Page 243 - ICSE Math 7

2
Area of the square = s = 11 cm × 11 cm = 121 cm 2
Clearly, the circle encloses more area.

Example 17: From a circular card sheet of radius 14 cm, two circles
of radius 3.5 cm and a rectangle of length 3 cm and
breadth 1 cm are removed. Two identical right triangles
whose legs are of length 3 cm and 4 cm are joined
(as shown). Find the area of the figure so obtained.
22
(Take p = )
7
Solution: Area of the circular card sheet = pR 2 (  R = 14 cm)
22
= × 14 cm × 14 cm = 616 cm 2
7
2
Area of 1 small circle = pr ( r = 3.5 cm)
22
= × 3.5 cm × 3.5 cm = 22 × 0.5 cm × 3.5 cm = 38.50 cm 2
7
2
Area of 2 small circles = 2 × 38.50 cm = 77 cm 2
Area of the rectangle = l × b = 3 cm × 1 cm = 3 cm 2 (  l = 3 cm, b = 1 cm)
1 1
Area of the right triangle = × base × height = × 3 cm × 4 cm = 6 cm 2
2 2
Area of 2 right triangles = 6 cm × 2 cm = 12 cm 2
Area of the required figure = Area of circular card sheet + Area of 2 right triangles –
(Area of 2 circles + Area of rectangle)
2
2
2
2
2
= 616 cm + 12 cm – (77 + 3) cm = 628 cm – 80 cm
= 548 cm 2
Example 18: Four circles of diameter 2 cm each are cut from a square piece of an
aluminium sheet of side 10 cm. What is the area of the aluminium
sheet left? (Take p = 3.14)
Diameter
Solution: ∴ Radius of circular cut-outs = = 1 cm
2
2
2
2
Area of a circle = pr = 3.14 × 1 cm = 3.14 cm 2 10 cm
2
Area of 4 circles = 4 × 3.14 cm = 12.56 cm 2
2
Area of aluminium sheet = (side) = 10 cm × 10 cm = 100 cm 2
Area of left over aluminium sheet = Area of aluminium sheet – Area of 4 circles
2
2
= 100 cm – 12.56 cm = 87.44 cm 2
Example 19: A circular flowerbed is surrounded by a path 4 m wide. The
diameter of the flowerbed is 66 m. What is the area of this path?
22 66 m
(Take p = )
7
Solution: Diameter of the circular flowerbed = 66 m
Radius of the flowerbed (r) = 66 m ÷ 2 = 33 m 4 m

Radius of the outer circle = R = 33 m + 4 m = 37 m

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