Question

a circle is inscribed in a quadrilateral in which angle b is equal to 90 degree if ad is equal to 23 CM ab is equal to 29 CM AC is equal to 5 cm find the radius r of circle

Answer 1

Since tangents drawn from an external point to a circle are equal.

DR = DS = 5cm

Now, AR = AD - DR = 23 - 5 = 18cm

But, AR=AQ

AQ = 18cm

Also, BQ = AB - AQ = 29- 18 = 11cm

But, BP = BQ

BP=11cm

Also, Q=P =90o.

In quadrilateral OQBP,

QOP+P+Q+B=360°

QOP = 360° - (P+Q+B)

= 360° - (90°+90°+90°) = 90°

Hence, OQBP is a square.

BQ=OQ=OP=BP=11cm

Hence, radius of the circle is 11cm.

Answer 2

Answer:

Step-by-step explanation:

AB, BC, CD and AD are tangents to the circle with centre O at Q, P, S and R respectively. Given that, AB = 30 cm, AD = 24, DS = 8 cm and ∠B = 90°.

The lengths of the tangents drawn from an external point to a circle are equal. DS = DR = 8 cm ∴ AR = AD – DR = 24−8 = 16 cm AQ = AR = 16 cm ∴ QB = AB – AQ = 30 − 16 = 14 cm QB = BP = 14 cm. ∠PBQ = 90° [Given] ∠OPB = 90° [Angle between the tangent and the radius at the point of contact is a right angle.] ∠OQB = 90° [Angle between the tangent and the radius at the point of contact is a right angle.] ∠POQ = 90° [Angle sum property of a quadrilateral.] So, OQBP is a square. ∴QB = BP = r = 14 cm ∴ the radius of the circle is 14 cm.