| 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}
=\frac{1}{2} \times 8 \times \sqrt{(6)^2-(\frac{8}{2})^2} \\
=\frac{1}{\cancel2} \times \cancel8 \times \sqrt{(36)-16} \\
=4 \sqrt{20} \\
=4 \sqrt{2 \times 2 \times 5} \\
=4 \times \sqrt{5} \\
=8 \sqrt{5} \text { Sq.cm (Ans.) }
(iv) Let a = 16cm, b = 14cm, c = 10cm.
s =\frac{a+b+c}{2} \\
=\frac{16+1+\therefore}{2} \\
=\frac{40}{2} \mathrm{~cm}=20 \mathrm{~cm} .
Area of \triangle B C D =\sqrt{s(s-a)(s-b)(s-c)} \\
=\sqrt{20(20-16)(20-14)(20-10)} \\
=\sqrt{20 \times 4 \times 6 \times 10} \\
=\sqrt{10 \times 2 \times 2 \times 2 \times 2 \times 3 \times 10} \\
=2 \times 2 \times 10 \sqrt{3} \\
=40 \sqrt{3} \mathrm{Sq} . \mathrm{cm}(\text { Ans. })
Area of \angle \mathrm{ABCD}=2 \times \text { area of } \triangle \mathrm{BCD} =2 \times 40 \sqrt{3} \text { Sq.cm. } \\
=80 \sqrt{3} \quad \text { (Ans.) }
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
Solution:
Let the depth of water be x.
\therefore \quad A C=x, A B=15 \mathrm{~cm} \text {. }
\therefore \quad B C=(x+2) \mathrm{cm} \text {. }
We have,
\mathrm{BC}^2=\mathrm{AB}^2+\mathrm{AC}^2
or, (x+2)^2=(15)^2+(x)^2
or, (x)^2+2 \cdot x \cdot 2+(2)^2=225+x^2
or, \cancel{x^2}+4 x+4=225+\cancel{x^2}
or, 4 x=225-4
or, 4 \mathrm{x}=221
or, x=\frac{221}{4}
\therefore \quad \mathrm{x}=55.25
\thereforeDepth of water =55.25cm (Ans.)
3. The length of hypotenuse of an isosceles right-angle triangle is 12 \sqrt{2} \mathrm{~cm}, Let us write by calculating what wi be the area of that fied.
Solution:
\because \triangle A B C is an lsosceles right angled triangle.
\therefore \quad \mathrm{AB}=\mathrm{BC}
Let AB = BC = x
We know that,
\mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2
or, (x)^2+(x)^2=(12 \sqrt{2})^2
or, x^2+x^2=144 \times 2
or, \cancel2 x^2=144 \times \cancel2
or, x^2=144
\text { or, } x=\sqrt{144} \quad \therefore \quad x=12
\therefore \mathrm{AB}=\mathrm{BC}=12 \mathrm{~cm}.
Area of \triangle A B C=\frac{1}{2} \times B C \times A B
=\frac{1}{2} \times 12 \times 12 \mathrm{sq} \cdot \mathrm{cm}=72 \mathrm{Sq} \cdot \mathrm{cm}
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.
Solution:
\text { Let } a=65 m, b=70 m, c=75 m \\
s=\frac{a+b+c}{2} \\
=\frac{65+70+75}{2} \\
=\frac{210}{2}=105 \mathrm{~m}
Area of \triangle A B C =\sqrt{s(s-a)(s-b)(s-c)} \\
=\sqrt{105(105-65)(105-70)(105-75)} \\
=\sqrt{105 \times 40 \times 35 \times 30} \\
=\sqrt{35 \times 3 \times 10 \times 2 \times 2 \times 35 \times 3 \times 10} \\
=35 \times 3 \times 2 \times 10 \mathrm{Sq} . \mathrm{m} . \\
=2100 \mathrm{Sq} . \mathrm{m}
\therefore We also know that,
Area of \triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{CD} \\
\text { or, } 2100=\frac{1}{2} \times 75 \times C D \\
\text { or, } 75 \mathrm{CD}=4200 \\
\text { or, } C D=\frac{4200}{75} \\
\therefore \quad C D=56 \mathrm{~m} \\
\therefore \quad \text { Length of perpendicular }=56 \mathrm{~m} \text { (Ans.) } \\
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.
Solution: We have.
Ratio of the twa triangles = 3 : 4
\therefore Ratio of area of the two triangles = 4 : 3
\therefore \frac{\text { Area of } 1 \text { st. triangle }}{\text { Area of } 2 \text { nd. triangle }}=\frac{4}{3}
\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}
or, \frac{\text { Base of } 1 \text { st. triangle }}{\text { Base of } 2 \text { nd. triangle }} \times \frac{3}{4}=\frac{4}{3}
\therefore \quad \frac{\text { Base of } 1 \text { st. triangle }}{\text { Base of } 2 \text { nd triangle }} \times \frac{16}{9}
\therefore Ratio of two bases = 16 : 9 (Ans.)
Let us work out – 15.1
1. I see the house of Kamal and let us find the answers.
(i) Let us write by calculating the area of Kamal’s garden.
(ii) Let us write by calculating how much cost is required to repair the floor of Kamal’s verandah at the rate of Rs. 30/m.
(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 the floor of the reading room with the size of tiles 25cm. × 25cm.
Solution:
(i) Area of Kamal’s garden =20 \mathrm{~m} \times 20 \mathrm{~m}=400 \mathrm{Sq} . \mathrm{m}. (Ans)
(ii) Area of Kamal’s varandah =10 \mathrm{~m} \times 5 \mathrm{~m}=50 \mathrm{sq} . \mathrm{m}.
\therefore \quad Cost of repair the floor of Kamal’s verandah = Rs. 30 \times 50
=\mathrm{Rs} .1500(\mathrm{Ans} \text {.) }
(iii) Length of reading room =6 \mathrm{~m}=600 \mathrm{~cm}.
Breadth \text{ " " " " " } =5 \mathrm{~m}=500 \mathrm{~cm}.
Size of one tile =25 \mathrm{~cm} . \times 25 \mathrm{~cm}.
No. of tiles =\frac{600 \times 500}{25 \times 25}=480 (Ans.)
2. Let us see the following pictures and calculate the area of its coloured part.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Solution:
(i) 
Area of ABCD = AB × AD
=12 \mathrm{~m} \times 8 \mathrm{~m} \\
=96 \text { sq. } \mathrm{m} \text {. } \\
\therefore \quad \mathrm{AE}=12 \mathrm{~m}-3 \mathrm{~m}=9 \mathrm{~m} \text {. } \\
AG = 8m – 3m = 5m
\text { Area of } \mathrm{AEFG}=\mathrm{AE} \times \mathrm{AG} \\
=9 \mathrm{~m} \times 5 \mathrm{~m} \\
=45 \text { sq. } \mathrm{m} \text {. } \\
Area of coloured part =(96-45) Sq.m.
=51 \mathrm{sq} \cdot \mathrm{m} \text {. (Ans.) }
(ii) 
\mathrm{AB}=14 \mathrm{~m}, \quad \mathrm{BC}=26 \mathrm{~m} \\
\text { Area of } \mathrm{ABCD}=\mathrm{AB} \times \mathrm{BC} \\
=14 \mathrm{~m} \times 26 \mathrm{~m} \\
=364 \mathrm{sq} . \mathrm{m}
\mathrm{DE}=\frac{26-3}{2}=\frac{23}{2} \mathrm{~m} \\
\mathrm{DG}=\frac{14-3}{2} \mathrm{~m}=\frac{11}{2} \mathrm{~m} \\
\text { Area of DEFG }=\mathrm{DE} \times \mathrm{DG} \\
=\frac{23}{2} \mathrm{~m} \times \frac{11}{2} \mathrm{~m} \\
=\frac{253}{4} \mathrm{Sq} . \mathrm{m}
\text { Area of } 4 \text { small square land }=4 \times \frac{253}{4} \text { Sq.m. } \\
=253 \mathrm{Sq} \cdot \mathrm{m} \text {. } \\
\text { Area of coloured part }=(364-253) \text { Sq.m } \\
=111 \text { Sq.m. (Ans.) } \\
(iii) 
\because \quad A B=16 \mathrm{~m}, \quad A D=9 \mathrm{~m} \text {. } \\
\mathrm{EF}=16 \mathrm{~m}+2 \times 4 \mathrm{~m} \\
=16 m+8 m=24 m \text {. } \\
\mathrm{HE}=9 \mathrm{~m}+2 \times 4 \mathrm{~m} \\
=9 m+8 m=17 m \text {. } \\
\text { Area of } \mathrm{ABCD}=\mathrm{AB} \times \mathrm{AD} \\
=16 \mathrm{~m} \times 9 \mathrm{~m} \\
=144 \mathrm{Sq} . \mathrm{m} \text {. } \\
\text { Area of } \mathrm{EFGH}=\mathrm{EF} \times \mathrm{HE} \\
=24 \mathrm{~m} \times 17 \mathrm{~m} \\
=408 \mathrm{Sq} . \mathrm{m} \text {. } \\
Area of coloured part =(408-144) \mathrm{Sqm}
=264 \text { Sq.m. (Ans.) }
(iv) 
\because \quad A B =28 \mathrm{~m}, \quad B C=20 \mathrm{~m} . \\
\mathrm{EF} =(28-2 \times 3) \mathrm{m} \\
=(28-6) \mathrm{m}=22 \mathrm{~m} \\
\mathrm{FG} =(20-2 \times 3) \mathrm{m}
=(20-6) \mathrm{m} \\
=14 \mathrm{~m} .
\text { Area of } \mathrm{ABCD} =\mathrm{AB} \times \mathrm{BC} \\
=28 \mathrm{~m} \times 20 \mathrm{~m} \\
=560 \mathrm{Sq} \cdot \mathrm{m} . \\
\text { Area of EFGH } =\mathrm{EF} \times \mathrm{FG} \\
=22 \mathrm{~m} \times 14 \mathrm{~m} \\
=308 \mathrm{Sq} . \mathrm{m} .
\text { Area of coloured part } =(560-308) \text { Sq.m. } \\
=252 \mathrm{Sq} \cdot \mathrm{m} . \text { (Ans.) }
(v) 
\because \quad \mathrm{BC}=120 \mathrm{~cm}, \quad \mathrm{CD}=90 \mathrm{~cm} . \\
\text { Area of } \mathrm{ABCD}=\mathrm{BC} \times \mathrm{CD} \\
=120 \mathrm{~cm} \times 90 \mathrm{~cm} . \\
=10800 \mathrm{Sq} . \mathrm{cm} . \\
\mathrm{II}=\mathrm{BC}=120 \mathrm{~cm}, \mathrm{IJ}=3 \mathrm{~cm} . \\
\text { Area of IJKL }=\mathrm{BC} \times \mathrm{IJ} \\
=120 \mathrm{~cm} \times 3 \mathrm{~cm}=360 \mathrm{Sq} . \mathrm{cm} . \\
\mathrm{EH}=3 \mathrm{~cm}, \quad \mathrm{FE}=\mathrm{BJ}=\frac{90-3}{2}=\frac{87}{2}
\mathrm{IL}=\mathrm{BC}=120 \mathrm{~cm}, \mathrm{IJ}=3 \mathrm{~cm}
\mathrm{EH}=3 \mathrm{~cm}, \quad \mathrm{FE}=\mathrm{BJ}=\frac{90-3}{2}=\frac{87}{2}
Area of \mathrm{EFGH}=\mathrm{EH} \times \mathrm{FE}
=3 \mathrm{~cm} \times \frac{87}{2} \mathrm{~cm}=\frac{261}{2} \text { Sq.cm }
\text { Area of } 4 \text { small rectangle } =4 \times \frac{261}{2} \text { Sq.cm } \\
=522 \mathrm{Sq} . \mathrm{cm} .
\text { Area of coloured part } =(360+522) \text { Sq.cm. } \\
=882 \text { Sq.cm.(Ans.) }
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.
Solution:
\text { Let length } =4 x, \\
\text { breadth } =3 x,
x, being the common ration.
\therefore \text { Perimeter } =2(4 x+3 x) \\
=2 \times 7 x \\
= 14x
According to the condition of the problem,
or, 14 \mathrm{x}=336
\text { or, } x=\frac{336}{14} \\
\therefore \quad \mathrm{x}=24 \\
\therefore \quad \text { Length }=4 x=4 \times 24=96 \mathrm{~m} \\
\text { Breadth }=3 \mathrm{x}=3 \times 24=72 \mathrm{~m} \\
\therefore \quad \text { Area of the rectangular field }=\text { Length } \times \text { Breadth } \\
=96 \mathrm{~m} \times 72 \mathrm{~m} \\
=6912 \text { Sq.m (Ans.) } \\
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.
Solution: The cost of farming a square land of Samar at the rate of Rs. 3.50 per sq.meter is Rs. 1400 .
\therefore \quad Area of the square land =(1400 \div 3.50) sq.m.
=\cancel{1400} \times \frac{100}{\cancel{350}} \text { Sq.m. } \\
= 400 \text { Sq.m. }
\text { Length of side of a square land } =\sqrt{\text { Area }} \\
=\sqrt{400} \quad=20 \mathrm{~m} .
Length of fencing around its four sides
=4 \dot{\times} \text { side } \\
=4 \times 20 \mathrm{~m}=80 \mathrm{~m} . \\
\text { Total cost of fencing }=\text { Rs. } 8.50 \times 80 \\
=\frac{850}{100} \times 80 \\
=\text { Rs. } 680 \text { (Ans.) }
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.
Solution: Let the length of the rectangular land be x
and ” Breadth ” ” ” ” ” ” ” ” ” y
\therefore Area of rectangular land = 500 sq.m
or, Length \times Breadth =500 \mathrm{sq} . \mathrm{m}
or, \mathrm{xy}=500…………. (i)
If length of land is decreased by 3m and breadth is increased by 2m, then the land formed a square.
\therefore \quad x-3=y+2
or, x=y+2+3
or, x=y+5……………. (ii)
Putting the value of x in equation
(y+5) y=500
or, y^2+5 y-500=0
or, y^2+25 y-20 y-500=0
or, y(y+25)-20(y+25)=0
or, (y+25)(y-20)=0
either,
y + 25 = 0
y – 20 = 0
y = -25
y = 20
Since distance can not be negative.
\therefore y = 20
From (ii),
x = y + 5
= 20 + 5 = 25
\therefore Length of the recangular field = 25m.
Breadth \text{ " " " " " " " " "} = 20m (Ans.)
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.
Solution: Side of a square land = 300m
Area \text{ " " " " " " " " "} =(3.00)^2 \text { Sq.m. } \\
=90000 \text { Sq.m. } \\
\text { Wide of the wall }=3 \mathrm{dcm} . \quad=0.3 \mathrm{~m} \text {. } \\
\text { Length of the square with wide }=(300+2 \times 0.3)^2 \mathrm{~m} \\
=(300+0.6) \mathrm{m} \\
=300.6 \mathrm{~m} \\
=(300.6)^2 \\
=90360.36 \mathrm{Sq} \cdot \mathrm{m} \\
\text { Area of } 4 \text { walls }=(90360.36-90000) \text { Sq.m. } \\
=360.36 \text { Sq.m. } \\
Cost of the wall 100 Sq.m at the rate of Rs. 5,000
\text { " " " } 1 \text { " } 1 \text { " } 360.36 \text { sq.m:" } "
=\frac{5,0\cancel{00}}{1\cancel{00}}=\text { Rs. } 50 \\
\text { " " " "360.36sq.m. " "" " " } =\text { Rs. } 360.36 \times 50 \\
=\text { Rs. } \frac{36036}{100} \times 50 \\
=\text { Rs. } 18018 \text { (Ans.) }
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.
Solution:
Length of rectangular garden = 14m
Breadth ” ” ” ” ” ” ” = 12m
Let the width of the park = x
\text { Length of rectangular garden } =14-2 × x \\
= 14 – 2x
Breadth ” ” ” ” ” ” = 12 – 2 × x
= 12 – 2x
Cost of constructing of wide path of 20 sq.m = Rs. 1380
\text{ " " " " " " " " " " " " "} 1 \ sqm. =Rs\frac{1380}{20} \\
=\text { Rs. } 69
According to the condition of the problem
(14-2 x)(12-2 x)=69 \\
\text { or, } 168-28 x-24 x+4 x^2=69 \\
\text { or, } 4 x^2-52 x+168-69=0 \\
\text { or, } 4 x^2-52 x+69=0 \\
\text { or, } 4 x^2-46 x-6 x+69=0 \\
\text { or, } 2 x(2 x-23)-3(2 x-23)=0 \\
\text { or, }(2 x-23)(2 x-3)=0
either,
2 x-23=0 \quad \text { or, } \quad 2 x-3=0
or, 2 x=23 \quad \text { or, } 2 x=3
\therefore x=\frac{23}{2}=11.5 \quad \therefore x=\frac{3}{2}=1.5
\therefore \quad x = 11.5 is not possible because wide can not be greater or equal or nearest to recatangular length
\therefore x = 1.5
\therefore \quad Wide of the path = 1.5m (Ans.)
8. If the length of recatangular garden with area 1200 sq.cm. is 40cm. then let us write by calculating the area square field which is drawn on its diagonal.
Solution:
Length of rectangular garden = 40cm.
Area \text{ " " " " " " " " " " " " "} = 1200 Sq.cm.
Breadth \text{ " " " " " " " " " " " " "} = \frac{\cancel{1200}}{\cancel{40}} cm
= 30cm.
\therefore \text { Lenth of diagonal of rectangle } =\text { side of the square. } \\
=\sqrt{(40)^2+(30)^2} \\
=\sqrt{1600+900} \\
=\sqrt{2500} \\
=50 \mathrm{~cm} .
\therefore \text { Area of the square field } =(\text { Side })^2 \\
=(50)^2 \\
=2500 \text { Sq.cm. } \quad \text { (Ans.) }
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.
Solution: Length = 4m, Breadth = 6cm, height = 4m :
\text { Area of one door } =1.5 \times 1 \mathrm{~m} =1.5 \mathrm{Sq} . \mathrm{m} . \\
\text { Area of three doors } =3 \times 1.5 \mathrm{Sq} . \mathrm{m} =4.5 \mathrm{Sq} . \mathrm{m} . \\
\text { Area of one window } =1.2 \mathrm{~m} \times 1 \mathrm{~m} =1.2 \mathrm{Sq} . \mathrm{m} \\
” ” ” ” \quad 4 \text { Windows } =1.2 \times 4 \mathrm{Sq} . \mathrm{m} . =4.8 \mathrm{Sq} . \mathrm{m} .
Area of the walls with doors and windows
=2 \text { (Length }+ \text { Breadth) } \times \text { height. } \\
=2(4 \mathrm{~m}+6 \mathrm{~m}) \times 4 \mathrm{~m} \\
=2 \times 10 \mathrm{~m} \times 4 \mathrm{~m} . \\
=80 \mathrm{Sq} . \mathrm{m} .
Area of the walls without doors and windows.
=\{80-(4.5+4.8)\} \text { Sq.m. } \\
=(80-9.3) \mathrm{Sq} \cdot \mathrm{m} \\
=70.7 \mathrm{Sq} \cdot \mathrm{m} .
\text { Cost of colouring } =\text { Rs. } 70 \times 70.7 \\
=\text { Rs. } 4949 \quad \text { (Ans.) }
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.
Solution :
Length of room = 4m
Area of the floor = 12 \mathrm{Sq} \cdot \mathrm{m}
or, Length × Breadth = 12
\text { or, Breadth }=\frac{12}{\text { Length }} \\
=\frac{12}{4}=3 \mathrm{~m} \text {. } \\
\therefore Area of four walls of a room = 42 sq.cm
or, 2 (Length + Breadth) \times height =42
or, 2(4+3) \times height =42
or, 2 \times 7 \times height =42
or, 14 \times height =42
or, height =\frac{\cancel{42}}{\cancel{14}}
\therefore height =3 \mathrm{~m}.
\therefore Height of room =3 \mathrm{~m} (Ans)
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.
Solution : Let the length of the rectangular picture be x
and ” breadth” \text{ " " " " " " " " " " " " "} y
\therefore \quad Area = 84 Sq.cm.
or, x \times y=84
or, xy = 84…………… (i)
According to the condition of the problem,
x – y = 5
\text { or, } \quad x=5+y………. (ii)
Putting the value of x in equation ……………… (i)
(5 + y) y = 84
\text { or, } 5 y+y^2=84 \\
\text { or, } y^2+5 y-84=0 \\
\text { or, } y^2+12 y-7 y-84=0 \\
\text { or, } y(y+12)-7(y+12)=0 \\
\text { or, }(y+12)(y-7)=0
Either,
y + 12 = 0
or, y = -12
or, y – 7 = 0
\therefore \quad y = 7
\because Length can’t be negative
\therefore y=7
From (ii),
x =5 + y
= 5 + 7 = 12
\therefore Length of the rectangular picture = 12m.
\text { Breadth " " " " " } = 7m.
\therefore Perimeter ” ” ” ” = 2(12+7)m
=2 \times 19 \mathrm{~m}
=38 \mathrm{~m} \text {. (Ans.) }
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.
Solution : Let the side of the square garden be x There is a 2.5m wide path around the square.
Side of square garden with path
=(x+2 \times 25) \quad =(x+5)
Accodring to the condition of the problem,
(x+5)^2-(x)^2=165
or, (x)^2+2 \cdot x .5+(5)^2-x^2=165
or, \cancel{x^2}+10 x+25- \cancel{x^2}=165
or, 10 \mathrm{x}=165-25
or, 10 \mathrm{x}=140
\therefore \quad \mathrm{x}=14
\therefore Side of the square garden
\text { Side of the square garden } =14 \mathrm{~m} \\
\text { Area" } =(\text { side })^2 \\
=(14)^2 \\
=196 \text { Sq.m. }
\text { Length of diagonal } =\text { side } \sqrt{2} \mathrm{~m} \\
=14 \sqrt{2} \mathrm{~m} (Ans.)
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 20 \sqrt{2} meter. Let us write by calculating how much cost will be for planting grass at the rate of Rs. 20 per sq. meter.
Solution: Length of the diagonal of the square land = 20 \sqrt{2} \mathrm{~m}
\therefore \quadSide of the square land = 20m
Length of wall for walling outside round the square field = perimeter of the square.
= 4 × side
= 4 × 20m = 80m
\therefore \quad Area of the square land = (\text { side })^2
=(20)^2 \\
=400 \text { Sq. } \cdot \mathrm{m}
\therefore Cost for planting grass at the rate of Rs. 20 per Sq.m
=\text { Rs. } 20 \times 400 \\
=\text { Rs. } 8000 \quad \text { (Ans.) }
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.
Solution : Length of the rectangular garden = 12m.
\text { Breadth" " " " " " " " " " " " } = 7m
\text { Length of diagonal } =\sqrt{(\text { Length })^2+(\text { breadth })^2} \\
=\sqrt{(12)^2+(7)^2}
=\sqrt{193 \mathrm{~m}} \\
\text { Perimeter of the triangle } =12 \mathrm{~m}+7 \mathrm{~m}+\sqrt{193 \mathrm{~m}} \\
= (19+\sqrt{193}) \mathrm{m} \text { (Ans.) }
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.
Solution:
Let length of rectangle be 9x
\text { Breadth" " " " " " " " " " " " } 5 x
\therefore Perimeter of rectangle = 140m
or, 2 (Length + Breadth )= 140
or, 2(9x + 5x) = 140
or, x =\frac{140}{4 \times \sqrt{4}}
\therefore \quad x = 5
\therefore \quad Length of rectangle = 9 \mathrm{x}=9 \times 5=45 \mathrm{~m}
Breadth of rectangle =5 \mathrm{x}=5 \times 5=25 \mathrm{~m}
Area of rectangle = length \times breadth
=45 \mathrm{~m} \times 25 \mathrm{~m} \\
=1125 \text { Sq.m }
\text { Dimension of one tile } =25 \mathrm{~cm} \times 20 \mathrm{~cm} \\
=\frac{25}{100} \times \frac{20}{100} \text { Sq.m. } \\
=\frac{1}{20}
\text { No. of tiles } =1125 \div \frac{1}{20} \\
=1125 \times 20=22500
Rate of each 100 tiles = Rs. 500
\text { " " " " } 1tile = Rs. \frac{500}{100} \\
\text { " " " } 22500 Rs. \frac{500}{100} \times 22500 \\
=\mathrm{Rs}: 112500 \text { (Ans.) }
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.
Solution : Length of Rectangle = 18m
Let breadth of Rectangle be x
Total cost = Rs. 2160
Area of Rectangle = 18 \times x=18 \times Sq.m.
If the breadth of the floor would be 4 meter less.
\therefore Breadth of Rectangle = x – 4
Area of Rectangle =18(x-4) \mathrm{Sq} . \mathrm{m}
Cost of carpeting of 18x Sq. m. = Rs .2160
\text {" " " " " " " " " " " " }18(x – 4) Sq.m = Rs. 1620
Cost of carpeting of 18x – 18(x – 4)=Rs. (2160-1620)
\text {" " " " " " " " " " " " }18x – 18x + 72 = Rs. 540
\text {" " " " " " " " " " " " } 72 \mathrm{Sq} . \mathrm{m} .=Rs. 540
If Rs. 540, cost of carpeting of 72 Sq.m
” 1 \text {" " " " " " " " " " " " } \frac{72}{540} \mathrm{Sq} \cdot \mathrm{m}
” 1 \text {" " " " " " " " " " " " } \frac{72 \times 2160}{540} \mathrm{Sq} . \mathrm{m} \\
=288 \mathrm{Sq} . \mathrm{m}
\therefore \quad Area of Rectangle =288 Sq.m
or, 18x = 288
or, x =\frac{288}{18}
\therefore \quad x = 16
\therefore \quad \text { Breadth of Rectangle }=16 \mathrm{~m} \\
\text { Perimeter " " " " " " "} =2(\text { Length }+ \text { Breadth }) \\
=2(18 \mathrm{~m}+16 \mathrm{~m}) \\
=2 \times 34 \mathrm{~m}=68 \mathrm{~m}.\\
Area of Rectangle = 288 Sq.m. (Ans.)
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.
Solution :
Let the length of the rectangular land be x
and Breadth ” ” ” ” ” ” y
According to the condition of the problem,
\sqrt{x^2+y^2}=15
or, x^2+y^2=225
And x – y = 3
or, x = 3 + y
Putting the value of x in equation (i)
(3+y)^2+y^2=225
or, (3)^2+2 \cdot 3 \cdot y+y^2+(y)^2=225
or, 9+6 y+y^2+y^2-225=0
or, 2 y^2+6 y-216=0
or, y^2+3 y-108=0
or, y^2+12 y-9 y-108=0
or, y(y+12)-9(y+12)=0
or, (y+12)(y-9)=0
either, y + 12 = 0
y = -12
or, y – 9 = 0
y = 9
\therefore Length can not be negative.
\therefore y = 9
From (ii),
x = 3 + y
= 3 + 9 = 12
\therefore Length =12 \mathrm{~m} . \quad Breadth =9 \mathrm{~m}.
\therefore Perimeter of the rectangular land
=2 \text { (Length }+ \text { Breadth) } \\
=2(12 \mathrm{~m}+9 \mathrm{~m}) \\
=2 \times 21 \mathrm{~m} \\
=42 \mathrm{~m} .
Area of the rectangular land
=\text { length } \times \text { breadth } \\
=12 \mathrm{~m} \times 9 \mathrm{~m}=108 \text { Sq.m (Ans.) }
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.
Solution: First we find the H.C.F of 385 and 60
\therefore \quad H.C.F of 385 and 60 = 5
Side of the square tile = 5m
Area of rectangular courtyard = 385 \mathrm{~m} \times 60 \mathrm{~m}
Area of one square tile = (\text { Side })^2
=(5)^2 \\
=25 \mathrm{Sq} \cdot \mathrm{m} .
\therefore \quad \text { No. of tiles } =\frac{385 \times 60}{25} \\
=924 \text { (Ans.) }
19. (M.C.Q) :
(i) The length of diagonal of square is 12 \sqrt{2} \mathrm{~m}. The area of square is
(a) 288 \mathrm{sq} \cdot \mathrm{m}
(b) 144 \mathrm{~m}^2
(c) 72 \mathrm{~m}^2
(d) 18 \mathrm{~m}^2
Solution : Length of diagonal of square 12 \sqrt{2} \mathrm{~m}
\therefore \quad \text { Side } \sqrt{2}=12 \sqrt{2} \mathrm{~m}.
\therefore \quad \text { Side of a square } =12 \mathrm{~m} \\
\therefore \quad \text { Area of square } =(\text { Side })^2 \\
=(12)^2 \\
=144 \mathrm{~m}^2
\therefore (b) is correct option
(ii) If the area of square is A_1 \mathrm{sq}. units and the area of square drawn on the diagonal of that square is A_2 sq. unit, then the ratio of A_1: A_2 is
(a) 1: 2
(b) 2: 1
(c) 1: 4
(d) 4: 1
Solution : \quad Let A_1=a^2
and A_2=(a \sqrt{2})^2=2 a^2
\therefore A_1: A_2=1: 2
\therefore (a) is correct option
(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
(a) 1200
(b) 2400
(c) 600
(d) 1800
Solution : Area of Rectangle = 6 \mathrm{~m} \times 4 \mathrm{~m}
=24 \mathrm{~m}^2
\text { Area of square }=2 \mathrm{dm} \times 2 \mathrm{dm} \\
=\frac{2}{10} \times \frac{2}{10} \mathrm{~m}^2 \\
\text { No. of tiles }=\frac{24}{\frac{2}{10} \times \frac{2}{10}}=\frac{24 \times 10 \times 10}{2 \times 2}=600
No. of tiles =\frac{24}{\frac{2}{10} \times \frac{2}{10}}=\frac{24 \times 10 \times 10}{2 \times 2}=600
\therefore \quad(c) is correct option
(iv) If a square and a rectangle having the same perimeter and their areas are S and R respectively then
(a) S = R
(b) \mathrm{S}>\mathrm{R}
(c) \mathrm{S}<\mathrm{R}
Solution :
\therefore (b) is correct option
(v) If the length of diagonal of a rectangle is 10cm. and area is 62.5 sq.cm., then the sum of their length and breadth is
(a) 12 \mathrm{~cm}.
(b) 15 \mathrm{~cm}.
(c) 20 \mathrm{~cm}.
(d) 25 \mathrm{~cm}.
Solution : Given,
Length of diagonal = 10
\sqrt{(\text { Length })^2+(\text { Breadth })^2}=10
or, (\text { Length }+ \text { Breadth })^2-2.length \times breadth =100
or, (\text { Length }+ \text { Breadth })^2-2 \times 62.5=100
or, (\text { Length }+ \text { Breadth) }^2-125=100
or, (\text { Length }+ \text { Breadth })^2=100+125=225
or, (\text { Length }+ \text { Breadth })^2=(15)^2
\therefore Length + breadth = 15cm.
\therefore (b) is correct option
20. Short answer type :
(i) If the length of square is increased by 10%, then what percent of the area of square will be increased?
Solution : Let the side of square be a
\therefore Area of square =\mathrm{a}^2
If the length of square is increased by 10%
Side of square = a + 10% of a
= a+\frac{10}{100} \times a \\
= a+\frac{a}{10}=\frac{11 a}{10} \\
\text { Area of square } =\left(\frac{11 a}{10}\right)^2 \\
=\frac{121 a^2}{100} \\
\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}
Percentage of the area of square will be increased
=\frac{21 a^2}{\frac{100}{a^2}} \times 100 \\
=21 \% \text { (Ans.) }
(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?
Solution: Let’ l ‘ be the length of rectangle.
and ‘b'” ” breadth ” “
Area of rectangle = l \times b
If length is increased by 10% and breadth is decreased by 10%
\therefore \text { Increased length } =l+10 \% l \\
=l+\frac{10}{100} \times l \\
=l+\frac{l}{10}=\frac{11 l}{10}
Decreased beadth =\mathrm{b}-10 \% of \mathrm{b}
=b-\frac{10}{100} \times b \\
=b+\frac{b}{10}=\frac{9 b}{10} \\
\text { Area }=\frac{11 l}{10} \times \frac{9 b}{10}=\frac{99 l b}{100}
\text { Decreased area } =m-\frac{99 l b}{100} \\
=\frac{l b}{100}
\text { Percentage of area will be decreased } =\frac{l b}{100} \\
= lb \times 100 \% \\
=1 \% \text { (Ans.) }
(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?
Solution :
From the figure,
\mathrm{AC} =5 \mathrm{~cm} \\
\mathrm{OE} =2 \mathrm{~cm}, \mathrm{OF}=2 \mathrm{~cm} \\
\therefore \quad \mathrm{EF}=\mathrm{BC} =2 \mathrm{~cm}+2 \mathrm{~cm} \\
= 4 \mathrm{~cm}
Let breadth of rectangle be b Then,
\sqrt{b^2+(4)^2}=5 \\
\text { or, } b^2+16=25 \\
\text { or, } b^2=25-16 \\
\text { or, } b^2=9 \\
\text { or, } b^2=3^2 \\
\therefore \quad b=3 \\
\therefore \quad \text { Length of breadth }=3 \mathrm{~cm}
(iv) If the length of perpendicular from the intersecting point between two diagonals on any side of square is 2 \sqrt{2} \mathrm{~cm}, then wila is inc ientgn of each diagonal of square?
Solution :
\therefore \mathrm{EG}=\mathrm{FG}=2 \sqrt{2} \mathrm{~cm}.
\mathrm{EF} =(2 \sqrt{2}+2 \sqrt{2}) \mathrm{cm} \\
=4 \sqrt{2} \mathrm{~cm} \\
\therefore \quad \mathrm{EF} =\mathrm{BC}=4 \sqrt{2} \mathrm{~cm}
\text { Length of diagonal }=\operatorname{Side} \sqrt{2}
=4 \sqrt{2} \times \sqrt{2}
=4 \times 2 \mathrm{~cm}=8 \mathrm{~cm} \text {. (Ans.) }
(v) The perimeter of a rectangle is 34cm. and area is 60sq. cm. What is the length of each diagonal?
Solution : Let ‘ l ‘ be the length and ‘ b ‘ be the breadth of the rectangle
Perimeter =34 \mathrm{~cm}.
or, 2(l+b)=34
or, l+\mathrm{b}=17
\therefore \quad l=17-b
\therefore \quad Area =60 Sq.cm.
l \times \mathrm{b}=60
or, (17-b) \times b=60
or, \quad 17 b-b^2=60
or, b^2-17 b+60=0
or, b^2-12 b-5 b+60=0
or, b(b-12)-5(b-12)=0
or, (b-12)(b-5)=0
either,
b – 12 = 0 \quad \text { or, } b-5=0 \\
b = 12 \quad b = 5
\therefore Breadth is always less than length
\therefore \quad Breadth = 5cm
\therefore \quad l=17-5=12 \\
\therefore \quad \text { Length }=12 \mathrm{~cm}
\text { Length of each diagonal } =\sqrt{l^2+\mathrm{b}^2} \\
=\sqrt{(12)^2+(5)^2} \mathrm{~cm} . \\
=\sqrt{144+25} \mathrm{~cm} . \\
=\sqrt{169} \mathrm{~cm} . \\
=13 \mathrm{~cm} . \quad \text { (Ans.) }
Let us work out – 15.2
1. Let us write by calculating the area of the following regions:
Solution :
\text { Area of } \triangle A B C=\frac{\sqrt{3}}{4} \times(10)^2 \\
=\frac{\sqrt{3}}{4} \times 100 \\
=25\sqrt{3} \text { sq.cm } \\ (Ans.)
Area of \triangle \mathrm{ABC} =\frac{1}{2} \times 8 \times \sqrt{(10)^2 - (\frac{8}{2} )^2}
=4 \times \sqrt{100-16} \\
=4 \times \sqrt{84} \text { Sq. units. } \\
=4 \times \sqrt{4 \times 21} \text { Sq. units. } \\
=4 \times 2 \sqrt{21} \text { Sq. units. } \\
=8 \sqrt{21 \mathrm{Sq} \text {. units. }} \text { (Ans.) }
Area of Trapazium \mathrm{ABCD}=\frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{CD}
=\frac{1}{2}(5+4) \times 3 \\
=\frac{1}{2} \times 9 \times 3 \\
=\frac{27}{2} \text { Sq. Units } \\
=13.5 \text { Sq. units. }
\text { Area of Trapazium }=\frac{1}{2}(\mathrm{AD}+\mathrm{CD}) \times \mathrm{AD}
=\frac{1}{2}(15+40) \times 9 \\
=\frac{1}{2} \times 55 \times 9 \\
=\frac{495}{2} \\
=247.5 \text { Sq.cm (Ans.) }
\mathrm{AC}=42 \mathrm{~cm}, \mathrm{CD}=38 \mathrm{~cm}
\because \angle \mathrm{ADC}=90^{\circ}
We know that,
A D^2 =A C^2-C D^2 \\
=(42)^2-(38)^2 \\
=(42+38(42-38) \\
=80 \times 4=320
\therefore \quad A D =\sqrt{320} \\
=\sqrt{4 \times 4 \times 2 \times 2 \times 5} \\
=4 \times 2 \sqrt{5}=8 \sqrt{5} \mathrm{~cm}
\therefore \quad \text { Area of } \mathrm{ABCD} =\mathrm{CD} \times \mathrm{AD} \\
=38 \times 8 \sqrt{5} \text { Sq.cm. } \\
=304 \sqrt{5} \text { Sq.cm. (Ans.) }
2. The perimeter of an equilateral triangle is 48cm. Let us write by calculating its area.
Solution:
\therefore The perimeter of an equilateral triangle = 48cm.
\therefore \quad Side of the equilateral triangle
=\frac{48}{3} \mathrm{~cm} . \\
=16 \mathrm{~cm} .
\therefore \quad Area of the equilateral triangle
=\frac{\sqrt{3}}{4} \times(16)^2 \\
=\frac{\sqrt{3}}{4} \times 256 \\
=44 \sqrt{3} \mathrm{~cm} . \quad \text { (Ans.) }
3. If the height of an equilateral triangle ABC is 5 \sqrt{3}. Let us write by calculating the area of this triangle.
Solution :
The height of an equilateral triangle \mathrm{ABC}=5 \sqrt{3} \mathrm{~cm}. We know that,
\therefore \frac{\sqrt{3}}{2} \times \text { side }=5 \sqrt{3}
or, \frac{\text { Side }}{2}=5
\therefore \quad Side = 10
\therefore \quad Side of an equilateral triangle = 10cm.
\therefore Area of an equilateral triangle.
=\frac{\sqrt{3}}{4} \times(10)^2 \\
=\frac{\sqrt{3}}{4} \times 100 \text { Sq.cm } \\
=25 \sqrt{3} \text { Sq.cm. }(Ans.)
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 \triangle \mathrm{ABC}.
Solution :
Each equal side of an isosceles triangle ABC = 10cm.
Length of base = 4cm.
\text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 4 \times \sqrt{(10)^{2}-(_{2}^{4})^{2}} \\
=2 \times \sqrt{100-4} \\
=2 \times \sqrt{96} \\
=2 \times \sqrt{4 \times 4 \times 6} \\
=2 \times 4 \sqrt{6} \\
=8 \sqrt{8} \text { Sq.cm. (Ans.) }
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.
Solution :
Each equal side of an isosceles triangle ABC = 10cm.
Length of base = 12cm.
\therefore \quad \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times \sqrt{(10)^{2}-(\frac{12}{2})^{2}} \\
=6 \times \sqrt{100-36} \\
=6 \times \sqrt{64} \\
=6 \times 8= 48 \text { Sq.cm. (Ans.) }
6. Perimeter of any isosceles triangle is 544cm. and length of each equal side is \frac{5}{6} th of length of base. Let us wirte by calculating the area of this triangle.
Solution: Let the length of base be x
\therefore Length of each side =\frac{5}{6} \times x=\frac{5 x}{6}
\therefore \quad Perimeter =544 \mathrm{~cm} . \therefore \quad Perimeter =544 \mathrm{~cm}.
or, x+\frac{5 x}{6}+\frac{5 x}{6}=544
or, \frac{6 x+5 x+5 x}{6}=544
or, \frac{16 x}{6}=544
\text { or, } \frac{x}{6}=34 \\
\therefore x = 204
\therefore Length of the base \quad = 204 \mathrm{~cm} .
\text { Length of each side } =\frac{5}{6} \times 204 \\
= 170cm
\text { Area of } \mathrm{ABC}=\frac{1}{2} \times 204 \times \sqrt{(170)^{2}- \frac{204}{2} } \\
=102 \times \sqrt{28900-10404} \\
=102 \times \sqrt{18496} \\
=102 \times 136 \\
=13872 \text { Sq.cm. (Ans.) }
7. If the length of hypotenuse of an isosceles right-angled triangle is 12 \sqrt{2} \mathrm{~cm}. Let us write by calculating the area of this triangle.
Solution:
Let the length of equal side be x
We have,
A B^{2}+B C^{2}=A C^{2}
or, x^{2}+x^{2}=(12 \sqrt{2})^{2}
or, 2 x^{2}=144 \times 2
or, x^{2}=144
or, x=\sqrt{144}
\therefore x = 12
\therefore Length of equal side be 12cm.
\therefore \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times 12 \text { Sq.cm. } \\
=72 \text { Sq.cm (Ans.) }
8. Pritha drew a parallelogram of which lengh of two diagonals are 6 \mathrm{~cm} \ and \ 8 \mathrm{~cm} and each angle between two diagonals is 90^{\circ}. Let us write the length of sides of parallelogram and what type of parallelogram it is?
Solution:
\text { Let } A C=6 \mathrm{~cm}, \quad B D=8 \mathrm{~cm} . \\
\therefore \quad \mathrm{AO}=\mathrm{OC}=3 \mathrm{~cm} \\
\mathrm{BO}=\mathrm{OD}=4 \mathrm{~cm} . \\
\therefore \quad \angle C O D=90^{\circ} \\
\triangle \mathrm{In} \triangle \mathrm{COD} \\
\mathrm{CD}^{2}=O C^{2}+\mathrm{OD}^{2} \\
=(3)^{2}+(4)^{2} \\
=9+16=25 \\
\therefore \quad \mathrm{or}, \quad \mathrm{CD}=\sqrt{25}=5 \mathrm{~cm} . \\
\therefore \quad A B=B C=C D=D A=5 \mathrm{~cm} .
Hence ABCD is rhombus
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.
(i) Let us write by calculating the area of the park.
(ii) Let us write by calculating how long is to be walked from opposite vertex of longest side to that side straightly.
Solution :
Let a = 2x
b = 3x
c = 4x
\therefore \quad Perimeter =216m.
or, a + b + c = 216m
or, 2x + 3x + 4x = 216
or, 9x = 216
or, x =\frac{216}{9}
\therefore \quad x =24 \\
\therefore \quad a =2 x=2 \times 24=48 \mathrm{~m} \\
b = 3 \mathrm{x}=3 \times 24=72 \mathrm{~m} \\
c = 4 \mathrm{x}=4 \times 24=96 \mathrm{~m} \\
\therefore \quad s =\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} \\
=\frac{48+72+96}{2}=\frac{216}{2}=108 \mathrm{~m}.
\text { Area of the park } =\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\
=\sqrt{108(108-48)(108-72)(108-96)} \\
=\sqrt{108 \times 60 \times 36 \times 12} \\
=\sqrt{36 \times 3 \times 12 \times 5 \times 36 \times 12} \\
=36 \times 12 \sqrt{5 \times 3} \\
=432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m} \\
\text { Area of the park } =432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m}
Area of the park =432 \sqrt{15} \mathrm{Sq} \cdot \mathrm{m}
or, \quad \frac{1}{2} \times \mathrm{BC} \times \mathrm{AD}=432 \sqrt{15}
or, \frac{1}{2} \times 96 \times \mathrm{AD}=432 \sqrt{15}
\text { or, } \mathrm{AD}=\frac{432 \sqrt{15}}{48}
\text { or, } \mathrm{AD}=9 \sqrt{15 \mathrm{~m}} \text { (Ans.) }
10. The length of three sided of a triangular field of village of Paholampur are 26 meter, 28 meter and 30 meter.
(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.
(ii) Let us write by calculating how much cost will be for 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.
Solution :
\text { Let } a=26 \mathrm{~m} \\
\mathrm{~b}=28 \mathrm{~m} \\
\mathrm{c}=30 \mathrm{~m} \\
\mathrm{~s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2} \\
=\frac{26+28+30}{2}=\frac{84}{2}=42 \mathrm{~m} . \\
Area of the triangular field =\sqrt{s(s-a)(s-b)(s-c)} =\sqrt{42(42-26)(42-28)(42-30)} \\
=\sqrt{42 \times 16 \times 14 \times 12} \\
=\sqrt{14 \times 3 \times 4 \times 4 \times 4 \times 14 \times 2 \times 2 \times 3} \\
=14 \times 4 \times 3 \times 3 \\=336 \text { Sq.m }
Cost of planting grass in the triangular field =336 \times 5
= Rs. 1680 (Ans.)
Perimeter of the triangular field =26 \mathrm{~m}+28 \mathrm{~m}+30 \mathrm{~m}=84 \mathrm{~m}
But leaving a space 5m for constructing entrance gate,
\therefore \quad \text { Length for fencing around three sides }=84 \mathrm{~m}-5 \mathrm{~m}
= 79m
\text { Total cost }=\text { Rs. } 79 \times 18=\text { Rs. } 1422 \text { (Ans.) }
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.
Solution :
Let ‘a’ be the side of an equilateral triangle.
Area of the triangle PQR =\frac{\sqrt{3}}{4} \times(\mathrm{a})^{2}
=\frac{\sqrt{3}}{4} \times a^{2}
According to the condition of the problem
\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
or, \frac{\sqrt{3}}{4} \times a^{2}=5 a+6 a+4 a
or, \frac{\sqrt{3}}{4} \times a^{2}=15 a
or, \frac{\sqrt{3}}{4} \times a=15 \quad[\because a \neq 0]
or, \sqrt{3} \mathrm{a}=60
or, a=\frac{60}{\sqrt{3}}
\text { or, }=\frac{20 \times \sqrt{3} \times \sqrt{3}}{\sqrt{3}} \\
\therefore \quad a=20 \sqrt{3}
\text { Area of } \triangle P Q R =\frac{\sqrt{3}}{4} \times(20 \sqrt{3})^{2} \\
=\frac{\sqrt{3}}{4} \times 20 \times 20 \times 3 \\
=300 \sqrt{3} \text { Sq.m (Ans.) }
12. The length of each equal side of an isosceles triangle is 20m and the angle included between them is 45^{\circ}. Let us write by calculating the area of triangle.
Solution:
In \triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}=20 \mathrm{~m} \ and \ \angle \mathrm{A}=45^{\circ},
CD is drawn perpendicular to AB
Now, if we take AB as base of the triangle,
Then its altitude is CD.
According to construction, \angle \mathrm{ACD}=45^{\circ}, \therefore \mathrm{AD}=\mathrm{CD}
In the right angled triangle ADC,
\mathrm{CD}^{2}+\mathrm{AD}^{2}=\mathrm{AC}{ }^{2}=(20)^{2} \mathrm{Sqm}=400 \mathrm{Sq} \cdot \mathrm{m} \\
\therefore \quad 2 \mathrm{CD}^{2}=400 \mathrm{Sq} \cdot \mathrm{m}[\because \mathrm{AD}=\mathrm{CD}]
\therefore \quad 2 \mathrm{CD}^{2}=200 \text { Sq.m } \therefore \mathrm{CD}=\sqrt{200} \\
\text { Area of the triangle }=\frac{1}{2} \times \mathrm{AB} \times \mathrm{CD} \\
=\frac{1}{2} \times 20 \times 10 \sqrt{2} \\
=100 \sqrt{2} \text { Sq } \mathrm{m} \text {. } \\
13. The length of each equal side of an isosceles triangle is 20cm, and the angle included between them is 30^{\circ}. Let us write by calculating the area of triangle.
Solution :
In \triangle A B C \\
AB = AC = 20cm
We draw perpendicular CD on AB
Then, \angle A D C=90^{\circ},
\angle \mathrm{CAD}=30^{\circ} (Given)
and, \angle \mathrm{ACD}=60^{\circ}
\therefore In any right angled triangle,
the three angles are 90^{\circ}, 60^{\circ} and \ 30^{\circ},
Then,\mathrm{CD}=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \times 20 \mathrm{~cm}=10 \mathrm{~cm}
\text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \mathrm{AB} \times \mathrm{CD} \\
=\frac{1}{2} 10 \times 10 \text { Sq.cm } \\
=100 \text { Sq.cm (Ans.) }
14. If the perimeter of an isosceles right-angled triangle is (\sqrt{2}+1) \mathrm{cm}. Let us write by calculating the length of hypotenuse and area of triangle.
Solution :
Let the equal sides be ‘ a ‘
And length of the base be ‘ b ‘
\therefore \quad Perimeter =\sqrt{2}+1
or, a+a+b=\sqrt{2}+1
or, 2 a+b=\sqrt{2}+1…………. (i)
We have,
A C^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}
or, b^{2}=a^{2}+a^{2}
or, b^{2}=2 a^{2}
or, b=\sqrt{2 a^{2}}
or, b=\sqrt{2} \mathrm{a}…………. (ii)
From (i),
2 a+b=\sqrt{2}+1
or, 2 a+\sqrt{2 a}=\sqrt{2}+1 ………….by (ii)
or, \sqrt{2} a(\sqrt{2}+a)=\sqrt{2}+1
or, \sqrt{2} \mathrm{a}=1
\therefore a=\frac{1}{\sqrt{2}}
From (ii),
b=\sqrt{2 a} \\
= \sqrt{2} \times \frac{1}{\sqrt{2}} \quad[\because a=1 \sqrt{2}] \\
= 1
\therefore \quad Length of the hypotenuse = 1cm.
\text { Area of the triangle } =\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB} \\
=\frac{1}{2} \times \mathrm{a} \times \mathrm{a} \\
=\frac{1}{2} \times \mathrm{a}^{2} \\
=\frac{1}{2} \times\left(\frac{1}{\sqrt{2}}\right)^{2} \\
=\frac{1}{2} \times \frac{1}{2} \\
=\frac{1}{4}=0.25 \text { Sq.cm }(Ans.)
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.
Solution :
Total distance = speed × time
=18 \times \frac{10}{60} \mathrm{Km} . \\
= 3km.
Perimeter of equilateral trianglar park = 3 km .
\text { length of each side }=\frac{3}{3} \mathrm{Km} .=1 \mathrm{~km} \text {. }
Distance between mid point of a side and its opposite vertex = AD
=\frac{\sqrt{3}}{2} \times \text { side }=\frac{\sqrt{3}}{2} \times 1 \mathrm{~km}=\frac{\sqrt{3}}{2} \mathrm{~km}
Time taken to go from opposite vertex to mid-point of a side
=\frac{\text { Distance }}{\text { Speed }}
=\frac{\sqrt{3}}{2 \times 18} hours
=\frac{\sqrt{3}}{2 \times 18} \times 60 minutes
=\frac{5 \sqrt{3}}{3} minutes(Ans.)
16. If the length of each side of an equilateral triangle is increased by 1 meter, then its area will be increased by \sqrt{3} sq. meter. Let us write by calculating the length of side of equilateral triangle.
Solution: Let ‘ x ‘ be the side of an equilateral triangle.
\text { Area of the equilateral triangle } =\frac{\sqrt{3}}{4} \times(x)^{2} \\
=\frac{\sqrt{3}}{4} x^{2}
If length of each side of an equilateral triangle is increased by 1 meter,
Then, side of an equilateral triangle (x + 1)
\therefore \text { Area of the equilateral triangle }=\frac{\sqrt{3}}{4} \times(x+1)^{2}
According to the condition of problem,
\frac{\sqrt{3}}{4} \times(x+1)^{2}-\frac{\sqrt{3}}{4} x^{2}=\sqrt{3}
or, \frac{\sqrt{3}}{4}\{(x+1)^{2}-x^{2}\}=\sqrt{3}
or, \frac{1}{4}(x+1+x)(x+1-x)=1
or, \frac{1}{4}(2 x+1)=1
or, 2x + 1 = 4
or, 2x = 4 – 1
or, 2x = 3
or, \mathrm{x}=\frac{3}{2} \quad \therefore \mathrm{x}=1.5
\therefore \quad Length of side equilateral triangle = 1.5m.
17. The area of an equilateral triangle and area of square are in the ratio \sqrt{3} : 2 . If the length of diagonal of square is 60cm. Let us write by calculating perimeter of an equilateral triangle.
Solution : Let ‘a’ be the side of a square
Then, diagonal of square = 60cm.
\therefore \quad \mathrm{a} \sqrt{2}=60 \mathrm{~cm}
or, a=\frac{60}{\sqrt{2}} \mathrm{~cm}
By the problem,
Area of equilateral triangle : area of square =\frac{\sqrt{3}}{2}
\text { or, } \frac{\text { Area of equilateral triangle }}{\text { Area of square }}=\frac{\sqrt{3}}{2} \\
\text { or, } \frac{\text { Area of equilateral triangle }}{(\frac{60}{\sqrt{2}})^{2}}=\frac{\sqrt{3}}{2} \\
\text { or, } \frac{\text { Area of equilateral triangle }}{\frac{3600}{2}}=\frac{\sqrt{3}}{2}
or, Area of equilateral triangle =900 \sqrt{3}
or, \frac{\sqrt{3}}{4}(\text { Side })^{2}=900 \sqrt{3}
or, (\text { side })^{2}=3600 \\
or, \text { Side }=\sqrt{3600} \\
=60 \mathrm{~m} \text {. } \\
Perimeter of an equilateral triangle
=3 \times \text { side }=3 \times 60 \mathrm{~m}=180 \mathrm{~m}(Ans.)
18. Length of hypotenuse and perpendicular of a right-angled triangle are 13cm and 30cm. Let us write by calculating the area of triangle.
Solution : Length of hypotenuse = 13cm.
Perimeter of right angled triangle = 30cm.
\therefore Sum of remaining sides other than
hypotenuse = (30 – 13)cm = 17cm.
Let ‘ p ‘ be the perpendicular and ‘ b ‘ be the base of a right angled triangle.
\therefore \mathrm{p}+\mathrm{b}=17 ……….. (i)
Also,
p^{2}+b^{2}=h^{2}
or, (p+b)^{2}-2 p b=(13)^{2}
or, (17)^{2}-2 \mathrm{pb}=(13)^{2}
or, 289-2 \mathrm{pb}=169
or, -2 \mathrm{pb}=169-289
\text { or, }+2 \mathrm{pb}=f 120 \\
\text { or, } \mathrm{pb}=\frac{120}{2} \\
\therefore \mathrm{pb}=60
Again,
p^{2}+b^{2}=h^{2}
or, (p-b)^{2}+2 p b=h^{2}
or, (p-b)^{2}+2 \times 60=(13)^{2}
or, (\mathrm{p}-\mathrm{b})^{2}+120=169
or, (p-b)^{2}=169-120
or, (p-b)^{2}=49
or, (p-b)=\sqrt{49}
\therefore p – b = 7 ……….. (ii)
\therefore \quad p + b = 17 ......... (i) \\ \underline{p-b =7 ........... (ii)} \\ 2 \mathrm{p}=24(by \ adding)
or, p=\frac{24}{2}
\therefore \mathrm{p}=12
Putting the value of p in equation (i)
p + b = 17
or, 12 + b = 17
\text { or, } b=17-12 \quad \therefore b=5
\text { Area of right angled triangle }
=\frac{1}{2} \times \mathrm{b} \times \mathrm{p} \\
=\frac{1}{2} \times 5 \times 12 \\
=30 \text { Sq. cm. (Ans.) }
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.
Solution:
Let \mathrm{AB}=12 \mathrm{~cm},
\mathrm{BC}=5 \mathrm{~cm}
We know that,
A C^{2}=A B^{2}+B C^{2}
or, \mathrm{AC}^{2}=(12)^{2}+(5)^{2}
or, \mathrm{AC}^{2}=144+25
or, \mathrm{AC}^{2}=169
or, \mathrm{AC}^{2}=\sqrt{169}
\therefore \quad \mathrm{AC}=13
\therefore Area of \triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}
=\frac{1}{2} \times 5 \times 12
= 30 Sq. \mathrm{cm}
\therefore \quad \text{Area of} \triangle \mathrm{ABC}=30 Sq.cm.
or, \frac{1}{2} \times \mathrm{BD} \times \mathrm{AC}=30
or, \frac{1}{2} \times \mathrm{BD} \times 13=30
or, 13 \mathrm{BD}=60
or, \mathrm{BD}=\frac{60}{13}
or, \mathrm{BD}=4.615 \mathrm{~cm} (Approx)
\therefore \quadLength of the perpendicular =4.615 \mathrm{~cm} (Approx)(Ans.)
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.
Solution:
Let \mathrm{AB}=4 \mathrm{~cm}, \mathrm{BC}=3 \mathrm{~cm}
\mathrm{AC}=5 \mathrm{~cm}
Let \mathrm{BD}=\mathrm{DE}=\mathrm{EF}=\mathrm{BF}=\mathrm{a}
\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
or \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
or, 12=(4-a) a+2 a^{2}+(3-a) a
or, 12=4 a-a^{2}+2 a^{2}+3 a-a^{2}
or, 12=7 a
or, a=\frac{12}{7}
\therefore \quad \text { Length of square } =\frac{12}{7} \mathrm{~cm} \\
=1 \frac{5}{7} \mathrm{~cm}(Ans.)
21. (M.C.Q):
(i) If each side of an equilateral triangle is 4cm, the measure of height is
(a) 4 \sqrt{3} \mathrm{~cm}
(b) 16 \sqrt{3} \mathrm{~cm}
(c) 8 \sqrt{3 \mathrm{~cm}}
(d) 2 \sqrt{3} \mathrm{~cm}
Solution: Height of an equilateral triangle
=\frac{\sqrt{3}}{2} \times(\text { side }) \\
=\frac{\sqrt{3}}{\cancel2} \times {\cancel 4} \\
=2 \sqrt{3} \mathrm{~cm}.
\therefore \quad (d) is correct option (Ans.)
(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
(a) (1+\sqrt{2}) a unit
(b) (2+\sqrt{2}) a unit
(c) 3a unit
(d) (3+2 \sqrt{2}) a unit.
Solution:
Let AB = a, BC = a
or, A C^2=a^2+a^2
or, \mathrm{AC}=\sqrt{2 \mathrm{a}^2}
or, \mathrm{AC}=\sqrt{2} \mathrm{a}
\therefore \quad \text{Perimeter of triangle} =a+a+\sqrt{2} a
=2 a+\sqrt{2} a
=(2+\sqrt{2}) a unit.
\therefore \quad (b) is correct option (Ans.)
(iii) If the area, perimeter and height of an equilateral triangle are a, s and h, then value of 2a/sh is
(a) 1
(b) \frac{1}{2}
(c) \frac{1}{3}
(d) \frac{1}{4}
Solution :
We have,
\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}} \\
= \frac{2}{4} \times \frac{2}{3}=\frac{1}{3}
\therefore (c) is correct option (Ans.)
(iv) The length of each equal side of an isosceles triangle is 5cm. and length of base is 6cm The area of triangle is
(a) 18 \mathrm{sq} . \mathrm{cm}
(b) 12 \mathrm{sq} . \mathrm{cm}
(c) 15 \mathrm{sq} . \mathrm{cm}.
(d) 30 \mathrm{sq} . \mathrm{cm}.
Solution :
\text { Area of triangle } =\frac{1}{2} \times 6 \times \sqrt{(5)^2-(\frac{6}{3})^2} \\
=\frac{1}{2} \times 6 \times \sqrt{25-9} \\
=3 \times \sqrt{16} \\
=3 \times 4=12 \text { Sq.cm. }
\therefore (b) is correct option (Ans.)
(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.
(a) 16sq.cm
(b) 24sq.cm.
(c) 15sq.cm.
(d) 30sq.cm
Solution :
Area of triangle ABC = 40 Sq.cm
or, \mathrm{AD}: \mathrm{DC}=3: 2
or, \triangle \mathrm{ABD}: \triangle \mathrm{BDC}
or, \frac{\triangle A B D}{\triangle B D C}=\frac{3}{2}
or, 2 \triangle \mathrm{ABD}=3 \triangle \mathrm{BDC}
or, \mathrm{ABD}=\frac{3}{2} \triangle \mathrm{BDC}
Also \triangle \mathrm{ABC}=\triangle \mathrm{ABD}+\triangle \mathrm{BDC}
or, \triangle \mathrm{ABC}={ }_2^3 \triangle \mathrm{BDC}+\triangle \mathrm{BDC}
or, \triangle \mathrm{ABC}=\frac{5}{2} \triangle \mathrm{BDC}
or, \triangle B D C={ }_5^2 \triangle A B C
= \frac{2}{\cancel{5}} \times \cancel{40} \mathrm{Sq} \cdot \mathrm{cm}
=16 \mathrm{Sqcm}.
\therefore (a) is correct option (Ans.)
(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
(a) 20 \sqrt{7} \mathrm{Sq} . \mathrm{cm}
(b) 10 \sqrt{14} Sq.cm
(c) 20 \sqrt{14} \mathrm{Sq} . \mathrm{cm}
(d) 140 \mathrm{Sq} . \mathrm{cm}
Solution :
Let, s – a = 8
s – b = 7
s – c = 5
adding these,
3s – (a + b + c) = 8 + 7 + 5
\text { or, } 3 s-2 s=20 \quad\left[\therefore s=\frac{a+b+c}{2}\right]
\therefore \mathrm{s}=20
Area of Triangle =\sqrt{s(s-a)(s-b)(s-c)} \\
=\sqrt{20 \times 8 \times 7 \times 5} \\
=\sqrt{5 \times 4 \times 4 \times 2 \times 7 \times 5} \\
=5 \times 4 \sqrt{2 \times 7} \\
=20 \sqrt{14} \mathrm{Sq} \cdot \mathrm{cm}
\therefore \quad (c) is correct option (Ans.)
22. Short answer type question:
(i) The numerical values of area and height of an equilateval triangle are equal. What is the length of side of triangle?
Solution :
By question,
\quad \frac{\sqrt{3}}{4}(\text { side })^2=\frac{\sqrt{3}}{2} \times(\text { side }) \\
\text { or, } \frac{\text { side }}{2}=1 \\
\text { or, side }=2 \\
\text { Length of side of triangle }=2 \text { units }
\therefore Length of side of triangle = 2 units (Ans.)
(ii) If length of each side of an equilateral triangle is doubled, what percent of area will be increased of this. triangle?
Solution:
Let ‘ a ‘ be the side of an equilateral triangle,
\therefore \quad \text { Area of the triangle }=\frac{\sqrt{3}}{4} \mathrm{a}^2
If length of each side of an equilateral triangle is doubled
\therefore \quad \text { Area of the triangle } =\frac{\sqrt{3}}{4}(2 a)^2 \\
=\frac{\sqrt{3}}{4} \times 4 a^2
Increase area of the triangle
=\frac{\sqrt{3}}{4} \times 4 a^2-\frac{\sqrt{3}}{4} a^2 \\
=\frac{\sqrt{3}}{4} \times 3 \mathrm{a}^2 \\
\text { Perimeter of increased in area }=\frac{\frac{\sqrt{3}}{4} \times 3 a^2}{\frac{\sqrt{3}}{4} \times a^2} \times 100 \% \\
=300 \% \text { (Ans.) } \\
(iii) If the length of each side of an equilateral is trippled.
What percent of area will be increased of this triangle?
Solution :
Let ‘ a ‘ be the side of an equilateral triangle,
\therefore \text { Area of the triangle }=\frac{\sqrt{3}}{4} \mathrm{a}^2
If length of each side of an equilateral triangle is trippled
\therefore \quad \text { Area of the triangle } =\frac{\sqrt{3}}{4}(3 a)^2 \\
=\frac{\sqrt{3}}{4} \times 9 a^2
Increased area of the triangle
=\frac{\sqrt{3}}{4} \times 9 a^2-\frac{\sqrt{3}}{4} a^2 \\
=\frac{\sqrt{3}}{4} \times 8 a^2 \\
=\sqrt{3} \times 2 a^2
\text { Perimeter of increased in area } =\frac{\sqrt{3} \times 2 \mathrm{a}^2}{\sqrt{3}} \times 100 \% \\
=8 \times 100 \% \\
=800 \% \text { (Ans.) }
(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?
Solution:
We have, (\text { hypotenuse} )^2=(\text { base })^2+(\text { Perpendicular })^2
(x+2)^2=(x-2)^2+(x)^2
or, (x+2)^2-(x-2)^2=(x)^2
or, 4 \cdot x \cdot 2=x^2
\text { or, } 8 x=x^2 \\
\text { or, } 8=x \quad [\therefore x \neq 0] \\
\therefore \quad x=8
\therefore \quad \text { Length of hypotenuse } =(x+2) \mathrm{cm} \\
=(8+2) \mathrm{cm} \\
=10 \mathrm{~cm} \text {. (Ans.) }
(v) A square drawn on height of equilateral triangle. What is the ratio of area of triangle and square?
Solution :
Area of triangle : Area of square
=\frac{\sqrt{3}}{4} \times(\text { side })^2:\left(\frac{\sqrt{3}}{2} \times(\text { side })\right)^2 \\
=\frac{\sqrt{3}}{4} \times(\text { side })^2: \frac{\sqrt{3} \times \sqrt{3}}{4} \times(\text { side })^2 \\
= 1: \sqrt{3} \text { (Ans.) }
Let us work out – 15.3
1. Ratul draws a parallelogram with a length of base 5cm. and height 4cm. Let us calculate the area of the parallelogram drawn by Ratul.
Solution :
\text { Length of base }=5 \mathrm{~cm} \\
\text { Height }=4 \mathrm{~cm}
\text { Area of the parallelogram } =\text { base } \times \text { height } \\
=5 \mathrm{~cm} \times 4 \mathrm{~cm} \\
=20 \text { sq.cm. (Ans.) }
2. The base of a parallelogram is twice its height. If the area of the shape of a parallelogram is 98 sq cm. then let us calculate the length and height of the parallelogram.
Solution:
Let ‘ x ‘ be the height of a parallelogram
\therefore Base =2 \times height =2 \mathrm{x}
\therefore \text{Area of parallelogram} =98 sq. \mathrm{cm}.
or, Base × height = 98
or, 2x × x = 98
or, 2x^2 = 98
or, x^2-\frac{98}{2}
or, x^2=49
or, \mathrm{x}=\sqrt{49}
\therefore \quad \mathrm{x}=7
Length of base = 2x
=2 \times 7=14 \mathrm{~cm} \text {. }
Height of the parallelogram = x =7cm (Ans.)
3. There is channel of parallelogram land beside our house of which lengths of adjacent sides are 15 meters and 13 meters. If the length of one diagonal is 14 meters, then let us calculate the area of shape of parallelogram land.
Solution:
Let, a =13 \mathrm{~cm} \\
b =14 \mathrm{~cm} \\
c =15 \mathrm{~cm}
s=\frac{a+b+c}{2}
=\frac{13 m+14 m+15 m}{2}=\frac{42}{2} m=21 m
\text { Area of } \triangle \mathrm{ABD} =\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\
=\sqrt{21(21-13)(21-14)(21-15)}
=\sqrt{21 \times 8 \times 7 \times 6} \\
=\sqrt{7 \times 3 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3} \\
=2 \times 2 \times 3 \times 7 \\
=84 \mathrm{Sq} . \mathrm{m}
\text { Area of parallelogram } \mathrm{ABCD} =2 \times \text { Area of } \triangle \mathrm{ABD} \\
=2 \times 84 \mathrm{Sq} \cdot \mathrm{m}=168 \mathrm{Sq} \cdot \mathrm{m}
4. Pritha draws a parallelogram of which adjacent sides are 25cm and 15cm and the length of the diagonal is 20cm. Let us write by calculating the height of the parallelogram which is drawn on the side of 25cm.
Solution:
Let, \mathrm{a}=25 \mathrm{~cm}
b=20 \mathrm{~cm} \\
c=15 \mathrm{~cm}
s=\frac{a+b+c}{2} \\
=\frac{25+20+15}{2}=\frac{60}{2}=30 \mathrm{~cm}
=\frac{25+20+15}{2}=\frac{60}{2}=30 \mathrm{~cm}
\text { Area of } \triangle A B D =\sqrt{s(s-a)(s-b)(s-c)} \\
=\sqrt{30(30-25)(30-20)(30-15)} \\
=\sqrt{30 \times 5 \times 10 \times 15} \\
=\sqrt{15 \times 2 \times 2 \times 5 \times 15} \\
=2 \times 5 \times 15 \\
=150 \text { Sq.cm }
Area of parallelogram = 2 \times Area of \triangle \mathrm{ABD}
=2 \times 150 \text { Sq.cm. } \\
=300 \text { Sq. } \\
\therefore \quad Base \times height =300 Sq. \mathrm{cm}.
or, 25 \times height =300
or, height ^2=\frac{300}{25}
\therefore height =12 \mathrm{~cm}(Ans.)
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.
Solution:
Let, A B=15 \mathrm{~cm}
\mathrm{BC}=12 \mathrm{~cm} \\
\mathrm{AE}=7.5 \mathrm{~cm} .
Area of the \triangle A B C
=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AE} \\
=\frac{1}{2} \times 12 \times 7.5 \mathrm{Sq} . \mathrm{cm} . \\
=45 \text { Sq.cm. }
Let the distance between the longer two sides be ‘ x ‘
\therefore \text { Area of } \triangle \mathrm{ACD}=\text { Area of } \triangle \mathrm{ABC}
or, \frac{1}{2} \times B C \times distance between the longer two sides = 45
or, \frac{1}{2} \times 15 \times x=45
or, \frac{15 x}{2}=45
or, \frac{x}{2}=3
\therefore \quad x = 6
\therefore \quad The distance between the longer two sides = 6cm (Ans.)
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.
Solution:
\text { Let, } \mathrm{AC} =15 \mathrm{~m} \\
\mathrm{BD} =20 \mathrm{~m}
\mathrm{AO}=\mathrm{OC}=\frac{15}{2} \mathrm{~m} \\
\mathrm{BO}=\mathrm{OD}=\frac{20}{2} \mathrm{~m}=10 \mathrm{~m}
Area of Rhombus ABCD
=\frac{1}{2} \mathrm{AC} \times \mathrm{BD} \\
=\frac{1}{2} 15 \times 20 \mathrm{~m} \\
=150 \text { Sq.m (Ans.) }
In \triangle \mathrm{OAB}, \quad \angle \mathrm{AOB}=90^{\circ}
A B^2=A O^2+O B^2
or, A B^2=\left(\frac{15}{2}\right)^2+(10)^2
or, A B^2=\frac{225}{4}+100^2
or, \mathrm{AB}^2=\frac{225+400}{4}
or, A B^2=\frac{625}{4}
\therefore \quad \mathrm{AB}=\sqrt{\frac{625}{4}}=\frac{25}{2} \mathrm{~m}
\therefore \quad \text{Perimeter of} \mathrm{ABCD}=4 \times \mathrm{AB}
=4 \times \frac{25}{2} \\
=50 \mathrm{~m} \text { (Ans.) }
\therefore \quad Area of Rhombus ABCD =150 Sq.m
\therefore \quad Base \times height =150
or, \frac{25}{2} \times height ^2=150
or, \frac{\text { height }}{2}=6
\therefore height =12 \mathrm{~m} (Ans.)
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.
Solution:
\text { Side of Rhombus } =\frac{\text { Perimeter }}{4} \\
=\frac{440}{4} \\
=110 \mathrm{~m}
\therefore Base \mathrm{CD} \text{of Rhombus} \mathrm{ABCD}=110 \mathrm{~m}
\therefore Distance between two parallel sides = Altitude AE
\therefore \quad \text { Area of the Rhombus } =\text { Base } \times \text { Altitude } \quad=22 \mathrm{~m} \\
=\mathrm{CD} \times \mathrm{AE} \\
=110 \times 22 \text { Sq.m } \\
=2420 \text { Sq.m (Ans.) }
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.
Solution:
\text { Perimeter of Rhombus }=20 \mathrm{~cm} \\
\text { Side of a rhombus } =\frac{20}{4} \mathrm{~cm} \\
=5 \mathrm{~cm} .
Length of its one diagonal = 6cm.
Let B D=6 \mathrm{~cm}.
\mathrm{OB}=\frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm} . \\
AB = 5cm
In \triangle \mathrm{OAB}
\mathrm{OA}^2 =\mathrm{AB}^2-\mathrm{OB}^2 \\
\mathrm{OA}^2 =(5)^2-(3)^2 \\
\text { or, } \mathrm{OA}^2 =25-9
or, \mathrm{OA}^2=16
or, \mathrm{OA}=\sqrt{16}
\therefore \mathrm{OA} =4 \mathrm{~cm} \\
\mathrm{AC} =2 \times \mathrm{OA} \\
=2 \times 4 \mathrm{~cm}=8 \mathrm{~cm}
\text { Area of rhombus } \mathrm{ABCD} =\frac{1}{2} \times \mathrm{BD} \times \mathrm{AC} \\
=\frac{1}{2} \times 6 \times 8 \text { Sq.cm. } \\
=24 \text { Sq.cm. (Ans.) }
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.
Solution:
\text { Let } \mathrm{AE}=20 \mathrm{dm} \\
\mathrm{AD}=3 \mathrm{x} \\
\mathrm{BC}=4 \mathrm{x} \\
\text { trapazium }=1400 \text { Sq.cm }
Area of trapazium = 1400 Sq.cm
\text { or } \quad \frac{1}{2}(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AE}=1400 \\
\text { or } \quad \frac{1}{2}(3 \mathrm{x}+4 \mathrm{x}) \times 20=1400
7x = 1400
x=\frac{140}{7} \\
\therefore \quad x = 20
\therefore \quad A D=3 x=3 \times 20 \mathrm{dcm} .=60 \mathrm{dcm} \\
\quad \mathrm{BC}=4 \mathrm{x}=4 \mathrm{x} \times 20 \mathrm{dcm} .=80 \mathrm{dcm} \text {. (Ans.) }
10. Let us write by calculating the area of regular hexag field of which length of sides is 8cm
Solution:
If we draw diagonals we get equal six equilater triangles.
Area of \triangle A O B
=\frac{\sqrt{3}}{4}(\text { Side })^2 \\
=\frac{\sqrt{3}}{4}(8)^2 \\
=\frac{\sqrt{3}}{4} 64 \\
=16 \sqrt{3} \mathrm{Sq} \cdot \mathrm{cm} .
\text { Area of Hexagon } \mathrm{ABCDEF} =6 \times 16 \sqrt{3} \\
=96 \sqrt{3} \text { Sq. } \mathrm{cm} \text {. }(Ans.)
11. In a quadrilateral ABCD, AB = 5 meter, BC = 12 meter, DA = 15 meter and \angle A B C=90^{\circ}, Let us write by calculating the area of quadrilateral shape of field.
Solution:
\text { Let } A B=5 \mathrm{~m}, B C=12 \mathrm{~m} \text {, } \\
C D=14 \mathrm{~m}, \mathrm{DA}=15 \mathrm{~m} \text {, } \\
\therefore \angle A B C=90^{\circ} \\
\text { or, } A C^2=\sqrt{\mathrm{AB}^2+B C^2} \\
=\sqrt{(5)^2+(12)^2} \text {. } \\
=\sqrt{25+144} \\
=\sqrt{109} \\
=13 \mathrm{~m} \\
\therefore \text { Area of } \triangle \mathrm{ABC} =\frac{1}{2} \times 12 \times 5 \\
=30 \mathrm{Sq} . \mathrm{cm}
\text { Let } a=13 \mathrm{~m}, b=14 \mathrm{~m}, \mathrm{c}=15 \mathrm{~m}
s=\frac{a+b+c}{2} \\
s=\frac{13+14+15}{2}=\frac{42}{2}=21 \mathrm{~m}
Area of \triangle A C D=\sqrt{s(s-a)(s-b)(s-c)}
=\sqrt{21(21-13)(21-14)(21-15)} \\
=\sqrt{21 \times 8 \times 7 \times 6} \\
=\sqrt{3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 2 \times 3} \\
=2 \times 2 \times 3 \times 7 \mathrm{Sq} \cdot \mathrm{m} \\
=84 \text { Sq.cm. }
\text { Area of Quadrilateral } A B C D=(3+84) \text { Sq.m. } \\
=114 \text { Sq.m (Ans.) } \\
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.
Solution:
\text { Let, } \mathrm{BD} =11 \mathrm{~cm}, \\
\mathrm{AE} =5 \mathrm{~cm} \\
\mathrm{CF} =11 \mathrm{~cm}
\text { Area of } \triangle \mathrm{ABD}=\frac{1}{2} \times \mathrm{BD} \times \mathrm{AE} \\
=\frac{1}{2} \times 11 \times 5 \\
=\frac{\jmath \rho}{2} \text { Sq.cm } \\
\text { Area of } \triangle B C D=\frac{1}{2} \times B D \times C F \\
= \frac{1}{2} \times 11 \times 11 \\
=\frac{121}{2} \text { Sq.cm } \\
Area of Trapezium ABCD
=\text { Area of } \triangle A B D+\text { Area of } \triangle B C D
=(\frac{55}{2}+\frac{121}{2}) \text { Sq.cm } \\
=\frac{176}{2} \\
=88 \mathrm{Sq} . \mathrm{cm} \text { (Ans.) }
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 =\frac{4}{9} PE, Let us write by calculating the area of \ABCDE in shape of pentagon.
Solution:
\text { Let } P Q=\frac{4}{9}+PE= \frac{4}{9} \times 9cm \\
= 4cm
\therefore \quad \mathrm{QE}=(9-4) \mathrm{cm} .=5 \mathrm{~cm} \\
\text { Now, } \because A D \| B C \text { and } E P \perp B C \\
\therefore \mathrm{EQ} \perp \mathrm{AD} \\
\therefore \text { Area of pentagon } A B C D E \\
=\text { Area of } \triangle \mathrm{ADE}+\text { Area of trapezium } \mathrm{ABCD} \\
=\frac{1}{2} \times \mathrm{AD} \times \mathrm{QE}+\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC} ) \times \mathrm{PQ} \text { Sq.unit. } \\
=\frac{1}{2} \times 13 \times 5+\frac{1}{2} \times(13+7) \times 4 \text { Sq.unit. } \\
=(32.5+40) \cdot \mathrm{Sq} \cdot \mathrm{cm} \\
=72.5 \text { Sq.cm } \\
14. The length of rhombus is equal to length of a square and is 40 \sqrt{2} \mathrm{~cm}. 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
Solution:
Let ‘ a ‘ be the side of a square and also a side of a rhombus. The side of a rhombus.
The length of diagonal = 40 \sqrt{2} \mathrm{~cm}. or, a \sqrt{2}=40 \sqrt{2}\\
\therefore \quad a=40, \therefore \mathrm{CD}=40 \mathrm{~cm} \text {. }\\
Let \mathrm{AC}=3 \mathrm{x}, \mathrm{BD}=4 \mathrm{x}\\
\mathrm{OC}=\frac{3 \mathrm{x}}{2}, \mathrm{OD}=\frac{4 \mathrm{x}}{2} 2 \mathrm{x}\\
In \triangle \mathrm{COD}\\
O C^2+O D^2=C D^2\\
(\frac{3 x}{2})^2+(2 x)^2=(40)^2\\
or, \frac{9 x^2}{4}+4 x^2=1600\\
or, \frac{9 x^2+16 x^2}{4}=1600\\
or, \frac{25 x^2}{4}{ }^2=1600\\
or, 25 \mathrm{x}^2=1600 \times 4\\
or, x^2=\frac{1660 \times 4}{25}{ }^2\\
or, x^2=64 \times 4\\
\therefore x^2 =\sqrt{64 \times 4} \\=8 \times 2=16\\
\therefore \quad x =16\\
\mathrm{AC}=3 x=3 \times 16 \mathrm{~cm}=48 \mathrm{~cm} \\
\mathrm{BD}=4 \mathrm{x}=4 \times 16 \mathrm{~cm}=64 \mathrm{~cm} .\\
\text { Area of Rhombus }\mathrm{ABCD} =\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD} \\
=\frac{1}{2} \times 48 \times 64 \\
=1536 \text { Sq.cm. } (Ans.)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.
Solution:
Let \mathrm{AB}=\mathrm{CD}=10 \mathrm{~cm}
A D=5 \mathrm{~cm} \\
B C=17 \mathrm{~cm} \\
A D \| B C \text { and } A B=C D \\
A D=E F=5 \mathrm{~cm}
\therefore A D \| B C and A B=C D
\therefore A D =E F=5 \mathrm{~cm} . \\
\therefore \mathrm{BF} =\mathrm{CE}=\frac{17-5}{2}=\frac{12}{2} \mathrm{~cm}=6 \mathrm{~cm}
In \triangle \mathrm{ABF}
\mathrm{AF} =\sqrt{\mathrm{AB}^2-\mathrm{BF}^2} \\
=\sqrt{(10)^2-(6)^2} \\
=\sqrt{100-36} \\
=\sqrt{64} \\
=8 \\
\therefore \quad A F =8 \mathrm{~cm} . \quad \therefore \quad \mathrm{DE}=\mathrm{AF}=8 \mathrm{~cm}.
And BE = BF + EF = (6 + 5)cm =11cm (Ans.)
Area of Trapazium ABCD
=\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC}) \times \mathrm{AF} \\
=\frac{1}{2} \times(5+17) \times 8 \\
=\frac{1}{2} \times 22 \times 8 \\
=88 \text { Sq. } \mathrm{cm}
In \triangle \mathrm{BED}
\mathrm{BD}=\sqrt{\mathrm{DE}^2+\mathrm{BE}^2}
or, \mathrm{BD}=\sqrt{(8)^2+(11)^2}
or, \mathrm{BD}=\sqrt{64+121}
or, \mathrm{BD}=\sqrt{64+121}
\therefore \mathrm{BD}=\sqrt{185} cm.
\therefore Length of diagonal =\sqrt{185 cm}. (Ans.)
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.
Solution:
Let \mathrm{AD}=9 cm
\mathrm{BC} =19 cm \\
\mathrm{AB} =8 cm \\
\mathrm{CD} =6 cm \\
\mathrm{BC} =\mathrm{BF}+\mathrm{FE}+\mathrm{CE} \\
=\mathrm{BF}+\mathrm{CE}+9 \\
\mathrm{BF} =19-9-\mathrm{CE} \\
=10-\mathrm{CE}
In \triangle \mathrm{ABF},
A B^2=A F^2+B F^2
or, (8)^2=A F^2+(10-C E)^2
In \triangle \mathrm{CDE}
\mathrm{CD}^2=\mathrm{DE}^2+\mathrm{CE}^2
or, (6)^2=\mathrm{AF}^2+\mathrm{CE}^2[\because \mathrm{DE}=\mathrm{AF}]
or, (6)^2=(8)^2-(10-C E)^2+C E^2
or, 36=64-\{(10)^2-2.10 . \mathrm{CE}+(\mathrm{CE})^2\}-(\mathrm{CE})^2
or, 36=64-(10)^2+20 \mathrm{CE}-\mathrm{CE}^2+\mathrm{CE}^2
or, 36=64-100+20 \mathrm{CE}
or, 20 \mathrm{CE}=36+36
or, 20 \mathrm{CE}=72
or, \mathrm{CE}=\frac{72}{20}
\text { or, } \mathrm{CE} =\frac{36}{10}=3.6 cm \\
\therefore \mathrm{DE} =\sqrt{\mathrm{CD}^2-\mathrm{CE}^2} \\
=\sqrt{(6)^2-(3.6)^2} \\
=\sqrt{(6+3.6)(6-3.6))} \\
=\sqrt{9.6 \times 2.4} \\
=\sqrt{4.8 \times 4.8} \\
= 4.8 cm
Area of Trapazium AECE
=\frac{1}{2} \times(\mathrm{AD}+\mathrm{BC}) \times \mathrm{DE} \\
=\frac{1}{2} \times(9+19) \times 4.8 \\
=\frac{1}{2} \times 28 \times 4.8 \\
=67.2 \text { Sq.cm. (Ans.) }
17. (M.C.Q)
(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
(a) 4 cm.
(b) 8 cm.
(c) 16 cm
(d) 24 cm
Solution:
Area of parallelogram = 192 Sq.cm.
or, Base × height = 192
or, Base \times \frac{1}{3} base =192
or, (\text { Base })^2=476
or, Base =\sqrt{476}=24 cm.
\therefore \quad \text { Height } =\frac{1}{3} \times \text { base } \\
=\frac{1}{3} \times 24=8 cm .
\therefore (b) is correct option
(ii) If the length of one side of rhombus is 6 cm. and one angle is 60^{\circ}, then area of field in the shape of rhombus is
(a) 9 \sqrt{3} sq.cm
(b) 18 \sqrt{3} \mathrm{Sq} . \mathrm{cm}.
(c) 36 \sqrt{3} \mathrm{Sq} . \mathrm{cm}.
(d) 6 \sqrt{3} Sq.cm.
Solution:
\because \quad A B=B C=6 cm \\
\therefore \angle A B C = 60
\therefore \quad C is an equilateral triangle
\therefore \text { Area of } \triangle A B C=\frac{\sqrt{3}}{4} \times(6)^2 \\
=\frac{\sqrt{3}}{4} \times 36 \\
=9 \sqrt{3} \text { Sq.cm. } \\
\text { Area of rhombus } \mathrm{ABCD} =2 \times \triangle \mathrm{ABC} \\
=2 \times 9 \sqrt{3} \mathrm{Sq} \cdot \mathrm{cm} \\
=18 \sqrt{3} \mathrm{Sq} . \mathrm{sm}
\therefore \quad(b) is correct option
(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
(a) 8 cm
(b) 12 cm
(c) 16 cm
(d) 24 cm.
Solution:
Let 1st diagonal =x
2nd diagonal = 3x
Area of Rhombus = 96 Sq.cm.
\frac{1}{2} \times x \times 3 x=96
or, \frac{3 x^2}{2}=96
or, x^2=64
or, x=\sqrt{64}=8
Length of long diagonal = 3 \mathrm{x} =3 \times 8 cm=24 cm.
\therefore \quad (d) is correct option
(iv) A rhombus and a square on the same base. If the area of square is x^2 Sq. unit and area of field in the shape of rhombus is y sq. unit. then
(a) y>x^2
(b) v<x^2
(c) y=x^2
Solution:
\therefore (b) is correct option
(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
(a) 29 cm
(b) 31 cm
(c) 32 cm
(d) 33 cm.
Solution:
We have,
DE = 6 cm,
AB = ?, CD = 23 cm,
Area of trapazium =162 \mathrm{Sq} . \mathrm{cm}.
or, \frac{1}{2} \times(\mathrm{AB}+\mathrm{CD}) \times \mathrm{DE}=162
or, \frac{1}{2} \times(\mathrm{AB}+23) \times 6=162
or, \mathrm{AB}+23=54
or, \mathrm{AB}=54-23
\therefore \quad AB = 31cm.[/katex]
\therefore (b) is correct option
18. Short answer type:
(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.) }
