Page 137 - Maths Skills - 7

Perimeter and Area 135

Put this value in equation (iii)

12 + b = 17
⇒ b = 17 – 12 = 5
Thus, l = 12 cm and b = 5 cm
Example 2: The ratio of the length to the breadth is 3 : 2 and the area of the rectangle is 150 m . Find the cost
2
of fencing the field at the rate of ` 2.50 per metre.
Solution: Let the length and breadth be 3x and 2x.

We have, area = 150 m ⇒ (3x × 2x) = 150 m 2
2
⇒ 6x = 150 m 2
2
⇒ x = 150 = 25 m 2
2
6
⇒ x = 25m 2

⇒ x = 5 m
\ Length = 3x = 15 m and breadth = 2x = 10 m
Perimeter of the field = 2(l + b) = 2(15 m + 10 m) = 50 m
\ The cost of fencing at the rate of 2.50 per metre = ` (50 × 2.50) = ` 125

Example 3: The length of the diagonal of a square is 72 cm. Find the length of its side and also the area of
the square.
Solution: We have, diagonal of a square = 72 cm

⇒ a 2 = 72 cm, where a is a side of a square.
72
⇒ a = cm = 7 cm
2
⇒ a = 7 cm
Area of the square = a = (7 cm) = 49 cm 2
2
2
Example 4: The length, breadth and height of a room are 10 m, 12 m and 8 m respectively. Find the length of
the diagonal of the room.
Solution: Length = 10 m; Breadth = 12 m; Height = 8 m

Length of the diagonal = l 2 + b 2 + h 2

2
2
2
10
12
= () +() +() 8 m = 308 m =17 5 m.
Example 5: In Example 4, find the cost of whitewashing the four walls of it at the rate of ` 0.50 per m .
2
Solution: Area of the four walls = 2h(l + b) sq unit
= 2 × 8 × (10 + 12) m = 352 m 2
2
\ Cost of whitewashing at the rate of ` 0.50 per m = ` (352 × 0.50) = ` 176.
2
Exercise 8.2

1. Find the area of the squares whose sides are.

(i) 6.2 cm (ii) 4.8 cm (iii) 6.5 m