## Book: RS Aggarwal - Mathematics

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

#### Subject: Maths - Class 10th

##### Q. No. 50 of Exercise 18A

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

50
##### From a thin metallic piece in the shape of a trapezium ABCD in which AB ∥ CD and ∠BCD = 90°, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet.

Given:

AB CD

BCD = 90°

AB = BC = 3.5 cm = EC

DE = 2 cm

DC = DE + EC = 2 + 3.5 = 5.5 cm

Area of Trapezium = × Sum of Parallel Sides × h

= × (AB + DC) × BC

= × (3.5 + 5.5) × 3.5

= × 9 × 3.5

= 15.75 cm2

Area of Quadrant BFEC = × πr2 = × × 3.5 × 3.5

= 9.625 cm2

Thus, Area of remaining part of metal sheet

= Area of Trapezium – Area of Quadrant BFEC

= 15.75 – 9.625 = 6.125 cm2

Hence, the area of the remaining part of metal sheet is 6.125 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