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
Answer of 36 is wrongly done
What a good math
Thank you so much for your explanation
You welcome
I appreciate all your answers but the answer number 37 is wrong.
Ok will check