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A quadrilateral ABCD is drawn so that $∠ D = 90 °$ $BC = 38 cm$ and $CD = 25 cm$ . A circle is inscribed in the quadrilateral and it touches the side AB, BC, CD and DA at P, Q, R and S respectively. If $BP = 27 cm,$ find the radius of the inscribed circle.

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13 cm

16 cm

14 cm

15 cm

answer is B.

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Detailed Solution

Given that A quadrilateral ABCD is drawn so that

∠ D = 90 °

BC = 38 cm

and

CD = 25 cm

.
We have to find the radius of the circle.
The tangent of a circle at any point in a circle is perpendicular to the radius through the point of contact.
Consider the figure,
Question Image

As we know that the tangent of a circle at any point in a circle is perpendicular to the radius through the point of contact. So,

∠ O R D = ∠ O S D = 90 °

From the figure, OR = OS. So, ORDS is a square. Since tangents from an external point to a circle are equal in length. So,

B P = B Q, C Q = C R,

and

D R = D S

. It is given that, BP = 27cm.
So, BQ = 27cm.
Thus,

B C − C Q = 27 ⇒ 38 − C Q = 27 ⇒ C Q = 38 − 27 ⇒ C Q = 11

Also, CR = 11cm.
Now,

C D − D R = 11 ⇒ 25 − D R = 14 : C D = 25 cm ⇒ D R = 25 − 11 ⇒ D R = 14 cm

Since ORDS is a square.
So, OR =DR = 14cm.
Therefore, the radius is 14cm.
Hence, the correct option is 2.

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