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Question
In the given figure, ∆ABC is right-angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of ∆ABC.
Solution
Using Pythagoras' theorem for triangle ABC, we have:
CA2 + AB2 = BC2
CA` = sqrt("BC"^2 - "AB"^2)`
`=sqrt(100-36)`
`=sqrt(64)`
= 8 cm
Now, we must find the radius of the incircle. Draw OE, OD and OF perpendicular to AC, AB and BC, respectively.
Consider quadrilateral AEOD.
Here,
EO = OD(Both are radii.)
Because the circle is an incircle, AE and AD are tangents to the circle.
∠AEO = ∠ADO = 90°
Also,
∠A = 90°
Therefore, AEOD is a square.
Thus, we can say that AE = EO = OD = AD = r.
CE = CF = 8 - r
BF = BD = 6 - r
CF + BF = 10
⇒ (8 - r)+(6-r) = 10
⇒ 14 - 2r = 10
⇒ r = 2 cm
Area of the shaded part = Area of the triangle-- Area of the circle
`= {1/2xx6xx8} - {pixx2xx2}`
= 24 - 12.56
= 11.44 cm2
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