A circle is inscribed in a quadrilateral ABCD in which B = 90o. If AD = 23 cm, AB = 29 cm and DS = 5 cm. Find the radius of the circle.
Asked by Topperlearning User | 27th Jul, 2017, 01:09: PM
Since tangents drawn from an external point to a circle are equal.
DR = DS = 5cm
Now, AR = AD - DR = 23 - 5 = 18cm
AQ = 18cm
Also, BQ = AB - AQ = 29- 18 = 11cm
But, BP = BQ
Also, Q=P =90o.
In quadrilateral OQBP,
QOP = 360° - (P+Q+B)
= 360° - (90°+90°+90°) = 90°
Hence, OQBP is a square.
Hence, radius of the circle is 11cm.
Answered by | 27th Jul, 2017, 03:09: PM
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