## Book: RD Sharma - Mathematics (Volume 2)

### Chapter: 21. Areas of Bounded Regions

#### Subject: Maths - Class 12th

##### Q. No. 50 of Exercise 21.3

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

50
##### Find the area of the figure bounded by the curves y = |x – 1| and y = 3 – |x|.

To find the area of the figure bounded by

y = |x – 1|

Y = x – 1 is a straight line passing through A(1,0)

Y = 1 – x is a straight line passing through A(1,0) and cutting y - axis at B(0,1)

y = 3 – |x|

Y = 3 – x is a straight line passing through C(0,3)and O(3,0)

Y = 3 + x is a straight line passing through C(0,3)and D( – 3,0)

Point of intersection for

Y = x – 1

And y = 3 – x

We get

X – 1 = 3 – x

or, 2x – 4 = 0

or, x = 2

or, y = 2 – 1 = 1

Thus, point of intersection for y = x – 1 and y = 3 + x is B(2,1)

Point of intersection for

y = 1 – x

y = 3 + x

or,1 – x = 3 + x

or, 2x = – 2

or, x = – 1

or, y = 1 – ( – 1) = 2

Thus, point of intersection for y = 1 – x and y = 3 + x is D( – 1,2)

These are shown in the graph below:

Required area = Region ABCDA

= Region ABFA + Region AFCEA + Region CDEC

The area of the figure bounded by the curves y = |x – 1| and y = 3 – |x| is 4 sq. units

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
52