## Book: RS Aggarwal - Mathematics

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

#### Subject: Maths - Class 10th

##### Q. No. 57 of Exercise 18B

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57
##### In the given figure, ΔABC is right-angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and 0 is the centre of the incircle of ΔABC. [Take π = 3.14.]

Given AB = 6 cm, BC = 10 cm

Consider ∆ABC, BAC = 90°

102 = 62 + AC2 (putting given values)

100 = 36 + AC2

100 – 36 = AC2

AC2 = 64

AC = 8 cm

Join OB, OA, OC, OE, OF, OD

Here OE = OF = OD = radius of circle = r cm

OEC = ODB = OFB = 90° (angle at the point of contact of radius & tangent)

Area ∆ABC = Area of ∆OAC + Area of ∆OCB + Area of ∆OAB eqn1

Area of ∆OAC = 4r eqn2

Area of ∆OCB = 5r eqn3

Area of ∆OAB = 3r eqn4

Area of ∆ABC = 3×8

Area of ∆ABC = 24 cm2 eqn5

Putting all the values in equation we get;

24 = 4r + 5r + 3r

24 = 12r

r = 2 cm

Area of circle = πr2

Put the value of r, we get,

Area of circle = π×22

Area of circle = 3.14×4 (putting π = 3.14)

Area of circle = 12.56 cm2 eqn6

Area of shaded region = Area of triangle – Area of circle

Area of shaded region = 24 – 12.56 (from eqn5 and eqn6)

Area of shaded region = 11.44 cm2

Area of shaded region is 11.44 cm2.

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