Page 270 - Start Up Mathematics_8 (Non CCE)

fi Diameter of circle = 42 cm

42
fi Radius of circle (r) = = 22 cm
2
2
)
2
Area of circle = pr = 3.14 × (22 = 3.14 × 8 = 25.12 cm 2
2
2
Area of square = (side) = (4) = 16 cm 2
2
Area of shaded portion = Area of circle – Area of square = (25.12 – 16) cm = 9.12 cm 2
Example 5: The shape of a garden is rectangular in
the middle and semicircular at the ends 7 m
as shown in the figure. Find the area and
perimeter of this garden. (NCERT) 20 m
Solution: Diameter of circular ends = 7 m
7
fi Radius of circular ends r = m = 3.5 m
2
Now, length of the rectangle (l) = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
Breadth of the rectangle (b) = 7 m
Perimeter of the garden = 2l + Perimeter of the two circular ends
= 2l + pr + pr = 2l + 2pr

22
= 2 × 13 + 2 × × 3.5 = 26 + 22 = 48 m
7
Area of the garden = Area of rectangular part + Area of semicircular ends
1 1
2
2
= (l × b) + pr + pr = (l × b) + pr 2
2 2
22
= (13 × 7) + × 3.5 × 3.5 = 91 + 38.5 = 129.5 m 2
7
S W T
2
Example 6: The area of a square STEP is 36 cm . Find the area of the square
WING obtained by joining the midpoints of the sides of STEP.
Solution: Area of the square STEP = 36 cm 2 G I
\ Each side of the square = 36 = 6 cm
In D GPN,

1 1 P N E
PN = × PE = × 6 = 3 cm ( N is the midpoint of PE)
2 2
1 1
PG = × PS = × 6 = 3 cm ( G is the midpoint of PS)
2 2
2
2
2
2
2
\ GN = PN + PG = 3 + 3 = 18
GN = 18 = 32 cm
2
Hence, the area of square WING = (32 ) = 9 × 2 20 m
= 18 cm 2
Example 7: Mrs Kaushik has a square plot as shown in the figure. She 25 m 15 m
wants to construct a house in the middle of the plot. A garden
is to be developed around the house. Find the total cost of
developing the garden, if the cost of developing it is ` 55
per square metre. (NCERT) 25 m
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