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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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.
