Book: RS Aggarwal - Mathematics
Chapter: 18. Area of Circle, Sector and Segment
Subject: Maths - Class 10th
Q. No. 37 of Exercise 18A
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In the given figure, APB and CQD are semicircle of diameter 7 cm each, while ARC and BSD are semicircles of diameter 14 cm each. Find the (i) perimeter, (ii) area of the shaded region.
(i) Given:
Diameter of semicircles APB and CQD = 7 cm
⇒ Radius of semicircles APB and CQD = 
cm = r1
Diameter of semicircles ARC and BSD = 14 cm
⇒ Radius of semicircles ARC and BSD = 
cm = 7 cm = r2
Perimeter of APB = Perimeter of CQD
Area of APB = Area of CQD ………….. (i)
Perimeter of ARC = Perimeter of BSD
Area of ARC = Area of BSD ………….. (ii)
∵ Perimeter of semicircle = πr …………… (iii)
∴ Perimeter of APB = πr1
= 
× 
= 11 cm
Then, using (i), we have
Perimeter of CQD = 11 cm
Now, using (iii), we have
Perimeter of ARC = πr2
= 
× 7 = 22 cm
Then, using (ii), we have
Perimeter of BSD = 22 cm
Perimeter of shaded region
= (Perimeter of ARC + Perimeter of APB) + (Perimeter of BSD + Perimeter of CQD)
= (22 + 11) + (22 + 11) = 33 + 33 = 66 cm
Hence, the perimeter of the shaded region is 66 cm.
(ii) Now,
∵ Area of semicircle = 
πr2 …………. (iv)
∴ Area of APB = 
πr12
= 
× 
× 
× 
= 
cm2
Then, using (i), we have
Area of CQD = 
cm2
Now, using (iv), we have
Area of ARC = 
πr22
= 
× 
× 7 × 7 = 11 × 7 = 77 cm2
Then, by using (ii), we have
Area of BSD = 77 cm2
Area of shaded region
= (Area of ARC-Area of APB) + (Area of BSD- Area of CQD)
= (77 - 
) + (77 - 
)
= (
) + (
) = 
+ 
= 
= 115.5 cm2
Hence, the area of the shaded region is 115.5 cm2.
