Page 230 - ICSE Math 6

Example 22: Find the area of the shaded region.
5 cm

5 cm
(a) 7 cm (b) 7 m 10 m

5 m
7 cm 15 m
Solution: (a) Area of the outer square = 7 cm × 7 cm = 49 cm 2
Area of the inner square = 5 cm × 5 cm = 25 cm 2
\ Area of the shaded region = Area of the outer square – Area of the inner square
2
2
= 49 cm – 25 cm = 24 cm 2
(b) Area of the outer rectangle = 15 m × 10 m = 150 m 2
Area of the inner rectangle = 5 m × 7 m = 35 m 2
2
2
\ Area of shaded region = 150 m – 35 m = 115 m 2
Example 23: A wall to wall carpeting of a floor measuring 8 m by 6 m is to be done. What length of
the carpet is just enough if the roll of the carpet is 3 m wide?
Solution: Length of the floor = 8 m
Breadth of the floor = 6 m

Area to be carpeted = 8 × 6 = 48 sq. m
Let carpet of length l be sufficient to cover the floor.
Now, area of the carpet = area of the floor
l × 3 = 48 sq. m

⇒ l = 16 m
Example 24: The cost of cultivating a rectangular field at the rate of ` 3 per square metre is
` 1,728. If the breadth of the field is 18 m, find the cost of fencing the field at
` 8.5 per metre.
Total cost
Solution: Area of the rectangular field =
Cost per square metre

1,728
= = 576 sq. m
3
Now, Length × Breadth = 576 sq. m

576
∴ Length of the field = = 32 m
18
Perimeter of the field = 2(l + b)
= 2(32 + 18) = 100 m
Cost of fencing the field = ` 8.5 × 100 = ` 850
Example 25: A floor measuring 7 m by 6 m is to be covered by square tiles each of side 50 cm. Find
the number of tiles required to cover the floor. If the cost of one tile is ` 60, find the
total cost of tiles?

Solution: Area of the floor = Length × Breadth
= 7 m × 6 m = 42 sq. m

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