Question

In the figure, a circle is inscribed in a quadrilateral ABCD in which $\angle B={90}^o$. If $AD=23cm,AB=29cm$ and $DS=5cm$ find the radius of the circle.

Answer 1

In the figure. AB, BC, CD and DA are the tangents drawn to the circle at Q, P, S and R respectively.
$\therefore DS=DR$ (tangents drawn from a external point D to the circle).
but DS = 5 cm (given)
$\therefore$ DR = 5 cm
In the fig. AD = 23 cm, (given)
$\therefore$ AR = AD - DR = 23 - 5 = 18 cm
but AR = AQ
(tangents drawn from an external point A to the circle)
$\therefore$ 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)
$\angle QBP=\angle QOP={90}^o$ (given)
$\angle OQB=\angle OPB={90}^o$ (angle between the radius and tangent at the point of contact.)
$\therefore$ BQOP is a square.
$\therefore$ radius of the circle, OQ = 11 cm