Chapter – 15 : Area And Perimeter Of Triangle and Quadrilateral | Chapter Solution Class 9

Ganit Prakash Solution 9

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:

Perimeter $= 2(18+12) \mathrm{cm} \\$

$=2 \times 30 \mathrm{~cm}=60 \mathrm{~cm} \text {. (Ans) }$

Perimeter $= 4 \times 9 \\$

$= 36 \mathrm{~cm} \text {. (Ans) }$

Perimeter $= 8 \mathrm{~cm}+9 \mathrm{~cm}+15 \mathrm{~cm}+7 \mathrm{~cm} \\$

$=39 \mathrm{~cm} \text {. (Ans) }$

Perimeter $= 12 \mathrm{~cm}+12 \mathrm{~cm}+21 \mathrm{~cm}+21 \mathrm{~cm}$

$=66 \mathrm{~cm}. (Ans)$

Perimeter $= 5 \mathrm{~cm}+13 \mathrm{~cm}+12 \mathrm{~cm} \\$

$=30 \mathrm{~cm} \text {. (Ans) }$

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

$\therefore$ (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: