Page 236 - ICSE Math 7

Example 5: A door of length 2 m and breadth 1 m and a square window of dimension 0.3 m is
fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost
of whitewashing the wall, at the rate of ` 20 per sq. m.
Solution: Length (l) of the wall = 4.5 m
0.3
Breadth (b) of the wall = 3.6 m 1
Area of the wall = l × b = 4.5 m × 3.6 m = 16.20 m 2 2 3.6 m
2
Area of the door = l × b = 2 m × 1 m = 2 m
2
Area of the window = (side) = 0.3 m × 0.3 m = 0.09 m 2 4.5 m
∴ area to be whitewashed = Area of wall – (Area of door + Area of window)
2
2
= 16.20 m – (2 + 0.09) m = 14.11 m 2
Rate of whitewashing = ` 20 per m 2
∴ cost of whitewashing = ` 20 × 14.11 = ` 282.20

Example 6: A path 3 m wide runs outside and around a rectangular
park of length 120 m and breadth 65 m. Find the area
of the path in hectare. 65 m 3 m

Solution: Length of the rectangular park = 120 m 120 m
Breadth of the rectangular park = 65 m

Width of the path = 3 m
Length of the park alongwith path = 120 + 3 + 3 = 126 m

Breadth of the park alongwith path = 65 + 3 + 3 = 71 m
Area of the park = l × b = 120 m × 65 m = 7,800 m 2

Area of the park alongwith path = 126 m × 71 m = 8,946 m 2
2
2
Area of the path = 8,946 m – 7,800 m = 1,146 m 2
1,146
Area of the park in hectares = = 0.1146 hectares
10,000
Example 7: Through a rectangular field of length 80 m and 80 m
breadth 60 m, two roads are constructed which
are parallel to the sides and cut each other at right 60 m
angles through the centre of the field. If the width
of each road is 3 m, find:
(a) the area covered by the roads
(b) the cost of constructing the roads at the rate of ` 120 per m 2
(c) cost of levelling the field at the rate of ` 90 per m 2
Solution: (a) Length of the road along the length of the field = 80 m
Area of the road along the length = 80 m × 3 m = 240 m 2
Length of the road along the breadth of field = 60 m
Area of the road along the breadth = 60 m × 3 m = 180 m 2

Area common to the roads where they cross each other = 3 m × 3 m = 9 m 2
2
Total area of the roads = (240 + 180 – 9) m = 411 m 2

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