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Question
In the fig. a circle is inscribed in a quadrilateral ABCD in which ∠B = 90° if AD = 23cm,
AB = 29cm and DS = 5cm, find the radius of the circle.
Solution
Given AD = 23 cm
AB = 29 cm
∠B = 90°
DS = 5cm
From fig in quadrilateral POQB
∠OPB = ∠OQB = 90° = ∠B = ∠POQ
and PO = OQ. ∴ POQB is a square PB = BQ = r
We know that
Tangents drawn from external point to circle are equal in length.
We know that
Tangents drawn from external point to circle are equal in length.
From A, AR = AQ …. (i)
From B, PB = QB …. (ii)
From C, PC = CS …. (iii)
From D, DR = DS …. (iv)
(i) + (ii) + (iv) ⇒ AR + DB + DR = AQ + QB + DS
⇒ (AR + DR) + r = (AQ + QB) + DS
AD + r = AB + DS
⇒ 23 + r = 29 + 5
⇒ r = 34 – 23 = 11 cm
∴ radius = 11 cm
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