Page 247 - Start Up Mathematics_7

Perimeter of the square = 44 cm
4 × s = 44 cm (Perimeter of square = 4 × s)
44
∴ s = 4 = 11 cm

2
Area of the square = s = 11 × 11 = 121 cm 2
Clearly, the circle encloses more area.
Example 25: 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. (Take π = 22 )
7
Solution: We proceed stepwise as follows:
Step 1: ∴ area of the circular card sheet = πR 2 (  R = 14 cm)
22
= 7 × 14 × 14 = 616 cm 2
2
Step 2: ∴ area of 1 small circle = πr ( r = 3.5 cm)
22
= 7 × 3.5 × 3.5 = 22 × 0.5 × 3.5 = 38.50 cm 2
Area of 2 small circles = 2 × 38.50 = 77.00 cm 2
2
Step 3: Area of the rectangle = l × b = 3 × 1 = 3 cm ( l = 3, b = 1)
1 1
Step 4: Area of the right triangle = × base × height = × 3 × 4 = 6 cm 2
2 2
Area of 2 right triangles = 6 × 2 = 12 cm 2
Step 5: Area of the required figure = Area of circular card sheet + Area of
2 triangles – (Area of 2 circles + Area
of rectangle)

= 616 + 12 – (77 + 3) = 628 – 80 = 548 cm 2
Example 26: 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 π = 3.14)
Diameter
Solution: ∴ Radius of circular cut-outs = 2 = 1 cm

2
2
Area of a circle = πr = 3.14 × 1 = 3.14 cm 2 10 cm
Area of 4 circles = 4 × 3.14 = 12.56 cm 2
2
2
Area of aluminium sheet = (side) = 10 × 10 = 100 cm
Area of left over aluminium sheet = Area of aluminium sheet – Area of 4 circles
= 100 – 12.56 = 87.44 cm 2

Example 27: A circular flowerbed is surrounded by a path 4 m wide. The 66 m
diameter of the flowerbed is 66 m. What is the area of this path?
22
(Take π = ) (NCERT)
7
4 m

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