# Selina Concise Class 9 Maths Chapter 20 Area and Perimeter Of A Plane Figures Exercise 20B Solutions

## Selina Concise Class 9 Maths Chapter 20 Area and Perimeter Of A Plane Figures Exercise 20B Solutions Exercise 20B

(1) Find the area of a quadrilateral one of whose diagonals is 30 cm. long and the Perpendiculars from the other two vertices are 19 cm and 11 cm respectively.

Solution: Area (ABCD) = area (△ ABC) + area (△ ADC).

= 1/2 × AC × BM + 1/2 × AC × DN

= 1/2 × 30 × 19 + 1/2 × 30 × 11

= 30/2 × 19 + 30 / 2 × 11

= 15 (19 + 11)

= I5 (30)

= 450 cm2

The area of a quadrilateral is 450 cm2

(2) The diagonals of a quadrilateral are 16cm and 13cm. If they interest each other at right angles, find the area of the quadrilateral.

Solution: Area (ABCD) = area (△ABC) + area (△ADC)

= 1/2 × AC × BM + 1/2 × Ac × DM

= AC/2 [BM + DM]

= 16/2 [BD]

= 8 × BD

= 8 × 13

= 104 cm2

The area of the quadrilateral is 104 cm2.

(3) Calculate the area of quadrilateral ABCD, in which ABD = 90°, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.

Solution: In △ABD,

By Pythagoras Theorem,

(hypotenuse)2 = (Base)2 + (height)2

(26)2 = (24)2 + AB2

(26)2 – (24)2 = AB2

By using formula,

( a2 – b2) = (a + b) (a – b)

(26 + 24) (26 – 24) = AB2

50 × 2 = AB2

100 = AB2

Taking Square root on both Sides

√100 = √AB2

10 = AB

area (ABCD) = area (△ABD) + area (△BCD)

= 1/2 × BD × AB + √3/4 (24)2

= 1/2 × 24 × 10 + √3/4 × 24 × 24

= 120 + √3 × 6 × 24

= 120 + 6 × 24 √3

= 120 + 144 × 1.732

= 120 + 244. 41

= 369. 41 cm2

(4) Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24cm, A = 90° and BC = CD = 52 cm.

Solution: In △ABD,

By Pythagoras Theorem,

= (24)2 + (32)2

= 576 + 1024

BD2 = 1600

Taking Square root on both sides.

√BD2 = √1600

BD = 40 Cm

Area (ABCD) = area (△ABD) + area (△BCD)

= 1/2 × Base × height + b/4 √4a2 – b2

= 1/2 × AB × AD + 40 √4 × (52)2 – (40)2

= 1/2 × 32 × 24 + 10 √4 × 52 × 52 – 40 × 40

= 16 × 24 + 10 √4 × 4 × 13 × 52 – 4 × 10 × 10 × 4

= 384 + 10 √4 × 4 (13 × 52 – 100)

= 384 + 10 × 4 √676 – 100

= 384 + 40 √576

= 384 + 40 × 24

= 384 + 960

= 1344 cm2

The area of quadrilateral ABCD is 1344 Cm2

(5) The Perimeter of a rectangular field is 315 Km. If the length of the field is twice it’s width, find the area of the rectangle in Square meters.

Solution:

Let breath of rectangle is x α m and the length of rectangle is 2xm.

Perimeter = 3/5 Km

= 3/5 × 1000 m

= 3 × 200

Perimeter = 600 m.

2 (l + b) = 600

2 (2x + x) = 600

2 (3x) = 600

3x = 600/2

3x = 200

X = 300/3

X = 100

l = 2x = 2 × 100 = 200m

Area of rectangle = l × b

= 200 × 100

= 20,000 m2 or Sq. m.

Here is your solution of Selina Concise Class 9 Maths Chapter 20 Area and Perimeter Of A Plane Figures Exercise 20B

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Updated: February 17, 2023 — 1:24 pm