## Book: RS Aggarwal - Mathematics

### Chapter: 18. Area of Circle, Sector and Segment

#### Subject: Maths - Class 10th

##### Q. No. 37 of Exercise 18A

Listen NCERT Audio Books to boost your productivity and retention power by 2X.

37
##### 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 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 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.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51