Page 251 - Start Up Mathematics_7

= 5 – (1.5 + 1.5) = 5 – 3 = 2 cm

Area of the cardboard = l × b = 8 × 5 = 40 cm 2
Area of the painted portion = l × b = 5 cm × 2 cm = 10 cm 2
∴ Area of the margin = Area of cardboard – Area of painted portion
= 40 – 10 = 30 cm 2

Example 34: 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 angles through the centre of the field. If 60 m
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: Length of the field = 80 m
Breadth of the field = 60 m
Width of the road = 3 m
(a) Length of the road along the length of the field = 80 m
Area of the road along the length = 80 × 3 = 240 m 2
Length of the road along the breadth of field = 60 m

Area of the road along the breadth = 60 × 3 = 180 m 2
Area common to the roads where they cross each other = 3 × 3 = 9 m 2
Total area of the roads = 240 + 180 – 9 = 411 m 2
(b) Cost of constructing the road = ` 120 × 411 = ` 49,320

(c) Area of the field including the roads = 80 × 60 = 4,800 m 2
Area of field excluding the roads = 4,800 – 411 = 4,389 m 2
2
Cost of levelling the field at the rate of ` 90 per m = 90 × 4,389 = ` 3,95,010
Example 35: Shubhangi wrapped a cord around a
spherical ball of radius 4 cm and cut off
the length required to wrap it fully. Then 4 cm
she wrapped it around a square box of
side 4 cm. Did she have any cord left?
(Take π = 3.14) 4 cm
Solution: Radius of the circular ball = 4 cm
∴ circumference of the greatest circle around the ball = 2πr

= 2 × 3.14 × 4 = 25.12 cm
∴ length of the cord used = 25.12 cm
Side of the square box = 4 cm

∴ perimeter of the square box = 4 × side = 4 × 4 = 16 cm
Hence, length of the cord left = 25.12 – 16 = 9.12 cm
Shubhangi is left with 9.12 cm of cord.

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