Question

in the figure a circle is inscribed in a quadrilateral avcd in which angle b equal to 90° ad= 23cm ab= 29cm ​

Answer 1

Answer:

In the figure. AB, BC, CD and DA are the tangents drawn to the circle at Q, P, S and R respectively.

∴DS=DR   (tangents drawn from a external point D to the circle).

but DS = 5 cm        (given)

∴ DR = 5 cm

In the fig. AD = 23 cm,         (given)

∴ AR = AD - DR = 23 - 5 = 18 cm

but AR = AQ

(tangents drawn from an external point A to the circle)

∴ AQ = 18 cm

If AQ = 18 cm then       (given AB = 29 cm)

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

In quadrilateral BQOP,

BQ = BP            (tangents drawn from an external point B)

OQ = OP           (radii of the same circle)

∠QBP=∠QOP=90o     (given)

∠OQB=∠OPB=90o     (angle between the radius and tangent at the point of contact.)

∴ BQOP is a square.

∴ radius of the circle, OQ = 11 cm