# Selina Concise Class 7 Math Chapter 20 Mensuration Exercise 20B Solution

## EXERCISE 20B

(1) Area of the rectangle = (25 × 16) cm2 = 400 cm2

(2) Here, diagonal = 1 m = 100 cm

Let the breadth of rectangular broad be x cm.

Applying Pythagoras theorem,

∴ AC2 = AB2 + BC2

⇒ (100)2 = (96)2 + x2

⇒ x2 = (100)2 – (96)2

⇒ x2 = 10000 – 9216

⇒ x2 = 784

⇒ x = √784 = 28

Breadth of the board = 28 cm

∴ Its area = (96 × 28) cm2 = 2688 cm2

(3) Let the length and breadth will be 4x and 3x m

Its area = (4x × 3x) = 12x2 m2

Given area = 1728 m2

∴ 12x2 = 1728

⇒ x2 = 1728/12

⇒ x2 = 144

⇒ x = √144 = 12

Its perimeter = 2 × [(4 × 12) + (3 × 12)] m

= 2 × (48 + 36) = 2 × 84 = 168 m

(ii) Rate of fencing = Rs 40

∴ Total cost of fencing = Rs (168 × 40) = Rs 6720

(4) Area of the floor = (40 × 15) m2 = 600 m2

Length of each tile = 60 cm = 0.6 m

And breadth = 50 cm = 0.5 m

∴ Area of each tile = (0.6 × 0.5) = 0.3 m2

∴ Number of tiles = (600×10)/3=2000    (9) Area of the parallelogram = (20 × 16) = 320 cm2

(10) Let the height of the parallelogram = x and its base = x/3

∴ x × x/3 = 768

⇒ x2 = 768 × 3

⇒ x2 = 2304

⇒ x = √2304 = 48

Hence, the base = 48 cm and height = 48/3 = 16 cm  (15) Area of the rhombus = 84 cm2

Perimeter = 56 cm

∴ Its side = 56/4 = 14 cm

∴ Height = 84/14 = 6 cm

(16) Area of triangle = ½ (base × height)

= ½ × 30 × 18 = 270 cm2

(17) Let the height of the triangle be x cm.

∴ ½ × 18 × x = 270

⇒ 18x = (270 × 2) = 540

⇒ x = 540/18

⇒ x = 30 cm

(18) Length of the other leg = (160 × 2)/16 = 20 cm

(19) Let the other leg will be x cm.

Applying Pythagoras theorem,

AC2 = AB2 + BC2

⇒ (13)2 = (12)2 + x2

⇒ x2 = (13)2 – (12)2

⇒ x2 = 169 – 144

⇒ x = √25 = 5 cm

∴ Area = ½ × 12 × 5 = 6 × 5 = 30 cm2           Get next Exercise 21A solutions click here

Updated: March 29, 2019 — 12:04 pm

1. Answer of 36 is wrongly done

1. What a good math

2. Thank you so much for your explanation

1. You welcome

3. I appreciate all your answers but the answer number 37 is wrong.

1. Ok will check