 20 + 5 = 25</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Length of the recangular field = 25m.</p>
<p dir="auto">Breadth <span class="katex-eq" data-katex-display="false">\text{ " " " " " " " " "}</span> = 20m (Ans.)</p>
<p dir="auto"><strong>6. Each side of a square land of our village is 300 meter. We shall fence that square land by 3dcm. wide wall with same height around its four sides. Let us see that how much will it cost for the wall at the rate of Rs. 5,000 per 100 sq.meter.</strong></p>
<p dir="auto"><strong>Solution:</strong> Side of a square land = 300m</p>
<p dir="auto">Area <span class="katex-eq" data-katex-display="false">\text{ " " " " " " " " "} =(3.00)^2 \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=90000 \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Wide of the wall }=3 \mathrm{dcm} . \quad=0.3 \mathrm{~m} \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of the square with wide }=(300+2 \times 0.3)^2 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(300+0.6) \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=300.6 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(300.6)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=90360.36 \mathrm{Sq} \cdot \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } 4 \text { walls }=(90360.36-90000) \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=360.36 \text { Sq.m. } \\</span>
<p dir="auto">Cost of the wall 100 Sq.m at the rate of Rs. 5,000</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { " " " } 1 \text { " } 1 \text { " } 360.36 \text { sq.m:" } "</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{5,0\cancel{00}}{1\cancel{00}}=\text { Rs. } 50 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { " " " "360.36sq.m. " "" " " } =\text { Rs. } 360.36 \times 50 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Rs. } \frac{36036}{100} \times 50 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Rs. } 18018 \text { (Ans.) }</span>
<p dir="auto"><strong>7. The length and breadth of recatangular garden of Rehana are 14 meter and 12 meter. If the cost of constructing an equally wide path inside around the garden is Rs. 1,380 at the rate of Rs. 20 per sq.meter, then let us write by calculating how much wide is the path.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23536" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3284-300x204.jpg" alt="IMG 3284" width="470" height="319" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 34" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3284-300x204.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3284-1024x696.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3284-768x522.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3284-24x16.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3284-36x24.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3284-48x33.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3284.jpg 1167w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Length of rectangular garden = 14m</p>
<p dir="auto">Breadth " " " " " " " = 12m</p>
<p dir="auto">Let the width of the park = x</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of rectangular garden } =14-2 × x \\</span>
<p dir="auto">= 14 - 2x</p>
<p dir="auto">Breadth " " " " " " = 12 - 2 × x</p>
<p dir="auto">= 12 - 2x</p>
<p dir="auto">Cost of constructing of wide path of 20 sq.m = Rs. 1380</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text{ " " " " " " " " " " " " "} 1 \ sqm. =Rs\frac{1380}{20} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Rs. } 69</span>
<p dir="auto">According to the condition of the problem</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">(14-2 x)(12-2 x)=69 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 168-28 x-24 x+4 x^2=69 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 4 x^2-52 x+168-69=0 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 4 x^2-52 x+69=0 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 4 x^2-46 x-6 x+69=0 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 2 x(2 x-23)-3(2 x-23)=0 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, }(2 x-23)(2 x-3)=0</span>
<p dir="auto">either,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">2 x-23=0 \quad \text { or, } \quad 2 x-3=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">2 x=23 \quad \text { or, } 2 x=3</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore x=\frac{23}{2}=11.5 \quad \therefore x=\frac{3}{2}=1.5</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> x = 11.5 is not possible because wide can not be greater or equal or nearest to recatangular length</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore x = 1.5</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Wide of the path = 1.5m (Ans.)</p>
<p dir="auto"><strong>8. If the length of recatangular garden with area 1200 <a href="http://sq.cm/" target="_blank" rel="noopener" data-saferedirecturl="https://www.google.com/url?q=http://sq.cm&source=gmail&ust=1687514548185000&usg=AOvVaw0Feg3zj6x1hOr2dbwbQ-fF">sq.cm</a>. is 40cm. then let us write by calculating the area square field which is drawn on its diagonal.</strong></p>
<p dir="auto"><strong>Solution:</strong></p>
<p dir="auto">Length of rectangular garden = 40cm.</p>
<p dir="auto">Area <span class="katex-eq" data-katex-display="false">\text{ " " " " " " " " " " " " "} = 1200 Sq.cm.</span>
<p dir="auto">Breadth <span class="katex-eq" data-katex-display="false">\text{ " " " " " " " " " " " " "} = \frac{\cancel{1200}}{\cancel{40}} cm</span>
<p dir="auto">= 30cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Lenth of diagonal of rectangle } =\text { side of the square. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{(40)^2+(30)^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{1600+900} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{2500} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=50 \mathrm{~cm} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Area of the square field } =(\text { Side })^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(50)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2500 \text { Sq.cm. } \quad \text { (Ans.) }</span>
<p dir="auto"><strong>9. The length, breadth and height of a hall are 4 meter, 6 meter and 4 meter. There are three doors and four windows in the room. The measurement of each door is 1.5 meter × 1 meter and each window is 1.2 meter × 1 meter. How much it will cost for covering four walls by coloured paper at the rate of Rs. 70 per square meter.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> Length = 4m, Breadth = 6cm, height = 4m :</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of one door } =1.5 \times 1 \mathrm{~m} =1.5 \mathrm{Sq} . \mathrm{m} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of three doors } =3 \times 1.5 \mathrm{Sq} . \mathrm{m} =4.5 \mathrm{Sq} . \mathrm{m} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of one window } =1.2 \mathrm{~m} \times 1 \mathrm{~m} =1.2 \mathrm{Sq} . \mathrm{m} \\</span>
<p dir="auto">" " " " <span class="katex-eq" data-katex-display="false"> \quad 4 \text { Windows } =1.2 \times 4 \mathrm{Sq} . \mathrm{m} . =4.8 \mathrm{Sq} . \mathrm{m} .</span>
<p dir="auto">Area of the walls with doors and windows</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \text { (Length }+ \text { Breadth) } \times \text { height. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2(4 \mathrm{~m}+6 \mathrm{~m}) \times 4 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 10 \mathrm{~m} \times 4 \mathrm{~m} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=80 \mathrm{Sq} . \mathrm{m} .</span>
<p dir="auto">Area of the walls without doors and windows.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\{80-(4.5+4.8)\} \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(80-9.3) \mathrm{Sq} \cdot \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=70.7 \mathrm{Sq} \cdot \mathrm{m} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Cost of colouring } =\text { Rs. } 70 \times 70.7 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Rs. } 4949 \quad \text { (Ans.) }</span>
<p dir="auto"><strong>10. The area of four walls of a room is 42sq. meter and area of floor is 12 sq.meter. Let us write by calculating the height of room if the Iength of room is 4 meter.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto">Length of room = 4m</p>
<p dir="auto">Area of the floor <span class="katex-eq" data-katex-display="false">= 12 \mathrm{Sq} \cdot \mathrm{m}</span>
<p dir="auto">or, Length × Breadth = 12</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, Breadth }=\frac{12}{\text { Length }} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{12}{4}=3 \mathrm{~m} \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Area of four walls of a room = 42 <a href="http://sq.cm/" target="_blank" rel="noopener" data-saferedirecturl="https://www.google.com/url?q=http://sq.cm&source=gmail&ust=1687514548185000&usg=AOvVaw0Feg3zj6x1hOr2dbwbQ-fF">sq.cm</a></p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">2 (Length + Breadth) \times height =42</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">2(4+3) \times</span> height =42</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">2 \times 7 \times</span> height =42</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">14 \times</span> height =42</p>
<p dir="auto">or, height <span class="katex-eq" data-katex-display="false">=\frac{\cancel{42}}{\cancel{14}}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore height =3 \mathrm{~m}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore Height of room =3 \mathrm{~m} (Ans)</span>
<p dir="auto"><strong>11. Sujata will draw a rectangular picture un a paper with area 84sq. cm The difference of length and breadth of paper is 5cm. Let us calculate the perimeter of paper of Sujata.</strong></p>
<p dir="auto"><strong>Solution :</strong> Let the length of the rectangular picture be x</p>
<p dir="auto">and " breadth" <span class="katex-eq" data-katex-display="false">\text{ " " " " " " " " " " " " "}</span> y</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Area = 84 Sq.cm.</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x \times y=84</span>
<p dir="auto">or, xy = 84............... (i)</p>
<p dir="auto">According to the condition of the problem,</p>
<p dir="auto">x - y = 5</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \quad x=5+y</span>.......... (ii)</p>
<p dir="auto">Putting the value of x in equation .................. (i)</p>
<p dir="auto">(5 + y) y = 84</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 5 y+y^2=84 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } y^2+5 y-84=0 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } y^2+12 y-7 y-84=0 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } y(y+12)-7(y+12)=0 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, }(y+12)(y-7)=0</span>
<p dir="auto">Either,</p>
<p dir="auto">y + 12 = 0</p>
<p dir="auto">or, y = -12</p>
<p dir="auto">or, y - 7 = 0</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> y = 7</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\because</span> Length can't be negative</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> y=7</p>
<p dir="auto">From (ii),</p>
<p dir="auto">x =5 + y</p>
<p dir="auto">= 5 + 7 = 12</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Length of the rectangular picture = 12m.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Breadth " " " " " } = 7m.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Perimeter " " " " = 2(12+7)m</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 19 \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=38 \mathrm{~m} \text {. (Ans.) }</span>
<p dir="auto"><strong>12. There is a 2.5 meter wide path around the square garden of Shiraj's. The area of path is 165 Sq.meter. Let us calculate the area of garden and the length of diagonal.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> Let the side of the square garden be x There is a 2.5m wide path around the square.</p>
<p dir="auto">Side of square garden with path</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(x+2 \times 25) \quad =(x+5)</span>
<p dir="auto">Accodring to the condition of the problem,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">(x+5)^2-(x)^2=165</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(x)^2+2 \cdot x .5+(5)^2-x^2=165</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\cancel{x^2}+10 x+25- \cancel{x^2}=165</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">10 \mathrm{x}=165-25</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">10 \mathrm{x}=140</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{x}=14</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Side of the square garden</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Side of the square garden } =14 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area" } =(\text { side })^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(14)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=196 \text { Sq.m. }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of diagonal } =\text { side } \sqrt{2} \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=14 \sqrt{2} \mathrm{~m} (Ans.)</span>
<p dir="auto"><strong>13. Let us write by calculating that how much length of wall in meter is required for walling outside round the square field, whereas the length of diagonal of the square land is <span class="katex-eq" data-katex-display="false">20 \sqrt{2}</span> meter. Let us write by calculating how much cost will be for planting grass at the rate of Rs. 20 per sq. meter.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> Length of the diagonal of the square land = <span class="katex-eq" data-katex-display="false">20 \sqrt{2} \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span>Side of the square land = 20m</p>
<p dir="auto">Length of wall for walling outside round the square field = perimeter of the square.</p>
<p dir="auto">= 4 × side</p>
<p dir="auto">= 4 × 20m = 80m</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Area of the square land = <span class="katex-eq" data-katex-display="false">(\text { side })^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(20)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=400 \text { Sq. } \cdot \mathrm{m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Cost for planting grass at the rate of Rs. 20 per Sq.m</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Rs. } 20 \times 400 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Rs. } 8000 \quad \text { (Ans.) }</span>
<p dir="auto"><strong>14. We shall fence our rectangular garden diagonally. The length and breadth of rectangular garden are 12 meter and 7 meter. Let us calculate the length of fence. Also find the perimeter of the two triangles formed by this fence.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> Length of the rectangular garden = 12m.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Breadth" " " " " " " " " " " " }</span> = 7m</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of diagonal } =\sqrt{(\text { Length })^2+(\text { breadth })^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{(12)^2+(7)^2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{193 \mathrm{~m}} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Perimeter of the triangle } =12 \mathrm{~m}+7 \mathrm{~m}+\sqrt{193 \mathrm{~m}} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= (19+\sqrt{193}) \mathrm{m} \text { (Ans.) }</span>
<p dir="auto"><strong>15. A big hall of house of Mousumi is in the form of rectangle, of which length and breadth are in the ratio 9:5 an, perimeter is 140 meter. Mousumi wants to cover the floor of her hall with rectangular tiles of dimensions 25cm, 20cm. The rate of each 100 tiles is Rs.500. Let us calculate the cost for covering the floor with tiles.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto">Let length of rectangle be 9x</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Breadth" " " " " " " " " " " " }</span> 5 x</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Perimeter of rectangle = 140m</p>
<p dir="auto">or, 2 (Length + Breadth )= 140</p>
<p dir="auto">or, 2(9x + 5x) = 140</p>
<p dir="auto">or, x =<span class="katex-eq" data-katex-display="false">\frac{140}{4 \times \sqrt{4}}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> x = 5</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Length of rectangle <span class="katex-eq" data-katex-display="false">= 9 \mathrm{x}=9 \times 5=45 \mathrm{~m}</span>
<p dir="auto">Breadth of rectangle <span class="katex-eq" data-katex-display="false">=5 \mathrm{x}=5 \times 5=25 \mathrm{~m}</span>
<p dir="auto">Area of rectangle <span class="katex-eq" data-katex-display="false">= length \times breadth</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=45 \mathrm{~m} \times 25 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=1125 \text { Sq.m }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Dimension of one tile } =25 \mathrm{~cm} \times 20 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{25}{100} \times \frac{20}{100} \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{20}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { No. of tiles } =1125 \div \frac{1}{20} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=1125 \times 20=22500</span>
<p dir="auto">Rate of each 100 tiles = Rs. 500</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { " " " " }</span> 1tile = Rs. <span class="katex-eq" data-katex-display="false">\frac{500}{100} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { " " " }</span> 22500 Rs. <span class="katex-eq" data-katex-display="false">\frac{500}{100} \times 22500 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\mathrm{Rs}: 112500 \text { (Ans.) }</span>
<p dir="auto"><strong>16. The cost of carpeting a big hall of length 18 meter is Rs.2160. If the breadth of the floor would be 4 meter less, then the cost would have been Rs.1,620. Let us calculate perimeter and area of the hall.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> Length of Rectangle = 18m</p>
<p dir="auto">Let breadth of Rectangle be x</p>
<p dir="auto">Total cost = Rs. 2160</p>
<p dir="auto">Area of Rectangle <span class="katex-eq" data-katex-display="false">= 18 \times x=18 \times Sq.m.</span>
<p dir="auto">If the breadth of the floor would be 4 meter less.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Breadth of Rectangle = x - 4</p>
<p dir="auto">Area of Rectangle <span class="katex-eq" data-katex-display="false">=18(x-4) \mathrm{Sq} . \mathrm{m}</span>
<p dir="auto">Cost of carpeting of 18x Sq. m. = Rs .2160</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text {" " " " " " " " " " " " }</span>18(x - 4) Sq.m = Rs. 1620</p>
<p dir="auto">Cost of carpeting of 18x - 18(x - 4)=Rs. (2160-1620)</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text {" " " " " " " " " " " " }</span>18x - 18x + 72 = Rs. 540</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text {" " " " " " " " " " " " } 72 \mathrm{Sq} . \mathrm{m}</span> .=Rs. 540</p>
<p dir="auto">If Rs. 540, cost of carpeting of 72 Sq.m</p>
<p dir="auto">" 1 <span class="katex-eq" data-katex-display="false"> \text {" " " " " " " " " " " " } \frac{72}{540} \mathrm{Sq} \cdot \mathrm{m}</span>
<p dir="auto">" 1 <span class="katex-eq" data-katex-display="false">\text {" " " " " " " " " " " " } \frac{72 \times 2160}{540} \mathrm{Sq} . \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=288 \mathrm{Sq} . \mathrm{m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad Area of Rectangle =288 Sq.m</span>
<p dir="auto">or, 18x = 288</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x =\frac{288}{18}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> x = 16</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Breadth of Rectangle }=16 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Perimeter " " " " " " "} =2(\text { Length }+ \text { Breadth }) \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2(18 \mathrm{~m}+16 \mathrm{~m}) \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 34 \mathrm{~m}=68 \mathrm{~m}.\\</span>
<p dir="auto">Area of Rectangle = 288 Sq.m. (Ans.)</p>
<p dir="auto"><strong>17. The length of diagonal of a rectangular land is 15 meter and the difference of length and breadth is 3 meter. Let us calculate perimeter and area.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto">Let the length of the rectangular land be x</p>
<p dir="auto">and Breadth " " " " " " y</p>
<p dir="auto">According to the condition of the problem,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\sqrt{x^2+y^2}=15</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x^2+y^2=225</span>
<p dir="auto">And x - y = 3</p>
<p dir="auto">or, x = 3 + y</p>
<p dir="auto">Putting the value of x in equation (i)</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">(3+y)^2+y^2=225</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(3)^2+2 \cdot 3 \cdot y+y^2+(y)^2=225</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">9+6 y+y^2+y^2-225=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 2 y^2+6 y-216=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> y^2+3 y-108=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">y^2+12 y-9 y-108=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">y(y+12)-9(y+12)=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(y+12)(y-9)=0</span>
<p dir="auto">either, y + 12 = 0</p>
<p dir="auto">y = -12</p>
<p dir="auto">or, y - 9 = 0</p>
<p dir="auto">y = 9</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Length can not be negative.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> y = 9</p>
<p dir="auto">From (ii),</p>
<p dir="auto">x = 3 + y</p>
<p dir="auto">= 3 + 9 = 12</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore Length =12 \mathrm{~m} . \quad Breadth =9 \mathrm{~m}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Perimeter of the rectangular land</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \text { (Length }+ \text { Breadth) } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2(12 \mathrm{~m}+9 \mathrm{~m}) \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 21 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=42 \mathrm{~m} .</span>
<p dir="auto">Area of the rectangular land</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { length } \times \text { breadth } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=12 \mathrm{~m} \times 9 \mathrm{~m}=108 \text { Sq.m (Ans.) }</span>
<p dir="auto"><strong>18. Let us calculate what is the longest size of the square tile that can be used for paving the rectangular courtyard with measurement of 385 meter × 60 meter and also find the number of tiles.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> First we find the H.C.F of 385 and 60</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> H.C.F of 385 and 60 = 5</p>
<p dir="auto">Side of the square tile = 5m</p>
<p dir="auto">Area of rectangular courtyard <span class="katex-eq" data-katex-display="false">= 385 \mathrm{~m} \times 60 \mathrm{~m}</span>
<p dir="auto">Area of one square tile <span class="katex-eq" data-katex-display="false">= (\text { Side })^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(5)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=25 \mathrm{Sq} \cdot \mathrm{m} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { No. of tiles } =\frac{385 \times 60}{25} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=924 \text { (Ans.) }</span>
<h2 dir="auto"><span class="ez-toc-section" id="19_MCQ"></span>19. <span style="text-decoration: underline;">(M.C.Q) :</span><span class="ez-toc-section-end"></span></h2>
<p dir="auto"><strong>(i) The length of diagonal of square is <span class="katex-eq" data-katex-display="false">12 \sqrt{2} \mathrm{~m}.</span> The area of square is</strong></p>
<p dir="auto"><strong>(a) <span class="katex-eq" data-katex-display="false">288 \mathrm{sq} \cdot \mathrm{m}</span></strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">144 \mathrm{~m}^2</span></strong></p>
<p dir="auto"><strong>(c) <span class="katex-eq" data-katex-display="false">72 \mathrm{~m}^2</span></strong></p>
<p dir="auto"><strong>(d) <span class="katex-eq" data-katex-display="false">18 \mathrm{~m}^2</span></strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> Length of diagonal of square <span class="katex-eq" data-katex-display="false">12 \sqrt{2} \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Side } \sqrt{2}=12 \sqrt{2} \mathrm{~m}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Side of a square } =12 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Area of square } =(\text { Side })^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(12)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=144 \mathrm{~m}^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (b) is correct option</p>
<p dir="auto"><strong>(ii) If the area of square is <span class="katex-eq" data-katex-display="false">A_1 \mathrm{sq}</span>. units and the area of square drawn on the diagonal of that square is <span class="katex-eq" data-katex-display="false">A_2 sq.</span> unit, then the ratio of <span class="katex-eq" data-katex-display="false">A_1: A_2</span> is</strong></p>
<p dir="auto"><strong>(a) 1: 2</strong></p>
<p dir="auto"><strong>(b) 2: 1</strong></p>
<p dir="auto"><strong>(c) 1: 4</strong></p>
<p dir="auto"><strong>(d) 4: 1</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> <span class="katex-eq" data-katex-display="false">\quad Let A_1=a^2</span>
<p dir="auto">and <span class="katex-eq" data-katex-display="false">A_2=(a \sqrt{2})^2=2 a^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore A_1: A_2=1: 2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (a) is correct option</p>
<p dir="auto"><strong>(iii) If a rectangular place of which length and breadth are 6 meter and 4 meter is desired to pave it with 2dm. square tiles, then the numbers of tiles is to be required</strong></p>
<p dir="auto"><strong>(a) 1200</strong></p>
<p dir="auto"><strong>(b) 2400</strong></p>
<p dir="auto"><strong>(c) 600</strong></p>
<p dir="auto"><strong>(d) 1800</strong></p>
<p dir="auto"><strong>Solution :</strong> Area of Rectangle <span class="katex-eq" data-katex-display="false">= 6 \mathrm{~m} \times 4 \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=24 \mathrm{~m}^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of square }=2 \mathrm{dm} \times 2 \mathrm{dm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{2}{10} \times \frac{2}{10} \mathrm{~m}^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { No. of tiles }=\frac{24}{\frac{2}{10} \times \frac{2}{10}}=\frac{24 \times 10 \times 10}{2 \times 2}=600</span>
<p dir="auto">No. of tiles <span class="katex-eq" data-katex-display="false">=\frac{24}{\frac{2}{10} \times \frac{2}{10}}=\frac{24 \times 10 \times 10}{2 \times 2}=600</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span>(c) is correct option</p>
<p dir="auto"><strong>(iv) If a square and a rectangle having the same perimeter and their areas are S and R respectively then</strong></p>
<p dir="auto"><strong>(a) S = R</strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">\mathrm{S}>\mathrm{R}</span></strong></p>
<p dir="auto"><strong>(c) <span class="katex-eq" data-katex-display="false">\mathrm{S}<\mathrm{R}</span></strong></p>
<p dir="auto"><strong>Solution :</strong></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (b) is correct option</p>
<p dir="auto"><strong>(v) If the length of diagonal of a rectangle is 10cm. and area is 62.5 <a href="http://sq.cm/" target="_blank" rel="noopener" data-saferedirecturl="https://www.google.com/url?q=http://sq.cm&source=gmail&ust=1687598773375000&usg=AOvVaw3jYEIf97aXBcnA85Uy3rVO">sq.cm</a>., then the sum of their length and breadth is</strong></p>
<p dir="auto"><strong>(a) <span class="katex-eq" data-katex-display="false">12 \mathrm{~cm}.</span></strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">15 \mathrm{~cm}.</span></strong></p>
<p dir="auto"><strong>(c) <span class="katex-eq" data-katex-display="false">20 \mathrm{~cm}.</span></strong></p>
<p dir="auto"><strong>(d) <span class="katex-eq" data-katex-display="false">25 \mathrm{~cm}.</span></strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> Given,</p>
<p dir="auto">Length of diagonal = 10</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\sqrt{(\text { Length })^2+(\text { Breadth })^2}=10</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(\text { Length }+ \text { Breadth })^2-2.length \times breadth =100</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(\text { Length }+ \text { Breadth })^2-2 \times 62.5=100</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(\text { Length }+ \text { Breadth) }^2-125=100</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(\text { Length }+ \text { Breadth })^2=100+125=225</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(\text { Length }+ \text { Breadth })^2=(15)^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Length + breadth = 15cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (b) is correct option</p>
<h2 dir="auto"><span class="ez-toc-section" id="20_Short_answer_type"></span><strong>20. <span style="text-decoration: underline;">Short answer type :</span></strong><span class="ez-toc-section-end"></span></h2>
<p dir="auto"><strong>(i) If the length of square is increased by 10%, then what percent of the area of square will be increased?</strong></p>
<p dir="auto"><strong>Solution :</strong> Let the side of square be a</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Area of square <span class="katex-eq" data-katex-display="false">=\mathrm{a}^2</span>
<p dir="auto">If the length of square is increased by 10%</p>
<p dir="auto">Side of square = a + 10% of a</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= a+\frac{10}{100} \times a \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= a+\frac{a}{10}=\frac{11 a}{10} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of square } =\left(\frac{11 a}{10}\right)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{121 a^2}{100} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Increased area of square } =\frac{121 a^2}{100}-a^2=\frac{121 a^2-100 a^2}{100}=\frac{21 a^2}{100}</span>
<p dir="auto">Percentage of the area of square will be increased</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{21 a^2}{\frac{100}{a^2}} \times 100 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=21 \% \text { (Ans.) }</span>
<p dir="auto"><strong>(ii) If the length is increased by 10% and breadth is decreased by 10% of a rectangle, then what percent of area will be increased or decreased?</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> Let' l ' be the length of rectangle.</p>
<p dir="auto">and 'b'" " breadth " "</p>
<p dir="auto">Area of rectangle = <span class="katex-eq" data-katex-display="false">l \times b</span>
<p dir="auto">If length is increased by 10% and breadth is decreased by 10%</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Increased length } =l+10 \% l \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=l+\frac{10}{100} \times l \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=l+\frac{l}{10}=\frac{11 l}{10}</span>
<p dir="auto">Decreased beadth <span class="katex-eq" data-katex-display="false">=\mathrm{b}-10 \% of \mathrm{b}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=b-\frac{10}{100} \times b \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=b+\frac{b}{10}=\frac{9 b}{10} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area }=\frac{11 l}{10} \times \frac{9 b}{10}=\frac{99 l b}{100}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Decreased area } =m-\frac{99 l b}{100} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{l b}{100}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Percentage of area will be decreased } =\frac{l b}{100} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= lb \times 100 \% \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=1 \% \text { (Ans.) }</span>
<p dir="auto"><strong>(iii) The length of the diagonal of a rectangle is 5cm. The length of perpendicular on a breadth of rectangle from intersecting point between two diagonals is 2cm. What is the length of breadth?</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23576" src="https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2-300x225.jpeg" alt="47E27AAB 8A8F 4D7F 891C 65C59FAADCD2" width="470" height="353" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 35" srcset="https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2-300x225.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2-1024x770.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2-768x577.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2-1536x1154.jpeg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2-24x18.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2-36x27.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2-48x36.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/47E27AAB-8A8F-4D7F-891C-65C59FAADCD2.jpeg 1860w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">From the figure,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AC} =5 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{OE} =2 \mathrm{~cm}, \mathrm{OF}=2 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{EF}=\mathrm{BC} =2 \mathrm{~cm}+2 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= 4 \mathrm{~cm}</span>
<p dir="auto">Let breadth of rectangle be b Then,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\sqrt{b^2+(4)^2}=5 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } b^2+16=25 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } b^2=25-16 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } b^2=9 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } b^2=3^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad b=3 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Length of breadth }=3 \mathrm{~cm}</span>
<p dir="auto"><strong>(iv) If the length of perpendicular from the intersecting point between two diagonals on any side of square is <span class="katex-eq" data-katex-display="false">2 \sqrt{2} \mathrm{~cm},</span> then wila is inc ientgn of each diagonal of square?</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23577" src="https://flasheducation.online/wp-content/uploads/2023/06/76BFFA55-77C9-499E-816C-9FEDD975022F-300x278.jpeg" alt="76BFFA55 77C9 499E 816C 9FEDD975022F" width="470" height="435" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 36" srcset="https://flasheducation.online/wp-content/uploads/2023/06/76BFFA55-77C9-499E-816C-9FEDD975022F-300x278.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/76BFFA55-77C9-499E-816C-9FEDD975022F-1024x949.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/76BFFA55-77C9-499E-816C-9FEDD975022F-768x712.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/76BFFA55-77C9-499E-816C-9FEDD975022F-24x22.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/76BFFA55-77C9-499E-816C-9FEDD975022F-36x33.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/76BFFA55-77C9-499E-816C-9FEDD975022F-48x44.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/76BFFA55-77C9-499E-816C-9FEDD975022F.jpeg 1470w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{EG}=\mathrm{FG}=2 \sqrt{2} \mathrm{~cm}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{EF} =(2 \sqrt{2}+2 \sqrt{2}) \mathrm{cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \sqrt{2} \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{EF} =\mathrm{BC}=4 \sqrt{2} \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of diagonal }=\operatorname{Side} \sqrt{2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \sqrt{2} \times \sqrt{2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times 2 \mathrm{~cm}=8 \mathrm{~cm} \text {. (Ans.) }</span>
<p dir="auto"><strong>(v) The perimeter of a rectangle is 34cm. and area is 60sq. cm. What is the length of each diagonal?</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> Let ' l ' be the length and ' b ' be the breadth of the rectangle</p>
<p dir="auto">Perimeter <span class="katex-eq" data-katex-display="false">=34 \mathrm{~cm}.</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">2(l+b)=34</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">l+\mathrm{b}=17</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad l=17-b</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad Area =60 Sq.cm.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">l \times \mathrm{b}=60</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(17-b) \times b=60</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\quad 17 b-b^2=60</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">b^2-17 b+60=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">b^2-12 b-5 b+60=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">b(b-12)-5(b-12)=0</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(b-12)(b-5)=0</span>
<p dir="auto">either,</p>
<p dir="auto">b - 12 = 0 <span class="katex-eq" data-katex-display="false">\quad \text { or, } b-5=0 \\</span>
<p dir="auto">b = 12 <span class="katex-eq" data-katex-display="false">\quad</span> b = 5</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Breadth is always less than length</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Breadth = 5cm</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad l=17-5=12 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Length }=12 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of each diagonal } =\sqrt{l^2+\mathrm{b}^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{(12)^2+(5)^2} \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{144+25} \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{169} \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=13 \mathrm{~cm} . \quad \text { (Ans.) }</span>
<hr />
<h2 dir="auto" style="text-align: center;"><span class="ez-toc-section" id="Let_us_work_out_-_152"></span><span style="text-decoration: underline;"><em>Let us work out - 15.2</em></span><span class="ez-toc-section-end"></span></h2>
<p dir="auto"><strong>1. Let us write by calculating the area of the following regions:</strong></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23578" src="https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-300x189.jpeg" alt="98FAA98B E2ED 4EE0 A450 499AD041CCE0" width="480" height="303" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 37" srcset="https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-300x189.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-1024x646.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-768x485.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-1536x969.jpeg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-2048x1292.jpeg 2048w, https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-24x15.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-36x23.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/98FAA98B-E2ED-4EE0-A450-499AD041CCE0-48x30.jpeg 48w" sizes="(max-width: 480px) 100vw, 480px" /></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23581" src="https://flasheducation.online/wp-content/uploads/2023/06/D5F95B3F-C696-4E07-ABB3-A7BE0A1A2756-300x256.jpeg" alt="D5F95B3F C696 4E07 ABB3 A7BE0A1A2756" width="470" height="400" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 38" srcset="https://flasheducation.online/wp-content/uploads/2023/06/D5F95B3F-C696-4E07-ABB3-A7BE0A1A2756-300x256.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/D5F95B3F-C696-4E07-ABB3-A7BE0A1A2756-24x20.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/D5F95B3F-C696-4E07-ABB3-A7BE0A1A2756-36x31.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/D5F95B3F-C696-4E07-ABB3-A7BE0A1A2756-48x41.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/D5F95B3F-C696-4E07-ABB3-A7BE0A1A2756.jpeg 750w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle A B C=\frac{\sqrt{3}}{4} \times(10)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 100 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=25\sqrt{3} \text { sq.cm } \\</span> (Ans.)</p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23579" src="https://flasheducation.online/wp-content/uploads/2023/06/758B9250-3CDB-4ED0-A0F3-B02ED03708A9-283x300.jpeg" alt="758B9250 3CDB 4ED0 A0F3 B02ED03708A9" width="470" height="498" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 39" srcset="https://flasheducation.online/wp-content/uploads/2023/06/758B9250-3CDB-4ED0-A0F3-B02ED03708A9-283x300.jpeg 283w, https://flasheducation.online/wp-content/uploads/2023/06/758B9250-3CDB-4ED0-A0F3-B02ED03708A9-966x1024.jpeg 966w, https://flasheducation.online/wp-content/uploads/2023/06/758B9250-3CDB-4ED0-A0F3-B02ED03708A9-768x814.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/758B9250-3CDB-4ED0-A0F3-B02ED03708A9-24x24.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/758B9250-3CDB-4ED0-A0F3-B02ED03708A9-34x36.jpeg 34w, https://flasheducation.online/wp-content/uploads/2023/06/758B9250-3CDB-4ED0-A0F3-B02ED03708A9-45x48.jpeg 45w, https://flasheducation.online/wp-content/uploads/2023/06/758B9250-3CDB-4ED0-A0F3-B02ED03708A9.jpeg 1290w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\triangle \mathrm{ABC} =\frac{1}{2} \times 8 \times \sqrt{(10)^2 - (\frac{8}{2} )^2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times \sqrt{100-16} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times \sqrt{84} \text { Sq. units. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times \sqrt{4 \times 21} \text { Sq. units. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times 2 \sqrt{21} \text { Sq. units. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=8 \sqrt{21 \mathrm{Sq} \text {. units. }} \text { (Ans.) }</span>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23580" src="https://flasheducation.online/wp-content/uploads/2023/06/929860C1-79E9-4F16-8868-241347A04BEB-300x247.jpeg" alt="929860C1 79E9 4F16 8868 241347A04BEB" width="470" height="386" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 40" srcset="https://flasheducation.online/wp-content/uploads/2023/06/929860C1-79E9-4F16-8868-241347A04BEB-300x247.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/929860C1-79E9-4F16-8868-241347A04BEB-1024x842.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/929860C1-79E9-4F16-8868-241347A04BEB-768x632.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/929860C1-79E9-4F16-8868-241347A04BEB-24x20.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/929860C1-79E9-4F16-8868-241347A04BEB-36x30.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/929860C1-79E9-4F16-8868-241347A04BEB-48x39.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/929860C1-79E9-4F16-8868-241347A04BEB.jpeg 1452w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Area of Trapazium <span class="katex-eq" data-katex-display="false">\mathrm{ABCD}=\frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{CD}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2}(5+4) \times 3 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 9 \times 3 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{27}{2} \text { Sq. Units } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=13.5 \text { Sq. units. }</span>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23582" src="https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA-300x209.jpeg" alt="FD5751AE ACD3 4D1E A053 5F1BDEC982DA" width="470" height="327" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 41" srcset="https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA-300x209.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA-1024x712.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA-768x534.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA-1536x1068.jpeg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA-24x17.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA-36x25.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA-48x33.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/FD5751AE-ACD3-4D1E-A053-5F1BDEC982DA.jpeg 1614w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of Trapazium }=\frac{1}{2}(\mathrm{AD}+\mathrm{CD}) \times \mathrm{AD}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2}(15+40) \times 9 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 55 \times 9 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{495}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=247.5 \text { Sq.cm (Ans.) }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AC}=42 \mathrm{~cm}, \mathrm{CD}=38 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\because \angle \mathrm{ADC}=90^{\circ}</span>
<p dir="auto">We know that,</p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23575" src="https://flasheducation.online/wp-content/uploads/2023/06/008D1BD8-6EA8-4877-9B66-7C2D27D78340-300x222.jpeg" alt="008D1BD8 6EA8 4877 9B66 7C2D27D78340" width="470" height="347" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 42" srcset="https://flasheducation.online/wp-content/uploads/2023/06/008D1BD8-6EA8-4877-9B66-7C2D27D78340-300x222.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/008D1BD8-6EA8-4877-9B66-7C2D27D78340-1024x757.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/008D1BD8-6EA8-4877-9B66-7C2D27D78340-768x568.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/008D1BD8-6EA8-4877-9B66-7C2D27D78340-24x18.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/008D1BD8-6EA8-4877-9B66-7C2D27D78340-36x27.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/008D1BD8-6EA8-4877-9B66-7C2D27D78340-48x35.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/008D1BD8-6EA8-4877-9B66-7C2D27D78340.jpeg 1425w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A D^2 =A C^2-C D^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(42)^2-(38)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(42+38(42-38) \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=80 \times 4=320</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad A D =\sqrt{320} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{4 \times 4 \times 2 \times 2 \times 5} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times 2 \sqrt{5}=8 \sqrt{5} \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Area of } \mathrm{ABCD} =\mathrm{CD} \times \mathrm{AD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=38 \times 8 \sqrt{5} \text { Sq.cm. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=304 \sqrt{5} \text { Sq.cm. (Ans.) }</span>
<p dir="auto"><strong>2. The perimeter of an equilateral triangle is 48cm. Let us write by calculating its area.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> The perimeter of an equilateral triangle = 48cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Side of the equilateral triangle</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{48}{3} \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=16 \mathrm{~cm} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Area of the equilateral triangle</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times(16)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 256 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=44 \sqrt{3} \mathrm{~cm} . \quad \text { (Ans.) }</span>
<p dir="auto"><strong>3. If the height of an equilateral triangle ABC is <span class="katex-eq" data-katex-display="false">5 \sqrt{3}</span>. Let us write by calculating the area of this triangle.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23574" src="https://flasheducation.online/wp-content/uploads/2023/06/0B238916-7770-4B66-9F27-333C1447B737-300x244.jpeg" alt="0B238916 7770 4B66 9F27 333C1447B737" width="470" height="382" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 43" srcset="https://flasheducation.online/wp-content/uploads/2023/06/0B238916-7770-4B66-9F27-333C1447B737-300x244.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/0B238916-7770-4B66-9F27-333C1447B737-1024x833.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/0B238916-7770-4B66-9F27-333C1447B737-768x625.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/0B238916-7770-4B66-9F27-333C1447B737-24x20.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/0B238916-7770-4B66-9F27-333C1447B737-36x29.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/0B238916-7770-4B66-9F27-333C1447B737-48x39.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/0B238916-7770-4B66-9F27-333C1447B737.jpeg 1413w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">The height of an equilateral triangle <span class="katex-eq" data-katex-display="false">\mathrm{ABC}=5 \sqrt{3} \mathrm{~cm}.</span> We know that,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \frac{\sqrt{3}}{2} \times \text { side }=5 \sqrt{3}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{\text { Side }}{2}=5</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Side = 10</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Side of an equilateral triangle = 10cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Area of an equilateral triangle.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times(10)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 100 \text { Sq.cm } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=25 \sqrt{3} \text { Sq.cm. }(Ans.)</span>
<p dir="auto"><strong>4. If each equal side of an isosceles triangle ABC is 10cm. and length of base is 4cm. Let us write by calculating the area of <span class="katex-eq" data-katex-display="false"> \triangle \mathrm{ABC}.</span></strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23622" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3362-279x300.jpg" alt="IMG 3362" width="470" height="505" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 44" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3362-279x300.jpg 279w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3362-953x1024.jpg 953w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3362-768x825.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3362-22x24.jpg 22w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3362-34x36.jpg 34w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3362-45x48.jpg 45w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3362.jpg 1207w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Each equal side of an isosceles triangle ABC = 10cm.</p>
<p dir="auto">Length of base = 4cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 4 \times \sqrt{(10)^{2}-(_{2}^{4})^{2}} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times \sqrt{100-4} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times \sqrt{96} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times \sqrt{4 \times 4 \times 6} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 4 \sqrt{6} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=8 \sqrt{8} \text { Sq.cm. (Ans.) }</span>
<p dir="auto"><strong>5. If length of base of any isosceles triangle is 12cm and length of each equal side is 10cm. Let us write by calculating the area of that isosceles triangle.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23621" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3363-300x282.jpg" alt="IMG 3363" width="470" height="442" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 45" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3363-300x282.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3363-1024x963.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3363-768x723.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3363-24x24.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3363-36x34.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3363-48x45.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3363.jpg 1167w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Each equal side of an isosceles triangle ABC = 10cm.</p>
<p dir="auto">Length of base = 12cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times \sqrt{(10)^{2}-(\frac{12}{2})^{2}} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=6 \times \sqrt{100-36} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=6 \times \sqrt{64} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=6 \times 8= 48 \text { Sq.cm. (Ans.) }</span>
<p dir="auto"><strong>6. Perimeter of any isosceles triangle is 544cm. and length of each equal side is <span class="katex-eq" data-katex-display="false"> \frac{5}{6} th</span> of length of base. Let us wirte by calculating the area of this triangle.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> Let the length of base be x</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore Length of each side =\frac{5}{6} \times x=\frac{5 x}{6}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad Perimeter =544 \mathrm{~cm} . \therefore \quad Perimeter =544 \mathrm{~cm}.</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> x+\frac{5 x}{6}+\frac{5 x}{6}=544</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{6 x+5 x+5 x}{6}=544</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{16 x}{6}=544</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \frac{x}{6}=34 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> x = 204</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore Length of the base \quad = 204 \mathrm{~cm} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of each side } =\frac{5}{6} \times 204 \\</span>
<p dir="auto">= 170cm</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \mathrm{ABC}=\frac{1}{2} \times 204 \times \sqrt{(170)^{2}- \frac{204}{2} } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=102 \times \sqrt{28900-10404} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=102 \times \sqrt{18496} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=102 \times 136 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=13872 \text { Sq.cm. (Ans.) }</span>
<p dir="auto"><strong>7. If the length of hypotenuse of an isosceles right-angled triangle is <span class="katex-eq" data-katex-display="false"> 12 \sqrt{2} \mathrm{~cm}</span>. Let us write by calculating the area of this triangle.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23620" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364-300x271.jpg" alt="IMG 3364" width="470" height="424" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 46" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364-300x271.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364-1024x925.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364-768x693.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364-1536x1387.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364-24x22.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364-36x33.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364-48x43.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3364.jpg 1824w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let the length of equal side be x</p>
<p dir="auto">We have,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A B^{2}+B C^{2}=A C^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> x^{2}+x^{2}=(12 \sqrt{2})^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 2 x^{2}=144 \times 2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> x^{2}=144</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> x=\sqrt{144}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> x = 12</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Length of equal side be 12cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times 12 \text { Sq.cm. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=72 \text { Sq.cm (Ans.) }</span>
<p dir="auto"><strong>8. Pritha drew a parallelogram of which lengh of two diagonals are <span class="katex-eq" data-katex-display="false"> 6 \mathrm{~cm} \ and \ 8 \mathrm{~cm}</span> and each angle between two diagonals is <span class="katex-eq" data-katex-display="false"> 90^{\circ}</span>. Let us write the length of sides of parallelogram and what type of parallelogram it is?</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23619" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365-300x204.jpg" alt="IMG 3365" width="470" height="319" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 47" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365-300x204.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365-1024x695.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365-768x521.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365-1536x1043.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365-24x16.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365-36x24.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365-48x33.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3365.jpg 1884w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let } A C=6 \mathrm{~cm}, \quad B D=8 \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{AO}=\mathrm{OC}=3 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BO}=\mathrm{OD}=4 \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \angle C O D=90^{\circ} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\triangle \mathrm{In} \triangle \mathrm{COD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{CD}^{2}=O C^{2}+\mathrm{OD}^{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(3)^{2}+(4)^{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=9+16=25 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{or}, \quad \mathrm{CD}=\sqrt{25}=5 \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad A B=B C=C D=D A=5 \mathrm{~cm} .</span>
<p dir="auto">Hence ABCD is rhombus</p>
<p dir="auto"><strong>9. The ratio of the length of sides of a triangular park of our village is 2 : 3 : 4; perimeter of park is 216 meter. </strong></p>
<p dir="auto"><strong>(i) Let us write by calculating the area of the park. </strong></p>
<p dir="auto"><strong>(ii) Let us write by calculating how long is to be walked from opposite vertex of longest side to that side straightly.</strong></p>
<p dir="auto"><strong><em>Solution :</em> </strong></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23618" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366-300x238.jpg" alt="IMG 3366" width="470" height="373" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 48" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366-300x238.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366-1024x812.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366-768x609.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366-1536x1218.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366-24x19.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366-36x29.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366-48x38.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3366.jpg 1722w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let a = 2x</p>
<p dir="auto">b = 3x</p>
<p dir="auto">c = 4x</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Perimeter =216m.</p>
<p dir="auto">or, a + b + c = 216m</p>
<p dir="auto">or, 2x + 3x + 4x = 216</p>
<p dir="auto">or, 9x = 216</p>
<p dir="auto">or, x <span class="katex-eq" data-katex-display="false">=\frac{216}{9}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad x =24 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad a =2 x=2 \times 24=48 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">b = 3 \mathrm{x}=3 \times 24=72 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">c = 4 \mathrm{x}=4 \times 24=96 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad s =\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{48+72+96}{2}=\frac{216}{2}=108 \mathrm{~m}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of the park } =\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{108(108-48)(108-72)(108-96)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{108 \times 60 \times 36 \times 12} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{36 \times 3 \times 12 \times 5 \times 36 \times 12} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=36 \times 12 \sqrt{5 \times 3} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of the park } =432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m}</span>
<p dir="auto">Area of the park <span class="katex-eq" data-katex-display="false">=432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \quad \frac{1}{2} \times \mathrm{BC} \times \mathrm{AD}=432 \sqrt{15}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{1}{2} \times 96 \times \mathrm{AD}=432 \sqrt{15}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \mathrm{AD}=\frac{432 \sqrt{15}}{48}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \mathrm{AD}=9 \sqrt{15 \mathrm{~m}} \text { (Ans.) }</span>
<p dir="auto"><strong>10. The length of three sided of a triangular field of village of Paholampur are 26 meter, 28 meter and 30 meter. </strong></p>
<p dir="auto"><strong>(i) Let us write by calculating what will be the cost of planting grass in the triangular field at the rate of Rs.5 per sq. meter. </strong></p>
<p dir="auto"><strong>(ii) Let us write by calculating how much cost will be for </strong><strong>fencing around three sides at the rate of Rs 18 per meter leaving a space 5 meter for constructing entrance gate of that triangular field.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23617" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-300x174.jpg" alt="IMG 3367" width="470" height="273" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 49" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-300x174.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-1024x594.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-768x446.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-1536x891.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-2048x1188.jpg 2048w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-1200x700.jpg 1200w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-24x14.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-36x21.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3367-48x28.jpg 48w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let } a=26 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{~b}=28 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{c}=30 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{~s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{26+28+30}{2}=\frac{84}{2}=42 \mathrm{~m} . \\</span>
<p dir="auto">Area of the triangular field <span class="katex-eq" data-katex-display="false"> =\sqrt{s(s-a)(s-b)(s-c)} =\sqrt{42(42-26)(42-28)(42-30)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{42 \times 16 \times 14 \times 12} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{14 \times 3 \times 4 \times 4 \times 4 \times 14 \times 2 \times 2 \times 3} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=14 \times 4 \times 3 \times 3 \\=336 \text { Sq.m }</span>
<p dir="auto">Cost of planting grass in the triangular field <span class="katex-eq" data-katex-display="false"> =336 \times 5</span>
<p dir="auto">= Rs. 1680 (Ans.)</p>
<p dir="auto">Perimeter of the triangular field <span class="katex-eq" data-katex-display="false"> =26 \mathrm{~m}+28 \mathrm{~m}+30 \mathrm{~m}=84 \mathrm{~m}</span>
<p dir="auto">But leaving a space 5m for constructing entrance gate,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Length for fencing around three sides }=84 \mathrm{~m}-5 \mathrm{~m}</span>
<p dir="auto">= 79m</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Total cost }=\text { Rs. } 79 \times 18=\text { Rs. } 1422 \text { (Ans.) }</span>
<p dir="auto"><strong>11. Shakil draws an equilateral triangle PQR. I draw three perpendiculars from a point inside of that equilateral triangle on three sides, of which lengths are 10cm, 12cm . and 8cm. Let us write by calculating the area of the triangle.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23616" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368-300x250.jpg" alt="IMG 3368" width="470" height="391" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 50" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368-300x250.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368-1024x853.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368-768x639.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368-1536x1279.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368-24x20.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368-36x30.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368-48x40.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3368.jpg 1768w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let 'a' be the side of an equilateral triangle.</p>
<p dir="auto">Area of the triangle PQR <span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times(\mathrm{a})^{2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times a^{2}</span>
<p dir="auto">According to the condition of the problem</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\frac{\sqrt{3}}{4} \times \mathrm{a}^{2}=\frac{1}{2} \mathrm{a} \times 10+\frac{1}{2} \times a \times 12+\frac{1}{2} \times a \times 8</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{\sqrt{3}}{4} \times a^{2}=5 a+6 a+4 a</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{\sqrt{3}}{4} \times a^{2}=15 a</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{\sqrt{3}}{4} \times a=15 \quad[\because a \neq 0]</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \sqrt{3} \mathrm{a}=60</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> a=\frac{60}{\sqrt{3}}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, }=\frac{20 \times \sqrt{3} \times \sqrt{3}}{\sqrt{3}} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad a=20 \sqrt{3}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle P Q R =\frac{\sqrt{3}}{4} \times(20 \sqrt{3})^{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 20 \times 20 \times 3 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=300 \sqrt{3} \text { Sq.m (Ans.) }</span>
<p dir="auto"><strong>12. The length of each equal side of an isosceles triangle is 20m and the angle included between them is <span class="katex-eq" data-katex-display="false"> 45^{\circ}</span>. Let us write by calculating the area of triangle.</strong></p>
<p dir="auto"><strong><em>Solution</em>:</strong></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23615" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3369-296x300.jpg" alt="IMG 3369" width="470" height="477" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 51" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3369-296x300.jpg 296w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3369-1010x1024.jpg 1010w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3369-768x779.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3369-24x24.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3369-36x36.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3369-48x48.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3369.jpg 1284w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">In <span class="katex-eq" data-katex-display="false"> \triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}=20 \mathrm{~m} \ and \ \angle \mathrm{A}=45^{\circ},</span>
<p dir="auto">CD is drawn perpendicular to AB</p>
<p dir="auto">Now, if we take AB as base of the triangle,</p>
<p dir="auto">Then its altitude is CD.</p>
<p dir="auto">According to construction, <span class="katex-eq" data-katex-display="false"> \angle \mathrm{ACD}=45^{\circ}, \therefore \mathrm{AD}=\mathrm{CD}</span>
<p dir="auto">In the right angled triangle ADC,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{CD}^{2}+\mathrm{AD}^{2}=\mathrm{AC}{ }^{2}=(20)^{2} \mathrm{Sqm}=400 \mathrm{Sq} \cdot \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad 2 \mathrm{CD}^{2}=400 \mathrm{Sq} \cdot \mathrm{m}[\because \mathrm{AD}=\mathrm{CD}]</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad 2 \mathrm{CD}^{2}=200 \text { Sq.m } \therefore \mathrm{CD}=\sqrt{200} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of the triangle }=\frac{1}{2} \times \mathrm{AB} \times \mathrm{CD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 20 \times 10 \sqrt{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=100 \sqrt{2} \text { Sq } \mathrm{m} \text {. } \\</span>
<p dir="auto"><strong>13. The length of each equal side of an isosceles triangle is 20cm, and the angle included between them is <span class="katex-eq" data-katex-display="false"> 30^{\circ}</span>. Let us write by calculating the area of triangle.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23614" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-300x300.jpg" alt="IMG 3370" width="470" height="470" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 52" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-300x300.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-1024x1024.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-150x150.jpg 150w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-768x767.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-600x600.jpg 600w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-24x24.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-36x36.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370-48x48.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3370.jpg 1474w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">In <span class="katex-eq" data-katex-display="false">\triangle A B C \\</span>
<p dir="auto">AB = AC = 20cm</p>
<p dir="auto">We draw perpendicular CD on AB</p>
<p dir="auto">Then, <span class="katex-eq" data-katex-display="false"> \angle A D C=90^{\circ},</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\angle \mathrm{CAD}=30^{\circ}</span> (Given)</p>
<p dir="auto">and, <span class="katex-eq" data-katex-display="false"> \angle \mathrm{ACD}=60^{\circ}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> In any right angled triangle,</p>
<p dir="auto">the three angles are <span class="katex-eq" data-katex-display="false"> 90^{\circ}, 60^{\circ} and \ 30^{\circ},</span>
<p dir="auto">Then,<span class="katex-eq" data-katex-display="false">\mathrm{CD}=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \times 20 \mathrm{~cm}=10 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \mathrm{AB} \times \mathrm{CD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} 10 \times 10 \text { Sq.cm } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=100 \text { Sq.cm (Ans.) }</span>
<p dir="auto"><strong>14. If the perimeter of an isosceles right-angled triangle is <span class="katex-eq" data-katex-display="false"> (\sqrt{2}+1) \mathrm{cm}</span>. Let us write by calculating the length of hypotenuse and area of triangle.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23613" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3371-300x238.jpg" alt="IMG 3371" width="470" height="373" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 53" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3371-300x238.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3371-1024x812.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3371-768x609.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3371-24x19.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3371-36x29.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3371-48x38.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3371.jpg 1532w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let the equal sides be ' a '</p>
<p dir="auto">And length of the base be ' b '</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad Perimeter =\sqrt{2}+1</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> a+a+b=\sqrt{2}+1</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 2 a+b=\sqrt{2}+1</span>............. (i)</p>
<p dir="auto">We have,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A C^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> b^{2}=a^{2}+a^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> b^{2}=2 a^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> b=\sqrt{2 a^{2}}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> b=\sqrt{2} \mathrm{a}</span>............. (ii)</p>
<p dir="auto">From (i),</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">2 a+b=\sqrt{2}+1</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 2 a+\sqrt{2 a}=\sqrt{2}+1</span> .............by (ii)</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \sqrt{2} a(\sqrt{2}+a)=\sqrt{2}+1</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \sqrt{2} \mathrm{a}=1</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore a=\frac{1}{\sqrt{2}}</span>
<p dir="auto">From (ii),</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">b=\sqrt{2 a} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= \sqrt{2} \times \frac{1}{\sqrt{2}} \quad[\because a=1 \sqrt{2}] \\</span>
<p dir="auto">= 1</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Length of the hypotenuse = 1cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of the triangle } =\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times \mathrm{a} \times \mathrm{a} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times \mathrm{a}^{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times \frac{1}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{4}=0.25 \text { Sq.cm }(Ans.)</span>
<p dir="auto"><strong>15. Maria cycling at a speed of 18km per hour covers along the perimeter of an equilateral triangular field in 10 minutes. Let us write by calculating the time required for Maria to go directly to the mid point of the side of the field starting from its opposite vertex.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23612" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3372-300x264.jpg" alt="IMG 3372" width="470" height="414" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 54" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3372-300x264.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3372-1024x902.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3372-768x677.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3372-24x21.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3372-36x32.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3372-48x42.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3372.jpg 1481w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Total distance = speed × time</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=18 \times \frac{10}{60} \mathrm{Km} . \\</span>
<p dir="auto">= 3km.</p>
<p dir="auto">Perimeter of equilateral trianglar park = 3 km .</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { length of each side }=\frac{3}{3} \mathrm{Km} .=1 \mathrm{~km} \text {. }</span>
<p dir="auto">Distance between mid point of a side and its opposite vertex = AD</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{2} \times \text { side }=\frac{\sqrt{3}}{2} \times 1 \mathrm{~km}=\frac{\sqrt{3}}{2} \mathrm{~km}</span>
<p dir="auto">Time taken to go from opposite vertex to mid-point of a side</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\text { Distance }}{\text { Speed }}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{2 \times 18} hours</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{2 \times 18} \times 60 minutes</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{5 \sqrt{3}}{3} minutes(Ans.)</span>
<p dir="auto"><strong>16. If the length of each side of an equilateral triangle is increased by 1 meter, then its area will be increased by <span class="katex-eq" data-katex-display="false"> \sqrt{3} sq</span>. meter. Let us write by calculating the length of side of equilateral triangle.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em> Let ' x ' be the side of an equilateral triangle.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of the equilateral triangle } =\frac{\sqrt{3}}{4} \times(x)^{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} x^{2}</span>
<p dir="auto">If length of each side of an equilateral triangle is increased by 1 meter,</p>
<p dir="auto">Then, side of an equilateral triangle (x + 1)</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Area of the equilateral triangle }=\frac{\sqrt{3}}{4} \times(x+1)^{2}</span>
<p dir="auto">According to the condition of problem,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\frac{\sqrt{3}}{4} \times(x+1)^{2}-\frac{\sqrt{3}}{4} x^{2}=\sqrt{3}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{\sqrt{3}}{4}\{(x+1)^{2}-x^{2}\}=\sqrt{3}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{1}{4}(x+1+x)(x+1-x)=1</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \frac{1}{4}(2 x+1)=1</span>
<p dir="auto">or, 2x + 1 = 4</p>
<p dir="auto">or, 2x = 4 - 1</p>
<p dir="auto">or, 2x = 3</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{x}=\frac{3}{2} \quad \therefore \mathrm{x}=1.5</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> Length of side equilateral triangle = 1.5m.</p>
<p dir="auto"><strong>17. The area of an equilateral triangle and area of square are in the ratio <span class="katex-eq" data-katex-display="false"> \sqrt{3} : 2 </span>. If the length of diagonal of square is 60cm. Let us write by calculating perimeter of an equilateral triangle.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> Let 'a' be the side of a square</p>
<p dir="auto">Then, diagonal of square = 60cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{a} \sqrt{2}=60 \mathrm{~cm}</span>
<p dir="auto">or, a<span class="katex-eq" data-katex-display="false">=\frac{60}{\sqrt{2}} \mathrm{~cm}</span>
<p dir="auto">By the problem,</p>
<p dir="auto">Area of equilateral triangle : area of square <span class="katex-eq" data-katex-display="false"> =\frac{\sqrt{3}}{2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \frac{\text { Area of equilateral triangle }}{\text { Area of square }}=\frac{\sqrt{3}}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \frac{\text { Area of equilateral triangle }}{(\frac{60}{\sqrt{2}})^{2}}=\frac{\sqrt{3}}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \frac{\text { Area of equilateral triangle }}{\frac{3600}{2}}=\frac{\sqrt{3}}{2}</span>
<p dir="auto">or, Area of equilateral triangle <span class="katex-eq" data-katex-display="false">=900 \sqrt{3}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{\sqrt{3}}{4}(\text { Side })^{2}=900 \sqrt{3}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(\text { side })^{2}=3600 \\</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\text { Side }=\sqrt{3600} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=60 \mathrm{~m} \text {. } \\</span>
<p dir="auto">Perimeter of an equilateral triangle</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=3 \times \text { side }=3 \times 60 \mathrm{~m}=180 \mathrm{~m}(Ans.)</span>
<p dir="auto"><strong>18. Length of hypotenuse and perpendicular of a right-angled triangle are 13cm and 30cm. Let us write by calculating the area of triangle.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em> Length of hypotenuse = 13cm.</p>
<p dir="auto">Perimeter of right angled triangle = 30cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Sum of remaining sides other than</p>
<p dir="auto">hypotenuse = (30 - 13)cm = 17cm.</p>
<p dir="auto">Let ' p ' be the perpendicular and ' b ' be the base of a right angled triangle.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{p}+\mathrm{b}=17 </span>........... (i)</p>
<p dir="auto">Also,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">p^{2}+b^{2}=h^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> (p+b)^{2}-2 p b=(13)^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> (17)^{2}-2 \mathrm{pb}=(13)^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 289-2 \mathrm{pb}=169</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> -2 \mathrm{pb}=169-289</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, }+2 \mathrm{pb}=f 120 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \mathrm{pb}=\frac{120}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{pb}=60</span>
<p dir="auto">Again,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">p^{2}+b^{2}=h^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> (p-b)^{2}+2 p b=h^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> (p-b)^{2}+2 \times 60=(13)^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> (\mathrm{p}-\mathrm{b})^{2}+120=169</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(p-b)^{2}=169-120</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> (p-b)^{2}=49</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> (p-b)=\sqrt{49}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> p - b = 7 ........... (ii)</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false"> \therefore \quad p + b = 17 ......... (i) \\ \underline{p-b =7 ........... (ii)} \\ 2 \mathrm{p}=24(by \ adding) </span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> p=\frac{24}{2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{p}=12</span>
<p dir="auto">Putting the value of p in equation (i)</p>
<p dir="auto">p + b = 17</p>
<p dir="auto">or, 12 + b = 17</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } b=17-12 \quad \therefore b=5</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of right angled triangle }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times \mathrm{b} \times \mathrm{p} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 5 \times 12 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=30 \text { Sq. cm. (Ans.) }</span>
<p dir="auto"><strong>19. The lengths of the sides containing the right angle are 12cm and 5cm . Let us write by calculating the length of perpendicular drawn from vertex of right angle on hypotenuse.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23611" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3373-300x251.jpg" alt="IMG 3373" width="470" height="393" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 55" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3373-300x251.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3373-1024x857.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3373-768x643.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3373-24x20.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3373-36x30.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3373-48x40.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3373.jpg 1522w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let <span class="katex-eq" data-katex-display="false"> \mathrm{AB}=12 \mathrm{~cm},</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BC}=5 \mathrm{~cm}</span>
<p dir="auto">We know that,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A C^{2}=A B^{2}+B C^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \mathrm{AC}^{2}=(12)^{2}+(5)^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \mathrm{AC}^{2}=144+25</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \mathrm{AC}^{2}=169</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \mathrm{AC}^{2}=\sqrt{169}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{AC}=13</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Area of <span class="katex-eq" data-katex-display="false"> \triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 5 \times 12</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= 30 Sq. \mathrm{cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text{Area of} \triangle \mathrm{ABC}=30 Sq.cm.</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{1}{2} \times \mathrm{BD} \times \mathrm{AC}=30</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{1}{2} \times \mathrm{BD} \times 13=30</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 13 \mathrm{BD}=60</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \mathrm{BD}=\frac{60}{13}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> \mathrm{BD}=4.615 \mathrm{~cm} (Approx)</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span>Length of the perpendicular <span class="katex-eq" data-katex-display="false"> =4.615 \mathrm{~cm} (Approx)(Ans.)</span>
<p dir="auto"><strong>20. The largest square is cut-out from a right-angled triangular region with length of 3cm, 4cm and 5cm respectively in such a way that the one vertex of square lies on hypotenuse of triangle. Let us write by calculating the length of side of square.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23610" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3374-284x300.jpg" alt="IMG 3374" width="470" height="496" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 56" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3374-284x300.jpg 284w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3374-969x1024.jpg 969w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3374-768x811.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3374-24x24.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3374-34x36.jpg 34w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3374-45x48.jpg 45w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3374.jpg 1279w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let <span class="katex-eq" data-katex-display="false"> \mathrm{AB}=4 \mathrm{~cm}, \mathrm{BC}=3 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AC}=5 \mathrm{~cm}</span>
<p dir="auto">Let <span class="katex-eq" data-katex-display="false"> \mathrm{BD}=\mathrm{DE}=\mathrm{EF}=\mathrm{BF}=\mathrm{a}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle A B C= \text { Area of } \triangle A E F+ \text { Area of Square } BDEF + \text { Area of } \triangle D E C</span>
<p dir="auto">or <span class="katex-eq" data-katex-display="false"> \frac{1}{2} \times 3 \times 4=\frac{1}{2} \times(4-a) \times a+a^{2}+\frac{1}{2} \times(3-a) \times a</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 12=(4-a) a+2 a^{2}+(3-a) a</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 12=4 a-a^{2}+2 a^{2}+3 a-a^{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false"> 12=7 a</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">a=\frac{12}{7}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Length of square } =\frac{12}{7} \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=1 \frac{5}{7} \mathrm{~cm}(Ans.)</span>
<h2 dir="auto"><span class="ez-toc-section" id="21_MCQ"></span>21. <span style="text-decoration: underline;">(M.C.Q):</span><span class="ez-toc-section-end"></span></h2>
<p dir="auto"><strong>(i) If each side of an equilateral triangle is 4cm, the measure of height is</strong></p>
<p dir="auto"><strong>(a) <span class="katex-eq" data-katex-display="false"> 4 \sqrt{3} \mathrm{~cm}</span> </strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false"> 16 \sqrt{3} \mathrm{~cm}</span></strong></p>
<p dir="auto"><strong> (c) <span class="katex-eq" data-katex-display="false"> 8 \sqrt{3 \mathrm{~cm}}</span> </strong></p>
<p dir="auto"><strong>(d) <span class="katex-eq" data-katex-display="false"> 2 \sqrt{3} \mathrm{~cm}</span> </strong></p>
<p dir="auto"><strong>Solution: </strong>Height of an equilateral triangle</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{2} \times(\text { side }) \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{\cancel2} \times {\cancel 4} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \sqrt{3} \mathrm{~cm}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> (d) is correct option (Ans.)</p>
<p dir="auto"><strong>(ii) An isosecles right-angled triangle of which the length of each side of equal two sides is a unit. The perimeter of triangle is</strong></p>
<p dir="auto"><strong>(a) <span class="katex-eq" data-katex-display="false">(1+\sqrt{2})</span> a unit</strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">(2+\sqrt{2})</span> a unit</strong></p>
<p dir="auto"><strong>(c) 3a unit</strong></p>
<p dir="auto"><strong>(d) <span class="katex-eq" data-katex-display="false">(3+2 \sqrt{2})</span> a unit.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23645" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3407-300x270.jpg" alt="IMG 3407" width="470" height="422" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 57" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3407-300x270.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3407-1024x920.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3407-768x690.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3407-24x22.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3407-36x32.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3407-48x43.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3407.jpg 1211w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let AB = a, BC = a</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">A C^2=a^2+a^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{AC}=\sqrt{2 \mathrm{a}^2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{AC}=\sqrt{2} \mathrm{a}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text{Perimeter of triangle} =a+a+\sqrt{2} a</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 a+\sqrt{2} a</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(2+\sqrt{2})</span> a unit.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> (b) is correct option (Ans.)</p>
<p dir="auto"><strong>(iii) If the area, perimeter and height of an equilateral triangle are a, s and h, then value of <sup>2a</sup>/<sub>sh</sub> is</strong></p>
<p dir="auto"><strong>(a) 1</strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">\frac{1}{2}</span></strong></p>
<p dir="auto"><strong>(c) <span class="katex-eq" data-katex-display="false">\frac{1}{3}</span></strong></p>
<p dir="auto"><strong>(d) <span class="katex-eq" data-katex-display="false">\frac{1}{4}</span></strong></p>
<p dir="auto"><strong>Solution :</strong></p>
<p dir="auto">We have,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\frac{2 a}{\operatorname{sh}}=\frac{2 \times \frac{\sqrt{3}}{4} \times(\text { side })^{2}}{3 \times \operatorname{side} \times \frac{\sqrt{3}}{2} \times \text { side }}=\frac{\frac{2}{4}}{\frac{3}{2}} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= \frac{2}{4} \times \frac{2}{3}=\frac{1}{3}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (c) is correct option (Ans.)</p>
<p dir="auto"><strong>(iv) The length of each equal side of an isosceles triangle is 5cm. and length of base is 6cm The area of triangle is</strong></p>
<p dir="auto"><strong>(a) <span class="katex-eq" data-katex-display="false">18 \mathrm{sq} . \mathrm{cm}</span> </strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">12 \mathrm{sq} . \mathrm{cm}</span> </strong></p>
<p dir="auto"><strong>(c) <span class="katex-eq" data-katex-display="false">15 \mathrm{sq} . \mathrm{cm}.</span></strong></p>
<p dir="auto"><strong>(d) <span class="katex-eq" data-katex-display="false">30 \mathrm{sq} . \mathrm{cm}.</span></strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of triangle } =\frac{1}{2} \times 6 \times \sqrt{(5)^2-(\frac{6}{3})^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 6 \times \sqrt{25-9} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=3 \times \sqrt{16} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=3 \times 4=12 \text { Sq.cm. }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (b) is correct option (Ans.)</p>
<p dir="auto"><strong>(v) D is such a point on AC of triangle ABC so that AD : C = 3 : 2; If the area of triangle ABC is 40 sq.cm, the area of triangle BDC is.</strong></p>
<p dir="auto"><strong>(a) 16sq.cm</strong></p>
<p dir="auto"><strong>(b) 24sq.cm.</strong></p>
<p dir="auto"><strong>(c) 15sq.cm.</strong></p>
<p dir="auto"><strong>(d) 30sq.cm</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23644" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3408-300x212.jpg" alt="IMG 3408" width="470" height="332" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 58" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3408-300x212.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3408-1024x724.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3408-768x543.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3408-24x17.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3408-36x25.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3408-48x34.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3408.jpg 1225w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Area of triangle ABC = 40 Sq.cm</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{AD}: \mathrm{DC}=3: 2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\triangle \mathrm{ABD}: \triangle \mathrm{BDC}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{\triangle A B D}{\triangle B D C}=\frac{3}{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">2 \triangle \mathrm{ABD}=3 \triangle \mathrm{BDC}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{ABD}=\frac{3}{2} \triangle \mathrm{BDC}</span>
<p dir="auto">Also <span class="katex-eq" data-katex-display="false">\triangle \mathrm{ABC}=\triangle \mathrm{ABD}+\triangle \mathrm{BDC}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\triangle \mathrm{ABC}={ }_2^3 \triangle \mathrm{BDC}+\triangle \mathrm{BDC}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\triangle \mathrm{ABC}=\frac{5}{2} \triangle \mathrm{BDC}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\triangle B D C={ }_5^2 \triangle A B C</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= \frac{2}{\cancel{5}} \times \cancel{40} \mathrm{Sq} \cdot \mathrm{cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=16 \mathrm{Sqcm}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (a) is correct option (Ans.)</p>
<p dir="auto"><strong>(vi) The difference of length of each side of a triangle from its semiperimeter are 8cm, 7cm and 5cm respectively. The area of triangle is</strong></p>
<p dir="auto"><strong>(a) <span class="katex-eq" data-katex-display="false">20 \sqrt{7} \mathrm{Sq} . \mathrm{cm}</span></strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">10 \sqrt{14} Sq.cm</span></strong></p>
<p dir="auto"><strong>(c) <span class="katex-eq" data-katex-display="false">20 \sqrt{14} \mathrm{Sq} . \mathrm{cm}</span></strong></p>
<p dir="auto"><strong>(d) <span class="katex-eq" data-katex-display="false">140 \mathrm{Sq} . \mathrm{cm}</span></strong></p>
<p dir="auto"><strong>Solution :</strong></p>
<p dir="auto">Let, s - a = 8</p>
<p dir="auto">s - b = 7</p>
<p dir="auto">s - c = 5</p>
<p dir="auto">adding these,</p>
<p dir="auto">3s - (a + b + c) = 8 + 7 + 5</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 3 s-2 s=20 \quad\left[\therefore s=\frac{a+b+c}{2}\right]</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{s}=20</span>
<p dir="auto">Area of Triangle <span class="katex-eq" data-katex-display="false">=\sqrt{s(s-a)(s-b)(s-c)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{20 \times 8 \times 7 \times 5} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{5 \times 4 \times 4 \times 2 \times 7 \times 5} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=5 \times 4 \sqrt{2 \times 7} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=20 \sqrt{14} \mathrm{Sq} \cdot \mathrm{cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> (c) is correct option (Ans.)</p>
<h2 dir="auto"><span class="ez-toc-section" id="22_Short_answer_type_question"></span><strong>22. <span style="text-decoration: underline;">Short answer type question:</span></strong><span class="ez-toc-section-end"></span></h2>
<p dir="auto"><strong>(i) The numerical values of area and height of an equilateval triangle are equal. What is the length of side of triangle?</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto">By question,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\quad \frac{\sqrt{3}}{4}(\text { side })^2=\frac{\sqrt{3}}{2} \times(\text { side }) \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \frac{\text { side }}{2}=1 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, side }=2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of side of triangle }=2 \text { units }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Length of side of triangle = 2 units (Ans.)</p>
<p dir="auto"><strong>(ii) If length of each side of an equilateral triangle is doubled, what percent of area will be increased of this. triangle?</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto">Let ' a ' be the side of an equilateral triangle,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Area of the triangle }=\frac{\sqrt{3}}{4} \mathrm{a}^2</span>
<p dir="auto">If length of each side of an equilateral triangle is doubled</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Area of the triangle } =\frac{\sqrt{3}}{4}(2 a)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 4 a^2</span>
<p dir="auto">Increase area of the triangle</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 4 a^2-\frac{\sqrt{3}}{4} a^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 3 \mathrm{a}^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Perimeter of increased in area }=\frac{\frac{\sqrt{3}}{4} \times 3 a^2}{\frac{\sqrt{3}}{4} \times a^2} \times 100 \% \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=300 \% \text { (Ans.) } \\</span>
<p dir="auto"><strong>(iii) If the length of each side of an equilateral is trippled.</strong></p>
<p dir="auto"><strong>What percent of area will be increased of this triangle?</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto">Let ' a ' be the side of an equilateral triangle,</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Area of the triangle }=\frac{\sqrt{3}}{4} \mathrm{a}^2</span>
<p dir="auto">If length of each side of an equilateral triangle is trippled</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Area of the triangle } =\frac{\sqrt{3}}{4}(3 a)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 9 a^2</span>
<p dir="auto">Increased area of the triangle</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 9 a^2-\frac{\sqrt{3}}{4} a^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 8 a^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{3} \times 2 a^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Perimeter of increased in area } =\frac{\sqrt{3} \times 2 \mathrm{a}^2}{\sqrt{3}} \times 100 \% \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=8 \times 100 \% \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=800 \% \text { (Ans.) }</span>
<p dir="auto"><strong>(iv) The lenght of sides of a right-angle triangle are (x - 2) cm, x cm and (x + 2) cm. How much length of hypotenuse is?</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto">We have, <span class="katex-eq" data-katex-display="false">(\text { hypotenuse} )^2=(\text { base })^2+(\text { Perpendicular })^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">(x+2)^2=(x-2)^2+(x)^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(x+2)^2-(x-2)^2=(x)^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">4 \cdot x \cdot 2=x^2</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 8 x=x^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } 8=x \quad [\therefore x \neq 0] \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad x=8</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Length of hypotenuse } =(x+2) \mathrm{cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(8+2) \mathrm{cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=10 \mathrm{~cm} \text {. (Ans.) }</span>
<p dir="auto"><strong>(v) A square drawn on height of equilateral triangle. What is the ratio of area of triangle and square?</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto">Area of triangle : Area of square</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times(\text { side })^2:\left(\frac{\sqrt{3}}{2} \times(\text { side })\right)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times(\text { side })^2: \frac{\sqrt{3} \times \sqrt{3}}{4} \times(\text { side })^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= 1: \sqrt{3} \text { (Ans.) }</span>
<hr />
<h2 dir="auto" style="text-align: center;"><span class="ez-toc-section" id="Let_us_work_out_-_153"></span><em><span style="text-decoration: underline;">Let us work out - 15.3</span></em><span class="ez-toc-section-end"></span></h2>
<p dir="auto"><strong>1. Ratul draws a parallelogram with length of base 5cm. and height 4cm. Let us calculate the area of parallelogram drawn by Ratul.</strong></p>
<p dir="auto"><em><strong>Solution :</strong></em></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Length of base }=5 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Height }=4 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of the parallelogram } =\text { base } \times \text { height } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=5 \mathrm{~cm} \times 4 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=20 \text { sq.cm. (Ans.) }</span>
<p dir="auto"><strong>2. The base of a parallelogram is twice its height. If the area of shape of parallelogram is 98 sq.cm. then let us calculate the length and height of parallelogram.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto">Let ' x ' be the height of a parallelogram</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore Base =2 \times height =2 \mathrm{x}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text{Area of parallelogram} =98 sq. \mathrm{cm}.</span>
<p dir="auto">or, Base × height = 98</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">2x × x = 98</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">2x^2 = 98</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x^2-\frac{98}{2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x^2=49</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{x}=\sqrt{49}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{x}=7</span>
<p dir="auto">Length of base = 2x</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 7=14 \mathrm{~cm} \text {. }</span>
<p dir="auto">Height of the parallelogram = x =7cm (Ans.)</p>
<p dir="auto"><strong>3. There is chane of warallelogram land heside our house of which lengths of adjacent sides are 15 meter and 13 meter. If the length of one diagonal is 14 meter, then let us calculate the area of shape of parallelogram land.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23643" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409-300x193.jpg" alt="IMG 3409" width="470" height="303" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 59" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409-300x193.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409-1024x659.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409-768x494.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409-1536x989.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409-24x15.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409-36x23.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409-48x31.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3409.jpg 1538w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let, <span class="katex-eq" data-katex-display="false">a =13 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">b =14 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">c =15 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">s=\frac{a+b+c}{2}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{13 m+14 m+15 m}{2}=\frac{42}{2} m=21 m</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle \mathrm{ABD} =\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{21(21-13)(21-14)(21-15)}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{21 \times 8 \times 7 \times 6} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{7 \times 3 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 2 \times 3 \times 7 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=84 \mathrm{Sq} . \mathrm{m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of parallelogram } \mathrm{ABCD} =2 \times \text { Area of } \triangle \mathrm{ABD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 84 \mathrm{Sq} \cdot \mathrm{m}=168 \mathrm{Sq} \cdot \mathrm{m}</span>
<p dir="auto"><strong>4. Pritha draws a parallelogram of which adjacent sides are 25cm and 15cm and length of on diagonal is 20cm. Let us write by calculating the height of parallelogram which is drawn on the side of 25cm.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23642" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3410-300x183.jpg" alt="IMG 3410" width="470" height="287" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 60" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3410-300x183.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3410-1024x626.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3410-768x470.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3410-24x15.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3410-36x22.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3410-48x29.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3410.jpg 1305w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let, <span class="katex-eq" data-katex-display="false">\mathrm{a}=25 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">b=20 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">c=15 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">s=\frac{a+b+c}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{25+20+15}{2}=\frac{60}{2}=30 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{25+20+15}{2}=\frac{60}{2}=30 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle A B D =\sqrt{s(s-a)(s-b)(s-c)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{30(30-25)(30-20)(30-15)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{30 \times 5 \times 10 \times 15} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{15 \times 2 \times 2 \times 5 \times 15} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 5 \times 15 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=150 \text { Sq.cm }</span>
<p dir="auto">Area of parallelogram = <span class="katex-eq" data-katex-display="false">2 \times Area of \triangle \mathrm{ABD}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 150 \text { Sq.cm. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=300 \text { Sq. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad Base \times height =300 Sq. \mathrm{cm}.</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">25 \times height =300</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">height ^2=\frac{300}{25}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore height =12 \mathrm{~cm}(Ans.)</span>
<p dir="auto"><strong>5. The length of adjacent two sides are 15cm. and 12cm. of a parallelogram distance between two smaller sides is 7.5cm. Then let us calculate the distance between the longer two sides.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23641" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3411-300x241.jpg" alt="IMG 3411" width="470" height="378" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 61" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3411-300x241.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3411-1024x823.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3411-768x617.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3411-24x19.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3411-36x29.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3411-48x39.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3411.jpg 1054w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let, <span class="katex-eq" data-katex-display="false"> A B=15 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BC}=12 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AE}=7.5 \mathrm{~cm} .</span>
<p dir="auto">Area of the <span class="katex-eq" data-katex-display="false">\triangle A B C</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AE} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 12 \times 7.5 \mathrm{Sq} . \mathrm{cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=45 \text { Sq.cm. }</span>
<p dir="auto">Let the distance between the longer two sides be ' x '</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Area of } \triangle \mathrm{ACD}=\text { Area of } \triangle \mathrm{ABC}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{1}{2} \times B C \times</span> distance between the longer two sides = 45</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{1}{2} \times 15 \times x=45</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{15 x}{2}=45</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{x}{2}=3</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> x = 6</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> The distance between the longer two sides = 6cm (Ans.)</p>
<p dir="auto"><strong>6. If the measure of two diagonals of a rhombus are 15 meter and 20 meter, then let us write by calculating its perimeter, area and height.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23640" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-300x243.jpg" alt="IMG 3412" width="470" height="381" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 62" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-300x243.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-1024x830.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-768x622.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-24x19.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-36x29.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-48x39.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412.jpg 1272w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let, } \mathrm{AC} =15 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BD} =20 \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AO}=\mathrm{OC}=\frac{15}{2} \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BO}=\mathrm{OD}=\frac{20}{2} \mathrm{~m}=10 \mathrm{~m}</span>
<p dir="auto">Area of Rhombus ABCD</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \mathrm{AC} \times \mathrm{BD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} 15 \times 20 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=150 \text { Sq.m (Ans.) }</span>
<p dir="auto">In <span class="katex-eq" data-katex-display="false">\triangle \mathrm{OAB}, \quad \angle \mathrm{AOB}=90^{\circ}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A B^2=A O^2+O B^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">A B^2=\left(\frac{15}{2}\right)^2+(10)^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">A B^2=\frac{225}{4}+100^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{AB}^2=\frac{225+400}{4}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">A B^2=\frac{625}{4}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{AB}=\sqrt{\frac{625}{4}}=\frac{25}{2} \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text{Perimeter of} \mathrm{ABCD}=4 \times \mathrm{AB}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=4 \times \frac{25}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=50 \mathrm{~m} \text { (Ans.) }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad Area of Rhombus ABCD =150 Sq.m</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad Base \times height =150</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{25}{2} \times height ^2=150</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{\text { height }}{2}=6</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore height =12 \mathrm{~m} (Ans.)</span>
<p dir="auto"><strong>7. If perimeter of a rhombus is 440 meter and distance between two parallel sides are 22 meter, Let us write by calculating the area of shape of rhombus.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23655" src="https://flasheducation.online/wp-content/uploads/2023/06/0B3300B0-A01B-4AAD-BB90-319DE9D68D94-300x203.jpeg" alt="0B3300B0 A01B 4AAD BB90 319DE9D68D94" width="470" height="318" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 63" srcset="https://flasheducation.online/wp-content/uploads/2023/06/0B3300B0-A01B-4AAD-BB90-319DE9D68D94-300x203.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/0B3300B0-A01B-4AAD-BB90-319DE9D68D94-1024x693.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/0B3300B0-A01B-4AAD-BB90-319DE9D68D94-768x520.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/0B3300B0-A01B-4AAD-BB90-319DE9D68D94-24x16.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/0B3300B0-A01B-4AAD-BB90-319DE9D68D94-36x24.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/0B3300B0-A01B-4AAD-BB90-319DE9D68D94-48x32.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/0B3300B0-A01B-4AAD-BB90-319DE9D68D94.jpeg 1068w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Side of Rhombus } =\frac{\text { Perimeter }}{4} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{440}{4} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=110 \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore Base \mathrm{CD} \text{of Rhombus} \mathrm{ABCD}=110 \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Distance between two parallel sides = Altitude AE</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Area of the Rhombus } =\text { Base } \times \text { Altitude } \quad=22 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\mathrm{CD} \times \mathrm{AE} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=110 \times 22 \text { Sq.m } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2420 \text { Sq.m (Ans.) }</span>
<p dir="auto"><strong>8. If perimeter of a Rhombus is 20cm. and length of its one diagonal of its one diagonal is 6cm, then let us write by calculating the area of Rhombus.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23640" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-300x243.jpg" alt="IMG 3412" width="470" height="381" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 62" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-300x243.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-1024x830.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-768x622.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-24x19.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-36x29.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-48x39.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412.jpg 1272w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Perimeter of Rhombus }=20 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Side of a rhombus } =\frac{20}{4} \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=5 \mathrm{~cm} .</span>
<p dir="auto">Length of its one diagonal = 6cm.</p>
<p dir="auto">Let <span class="katex-eq" data-katex-display="false">B D=6 \mathrm{~cm}.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{OB}=\frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm} . \\</span>
<p dir="auto">AB = 5cm</p>
<p dir="auto">In <span class="katex-eq" data-katex-display="false">\triangle \mathrm{OAB}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{OA}^2 =\mathrm{AB}^2-\mathrm{OB}^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{OA}^2 =(5)^2-(3)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \mathrm{OA}^2 =25-9</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{OA}^2=16</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{OA}=\sqrt{16}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{OA} =4 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AC} =2 \times \mathrm{OA} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 4 \mathrm{~cm}=8 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of rhombus } \mathrm{ABCD} =\frac{1}{2} \times \mathrm{BD} \times \mathrm{AC} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 6 \times 8 \text { Sq.cm. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=24 \text { Sq.cm. (Ans.) }</span>
<p dir="auto"><strong>9. The area of field shaped in trapeziuim is 1400 sq.dcm. If the perpendicular distance between two parallel sides are 20dcm. and the length of two parallel sides are in the ratio 3 : 4, then let us write by calculating the lengths of two sides.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23639" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3414-300x189.jpg" alt="IMG 3414" width="470" height="297" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 65" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3414-300x189.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3414-1024x647.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3414-768x485.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3414-24x15.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3414-36x23.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3414-48x30.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3414.jpg 1167w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let } \mathrm{AE}=20 \mathrm{dm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AD}=3 \mathrm{x} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BC}=4 \mathrm{x} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { trapazium }=1400 \text { Sq.cm }</span>
<p dir="auto">Area of trapazium = 1400 Sq.cm</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or } \quad \frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AE}=1400 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or } \quad \frac{1}{2}(3 \mathrm{x}+4 \mathrm{x}) \times 20=1400</span>
<p dir="auto">7x = 1400</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">x=\frac{140}{7} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> x = 20</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad A D=3 x=3 \times 20 \mathrm{dcm} .=60 \mathrm{dcm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\quad \mathrm{BC}=4 \mathrm{x}=4 \mathrm{x} \times 20 \mathrm{dcm} .=80 \mathrm{dcm} \text {. (Ans.) }</span>
<p dir="auto"><strong>10. Let us write by calculating the area of regular hexag field of which length of sides is 8cm</strong></p>
<p dir="auto"><em><strong>Solution:</strong> </em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23648" src="https://flasheducation.online/wp-content/uploads/2023/06/E4D1ABFD-CED0-44B2-A9CD-806BD691FB95-300x284.jpeg" alt="E4D1ABFD CED0 44B2 A9CD 806BD691FB95" width="470" height="444" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 66" srcset="https://flasheducation.online/wp-content/uploads/2023/06/E4D1ABFD-CED0-44B2-A9CD-806BD691FB95-300x284.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/E4D1ABFD-CED0-44B2-A9CD-806BD691FB95-768x726.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/E4D1ABFD-CED0-44B2-A9CD-806BD691FB95-24x24.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/E4D1ABFD-CED0-44B2-A9CD-806BD691FB95-36x34.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/E4D1ABFD-CED0-44B2-A9CD-806BD691FB95-48x45.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/E4D1ABFD-CED0-44B2-A9CD-806BD691FB95.jpeg 879w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">If we draw diagonals we get equal six equilater triangles.</p>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\triangle A O B</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4}(\text { Side })^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4}(8)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} 64 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=16 \sqrt{3} \mathrm{Sq} \cdot \mathrm{cm} .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of Hexagon } \mathrm{ABCDEF} =6 \times 16 \sqrt{3} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=96 \sqrt{3} \text { Sq. } \mathrm{cm} \text {. }</span>(Ans.)</p>
<p dir="auto"><strong>11. In a quadrilateral ABCD, AB = 5 meter, BC = 12 meter, DA = 15 meter and <span class="katex-eq" data-katex-display="false">\angle A B C=90^{\circ}</span>, Let us write by calculating the area of quadrilateral shape of field.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23650" src="https://flasheducation.online/wp-content/uploads/2023/06/889E8FE8-342E-4E42-9BC8-36813B8546E5-300x215.jpeg" alt="889E8FE8 342E 4E42 9BC8 36813B8546E5" width="470" height="336" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 67" srcset="https://flasheducation.online/wp-content/uploads/2023/06/889E8FE8-342E-4E42-9BC8-36813B8546E5-300x215.jpeg 300w, https://flasheducation.online/wp-content/uploads/2023/06/889E8FE8-342E-4E42-9BC8-36813B8546E5-1024x733.jpeg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/889E8FE8-342E-4E42-9BC8-36813B8546E5-768x550.jpeg 768w, https://flasheducation.online/wp-content/uploads/2023/06/889E8FE8-342E-4E42-9BC8-36813B8546E5-24x17.jpeg 24w, https://flasheducation.online/wp-content/uploads/2023/06/889E8FE8-342E-4E42-9BC8-36813B8546E5-36x26.jpeg 36w, https://flasheducation.online/wp-content/uploads/2023/06/889E8FE8-342E-4E42-9BC8-36813B8546E5-48x34.jpeg 48w, https://flasheducation.online/wp-content/uploads/2023/06/889E8FE8-342E-4E42-9BC8-36813B8546E5.jpeg 1056w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let } A B=5 \mathrm{~m}, B C=12 \mathrm{~m} \text {, } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">C D=14 \mathrm{~m}, \mathrm{DA}=15 \mathrm{~m} \text {, } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \angle A B C=90^{\circ} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } A C^2=\sqrt{\mathrm{AB}^2+B C^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{(5)^2+(12)^2} \text {. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{25+144} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{109} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=13 \mathrm{~m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times 5 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=30 \mathrm{Sq} . \mathrm{cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let } a=13 \mathrm{~m}, b=14 \mathrm{~m}, \mathrm{c}=15 \mathrm{~m}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">s=\frac{a+b+c}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">s=\frac{13+14+15}{2}=\frac{42}{2}=21 \mathrm{~m}</span>
<p dir="auto">Area of <span class="katex-eq" data-katex-display="false">\triangle A C D=\sqrt{s(s-a)(s-b)(s-c)}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{21(21-13)(21-14)(21-15)} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{21 \times 8 \times 7 \times 6} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 2 \times 3 \times 7 \mathrm{Sq} \cdot \mathrm{m} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=84 \text { Sq.cm. }</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of Quadrilateral } A B C D=(3+84) \text { Sq.m. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=114 \text { Sq.m (Ans.) } \\</span>
<p dir="auto"><strong>12. Sahin draws a trapezium ABCD of which length of diagonal BD is 11cm, and draws two perpendiculars of which length are 5cm and 11cm respectively from the points A and C on the diagonal BD. Let us write by calculating the area of ABCD in the shape of trapezium.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23638" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416-300x175.jpg" alt="IMG 3416" width="470" height="275" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 68" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416-300x175.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416-1024x599.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416-768x449.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416-1200x700.jpg 1200w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416-24x14.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416-36x21.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416-48x28.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3416.jpg 1213w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let, } \mathrm{BD} =11 \mathrm{~cm}, \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AE} =5 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{CF} =11 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle \mathrm{ABD}=\frac{1}{2} \times \mathrm{BD} \times \mathrm{AE} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 11 \times 5 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\jmath \rho}{2} \text { Sq.cm } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of } \triangle B C D=\frac{1}{2} \times B D \times C F \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">= \frac{1}{2} \times 11 \times 11 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{121}{2} \text { Sq.cm } \\</span>
<p dir="auto">Area of Trapezium ABCD</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Area of } \triangle A B D+\text { Area of } \triangle B C D</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(\frac{55}{2}+\frac{121}{2}) \text { Sq.cm } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{176}{2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=88 \mathrm{Sq} . \mathrm{cm} \text { (Ans.) }</span>
<p dir="auto"><strong>13. ABCDE is a petagon of which side BC is parallel to diagonal AD, EP is perpendicular on BC and EP intersects AD at the point Q. BE = 7 cm, AD = 13cm, PE = 9cm. and if PQ =<span class="katex-eq" data-katex-display="false">\frac{4}{9}</span> PE, Let us write by calculating the area of \ABCDE in shape of pentagon.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23637" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3417-300x281.jpg" alt="IMG 3417" width="470" height="440" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 69" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3417-300x281.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3417-768x719.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3417-24x22.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3417-36x34.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3417-48x45.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3417.jpg 990w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Let } P Q=\frac{4}{9}+PE= \frac{4}{9} \times 9cm \\</span>
<p dir="auto">= 4cm</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \mathrm{QE}=(9-4) \mathrm{cm} .=5 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Now, } \because A D \| B C \text { and } E P \perp B C \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{EQ} \perp \mathrm{AD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Area of pentagon } A B C D E \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\text { Area of } \triangle \mathrm{ADE}+\text { Area of trapezium } \mathrm{ABCD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times \mathrm{AD} \times \mathrm{QE}+\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC} ) \times \mathrm{PQ} \text { Sq.unit. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 13 \times 5+\frac{1}{2} \times(13+7) \times 4 \text { Sq.unit. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=(32.5+40) \cdot \mathrm{Sq} \cdot \mathrm{cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=72.5 \text { Sq.cm } \\</span>
<p dir="auto"><strong>14. The length of rhombus is equal to length of a square and is <span class="katex-eq" data-katex-display="false">40 \sqrt{2} \mathrm{~cm}.</span> If the length of diagonals of a rhombus are in the ratio 3 : 4, Then let us write by calculating the area of a field in the shape of rhombus</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23640" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-300x243.jpg" alt="IMG 3412" width="470" height="381" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 62" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-300x243.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-1024x830.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-768x622.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-24x19.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-36x29.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412-48x39.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3412.jpg 1272w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let ' a ' be the side of a square and also side of a rhombus. Then side of a rhombus.</p>
<p dir="auto">The length of diagonal = <span class="katex-eq" data-katex-display="false">40 \sqrt{2} \mathrm{~cm}. or, a \sqrt{2}=40 \sqrt{2}\\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad a=40, \therefore \mathrm{CD}=40 \mathrm{~cm} \text {. }\\</span>
<p dir="auto">Let <span class="katex-eq" data-katex-display="false">\mathrm{AC}=3 \mathrm{x}, \mathrm{BD}=4 \mathrm{x}\\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{OC}=\frac{3 \mathrm{x}}{2}, \mathrm{OD}=\frac{4 \mathrm{x}}{2} 2 \mathrm{x}\\</span>
<p dir="auto">In <span class="katex-eq" data-katex-display="false">\triangle \mathrm{COD}\\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">O C^2+O D^2=C D^2\\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">(\frac{3 x}{2})^2+(2 x)^2=(40)^2\\</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{9 x^2}{4}+4 x^2=1600\\</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{9 x^2+16 x^2}{4}=1600\\</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{25 x^2}{4}{ }^2=1600\\</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">25 \mathrm{x}^2=1600 \times 4\\</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x^2=\frac{1660 \times 4}{25}{ }^2\\</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x^2=64 \times 4\\</span>
<span class="katex-eq" data-katex-display="false">\therefore x^2 =\sqrt{64 \times 4} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=8 \times 2=16\\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad x =16\\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AC}=3 x=3 \times 16 \mathrm{~cm}=48 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BD}=4 \mathrm{x}=4 \times 16 \mathrm{~cm}=64 \mathrm{~cm} .\\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of Rhombus }\mathrm{ABCD} =\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 48 \times 64 \\</span>
<span class="katex-eq" data-katex-display="false">=1536 \text { Sq.cm. } (Ans.)</span>
<p dir="auto"><strong>15. In a trapezium, the length of each slant sides is 10cm, and the length of parallel sides are 5cm. and 17cm. respectively. Let us write by calculating the area of field in shape of trapazium and its diagonal.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23635" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419-300x162.jpg" alt="IMG 3419" width="470" height="253" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 71" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419-300x162.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419-1024x552.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419-768x414.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419-1536x828.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419-24x13.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419-36x19.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419-48x26.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3419.jpg 1589w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let <span class="katex-eq" data-katex-display="false">\mathrm{AB}=\mathrm{CD}=10 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A D=5 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">B C=17 \mathrm{~cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A D \| B C \text { and } A B=C D \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A D=E F=5 \mathrm{~cm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore A D \| B C and A B=C D</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore A D =E F=5 \mathrm{~cm} . \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{BF} =\mathrm{CE}=\frac{17-5}{2}=\frac{12}{2} \mathrm{~cm}=6 \mathrm{~cm}</span>
<p dir="auto">In <span class="katex-eq" data-katex-display="false">\triangle \mathrm{ABF}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AF} =\sqrt{\mathrm{AB}^2-\mathrm{BF}^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{(10)^2-(6)^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{100-36} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{64} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=8 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad A F =8 \mathrm{~cm} . \quad \therefore \quad \mathrm{DE}=\mathrm{AF}=8 \mathrm{~cm}.</span>
<p dir="auto">And BE = BF + EF = (6 + 5)cm =11cm (Ans.)</p>
<p dir="auto">Area of Trapazium ABCD</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AF} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times(5+17) \times 8 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 22 \times 8 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=88 \text { Sq. } \mathrm{cm}</span>
<p dir="auto">In <span class="katex-eq" data-katex-display="false">\triangle \mathrm{BED}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BD}=\sqrt{\mathrm{DE}^2+\mathrm{BE}^2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{BD}=\sqrt{(8)^2+(11)^2}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{BD}=\sqrt{64+121}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{BD}=\sqrt{64+121}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{BD}=\sqrt{185} cm.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> Length of diagonal <span class="katex-eq" data-katex-display="false">=\sqrt{185 cm}. (Ans.)</span>
<p dir="auto"><strong>16. The length of parallel sides of a trapezium are 19cm. and 9cm. and length of slant sides are 8cm and 6cm. Let us calculate the area of the field in the shape of trapezium.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23634" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420-300x176.jpg" alt="IMG 3420" width="470" height="276" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 72" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420-300x176.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420-1024x601.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420-768x450.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420-1536x901.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420-24x14.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420-36x21.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420-48x28.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3420.jpg 1814w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let <span class="katex-eq" data-katex-display="false">\mathrm{AD}=9 cm</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BC} =19 cm \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{AB} =8 cm \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{CD} =6 cm \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BC} =\mathrm{BF}+\mathrm{FE}+\mathrm{CE} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\mathrm{BF}+\mathrm{CE}+9 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{BF} =19-9-\mathrm{CE} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=10-\mathrm{CE}</span>
<p dir="auto">In <span class="katex-eq" data-katex-display="false">\triangle \mathrm{ABF},</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">A B^2=A F^2+B F^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(8)^2=A F^2+(10-C E)^2</span>
<p dir="auto">In <span class="katex-eq" data-katex-display="false">\triangle \mathrm{CDE}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\mathrm{CD}^2=\mathrm{DE}^2+\mathrm{CE}^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(6)^2=\mathrm{AF}^2+\mathrm{CE}^2[\because \mathrm{DE}=\mathrm{AF}]</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(6)^2=(8)^2-(10-C E)^2+C E^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">36=64-\{(10)^2-2.10 . \mathrm{CE}+(\mathrm{CE})^2\}-(\mathrm{CE})^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">36=64-(10)^2+20 \mathrm{CE}-\mathrm{CE}^2+\mathrm{CE}^2</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">36=64-100+20 \mathrm{CE}</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">20 \mathrm{CE}=36+36</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">20 \mathrm{CE}=72</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{CE}=\frac{72}{20}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { or, } \mathrm{CE} =\frac{36}{10}=3.6 cm \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \mathrm{DE} =\sqrt{\mathrm{CD}^2-\mathrm{CE}^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{(6)^2-(3.6)^2} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{(6+3.6)(6-3.6))} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{9.6 \times 2.4} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\sqrt{4.8 \times 4.8} \\</span>
<p dir="auto">= 4.8 cm</p>
<p dir="auto">Area of Trapazium AECE</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC}) \times \mathrm{DE} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times(9+19) \times 4.8 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{2} \times 28 \times 4.8 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=67.2 \text { Sq.cm. (Ans.) }</span>
<hr />
<h2 dir="auto"><span class="ez-toc-section" id="17_MCQ"></span>17. <span style="text-decoration: underline;">(M.C.Q)</span><span class="ez-toc-section-end"></span></h2>
<p dir="auto"><strong>(i) The of parallelogram is 1 / 3 th of its base. If the area of field is 192 sq.cm. in the shape of parallelogram, the height is</strong></p>
<p dir="auto"><strong>(a) 4 cm.</strong></p>
<p dir="auto"><strong>(b) 8 cm.</strong></p>
<p dir="auto"><strong>(c) 16 cm</strong></p>
<p dir="auto"><strong>(d) 24 cm</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto">Area of parallelogram = 192 Sq.cm.</p>
<p dir="auto">or, Base × height = 192</p>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">Base \times \frac{1}{3} base =192</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">(\text { Base })^2=476</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">Base =\sqrt{476}=24 cm.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad \text { Height } =\frac{1}{3} \times \text { base } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{1}{3} \times 24=8 cm .</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (b) is correct option</p>
<p dir="auto"><strong>(ii) If the length of one side of rhombus is 6 cm. and one angle is <span class="katex-eq" data-katex-display="false">60^{\circ}</span>, then area of field in the shape of rhombus is</strong></p>
<p dir="auto"><strong>(a) <span class="katex-eq" data-katex-display="false">9 \sqrt{3} sq.cm</span></strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">18 \sqrt{3} \mathrm{Sq} . \mathrm{cm}.</span></strong></p>
<p dir="auto"><strong>(c) <span class="katex-eq" data-katex-display="false">36 \sqrt{3} \mathrm{Sq} . \mathrm{cm}.</span></strong></p>
<p dir="auto"><strong>(d) <span class="katex-eq" data-katex-display="false">6 \sqrt{3} Sq.cm.</span></strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"> <img loading="lazy" decoding="async" class="aligncenter wp-image-23846" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456-300x211.jpg" alt="IMG 3456" width="470" height="331" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 73" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456-300x211.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456-1024x721.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456-768x541.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456-1536x1081.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456-24x17.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456-36x25.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456-48x34.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3456.jpg 1625w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\because \quad A B=B C=6 cm \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \angle A B C</span> = 60</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> C is an equilateral triangle</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \text { Area of } \triangle A B C=\frac{\sqrt{3}}{4} \times(6)^2 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=\frac{\sqrt{3}}{4} \times 36 \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=9 \sqrt{3} \text { Sq.cm. } \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\text { Area of rhombus } \mathrm{ABCD} =2 \times \triangle \mathrm{ABC} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=2 \times 9 \sqrt{3} \mathrm{Sq} \cdot \mathrm{cm} \\</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">=18 \sqrt{3} \mathrm{Sq} . \mathrm{sm}</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span>(b) is correct option</p>
<p dir="auto"><strong>(iii) The length of one diagonal of rhombus is three times of another diagonal. If the area of field in the shape of rhombus is 96 sq.cm., then the length of long diagonal is</strong></p>
<p dir="auto"><strong>(a) 8 cm</strong></p>
<p dir="auto"><strong>(b) 12 cm</strong></p>
<p dir="auto"><strong>(c) 16 cm</strong></p>
<p dir="auto"><strong>(d) 24 cm.</strong></p>
<p dir="auto"><em><strong>Solution:</strong></em></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23845" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457-300x197.jpg" alt="IMG 3457" width="470" height="309" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 74" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457-300x197.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457-1024x674.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457-768x505.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457-1536x1011.jpg 1536w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457-24x16.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457-36x24.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457-48x32.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3457.jpg 1594w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">Let 1st diagonal =x</p>
<p dir="auto">2nd diagonal = 3x</p>
<p dir="auto">Area of Rhombus = 96 Sq.cm.</p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\frac{1}{2} \times x \times 3 x=96</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{3 x^2}{2}=96</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x^2=64</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">x=\sqrt{64}=8</span>
<p dir="auto">Length of long diagonal <span class="katex-eq" data-katex-display="false">= 3 \mathrm{x} =3 \times 8 cm=24 cm.</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> (d) is correct option</p>
<p dir="auto"><strong>(iv) A rhombus and a square on the same base. If the area of square is <span class="katex-eq" data-katex-display="false">x^2</span> Sq. unit and area of field in the shape of rhombus is y sq. unit. then</strong></p>
<p dir="auto"><strong>(a) <span class="katex-eq" data-katex-display="false">y>x^2</span></strong></p>
<p dir="auto"><strong>(b) <span class="katex-eq" data-katex-display="false">v<x^2</span></strong></p>
<p dir="auto"><strong>(c) <span class="katex-eq" data-katex-display="false">y=x^2</span></strong></p>
<p dir="auto"><strong>Solution:</strong></p>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore</span> (b) is correct option</p>
<p dir="auto"><strong>(v) Area of a field in the shape of trapezium is 162 sq.cm. and height is 6 cm. If length of one side is 23 cm, then the length of other side is</strong></p>
<p dir="auto"><strong>(a) 29 cm</strong></p>
<p dir="auto"><strong>(b) 31 cm</strong></p>
<p dir="auto"><strong>(c) 32 cm</strong></p>
<p dir="auto"><strong>(d) 33 cm.</strong></p>
<p dir="auto"><strong><em>Solution:</em></strong></p>
<p dir="auto"><img loading="lazy" decoding="async" class="aligncenter wp-image-23633" src="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3422-300x180.jpg" alt="IMG 3422" width="470" height="282" title="Chapter - 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9 75" srcset="https://flasheducation.online/wp-content/uploads/2023/06/IMG-3422-300x180.jpg 300w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3422-1024x615.jpg 1024w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3422-768x461.jpg 768w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3422-24x14.jpg 24w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3422-36x22.jpg 36w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3422-48x29.jpg 48w, https://flasheducation.online/wp-content/uploads/2023/06/IMG-3422.jpg 1435w" sizes="(max-width: 470px) 100vw, 470px" /></p>
<p dir="auto">We have,</p>
<p dir="auto">DE = 6 cm,</p>
<p dir="auto">AB = ?, CD = 23 cm,</p>
<p dir="auto">Area of trapazium <span class="katex-eq" data-katex-display="false">=162 \mathrm{Sq} . \mathrm{cm}.</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{1}{2} \times(\mathrm{AB}+\mathrm{CD}) \times \mathrm{DE}=162</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\frac{1}{2} \times(\mathrm{AB}+23) \times 6=162</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{AB}+23=54</span>
<p dir="auto">or, <span class="katex-eq" data-katex-display="false">\mathrm{AB}=54-23</span>
<p dir="auto"><span class="katex-eq" data-katex-display="false">\therefore \quad</span> AB = 31cm.
\therefore (b) is correct option (i) Area of field in the shape of parallelogram ABCD 96 sq.cm., length of diagonal BD is 12 cm: What is the perpendicular length drawn on diagonal BD from the point A? Solution: Area of parallelogram ABCD = 96 sq. cm. \mathrm{cm} \therefore Area of \bigtriangleup ABD = \frac{96}{2} Sq.cm.
=48 \text { Sq. } \mathrm{cm}
or, \frac{1}{2} \times \mathrm{AE} \times \mathrm{BD}=48
or, \frac{1}{2} \times \mathrm{AE} \times 12=48
\therefore \quad \mathrm{AE}=8 cm \quad (Ans.)
(ii) The length of adjacent sides of a parallelogram are 5 cm and 3 cm. If the distance between the longer side 2 cm. Find the distance between the smaller sides. Solution: \text { Let } AB \& = CD=3 cm . \\
AD = BC = 5 cm AF = 2 cm Area of \triangle \mathrm{ABD} =\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE} \\
=\frac{1}{2} \times 3 \times \mathrm{DE} \\
=\frac{3}{2} \mathrm{DE}
Area of parallelogram ABCD =2 \times \frac{3}{2} \times \mathrm{DE}=3 \times \mathrm{DE}
\therefore \text{Area of the parallelogram} \mathrm{ABCD} =\mathrm{BC} \times \mathrm{AF} \\
=5 cm \times 2 cm \\
=10 \mathrm{Sq} . \mathrm{cm} . \\
\therefore \quad 3 \times \mathrm{DE} =10 \\
\text { or, } \quad \mathrm{DE} =\frac{10}{3} \\
\therefore \quad \mathrm{DE} =3 \frac{1}{3} cm \text { (Ans.) }
(iii) Length of height of rhombus is 14 cm . and length of side is 5 cm. What is the area of field in the in the shape of rhombus? Solution: \text { Area of rhombus } =\text { Base } \times \text { Height } \\
=5 cm \times 14 cm . \\
=70 \mathrm{Sq} . \mathrm{cm} .
\text { (Ans.) }
(iv) Any adjacent parallel sides of trapeziun makes an angle 45^{\circ} and length of its slant side is 62 cm, What is the distance between two parallel sides? Solution: Let \mathrm{AB}=62, cm
\angle \mathrm{ABE}=\angle \mathrm{EAB}=45^{\circ}
AE = Height of trapezium We have, \mathrm{AE}^{2}+\mathrm{BE}^{2}=\mathrm{AB}^{2}
or, \mathrm{AE}^{2}+\mathrm{AE}^{2}=(62)^{2} \quad[\because \mathrm{BE}=\mathrm{AE}]
or, 2 \mathrm{AE}^{2}=(62)^{2}
or, \mathrm{AE}^{22}=\frac{(62)^{2}}{2}
\text { or, } \mathrm{AE} =\frac{(62)^{2}}{4} \times 2 \\
\text { or, } \mathrm{AE} =\sqrt{\frac{(62)^{2}}{4} \times 2} \\
\text { or, } \mathrm{AE} =\frac{62}{2} \times \sqrt{2} \\
=31 \sqrt{2} cm
\therefore Distance between two parallel sides = 31 \sqrt{2} cm (Ans.) (v) In parallelogram ABCD, AB = 4 cm, BC = 6 cm, and \angle A B C=30^{\circ} find the area of field in the shape of parallelograin ABCD. Solution: \mathrm{AB}=4 cm, \mathrm{BC}=6 cm
We draw perpendicular AE on BC Then, \angle B A E=60^{\circ} ,
\angle \mathrm{AEB}=90^{\circ}, \angle \mathrm{ABE}=30^{\circ}
In a right angled triangle, If the angles are 90^{\circ}, 60^{\circ} \ 30^{\circ}, then, \mathrm{AE} =\frac{1}{2} \mathrm{AB} \\
=\frac{1}{2} \times 4 cm=2 cm
\therefore \text { Area of parallelogram } =\text { Base } \times \text { altitude } \\
=\mathrm{BC} \times \mathrm{AE} \\
=6 \times 2 \mathrm{Sq} \cdot \mathrm{cm}. \\
=12 \mathrm{Sq} . \mathrm{cm} . \text { (Ans.) }
18. Short answer type: