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प्रश्न
In the given figure, a circle with center O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29cm, AD = 23cm, ∠B = 90° and DS=5cm then find the radius of the circle.
उत्तर
We know that tangent segments to a circle from the same external point are congruent
Now, we have
DS = DR, AR = AQ
Now AD = 23 cm
⇒ AR + RD = 23
⇒ AR =23- RD
⇒ AR = 23 -5 [ ∴ DS = DR = 5]
⇒ AR = 18 CM
Again , AB= 29 cm
⇒ AQ +QB = 29
⇒ QB = 29-AQ
⇒ QB = 29-18 [∵ AR = AQ = 18]
⇒QB = 11CM
Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ
Hence, BQOP is a square.
We know that all the sides of square are equal.
Therefore, BQ = PO = 11 cm
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