Question

A Circle with Centre 0 is inscibed in a quadrilateral
ABCD such that, it touches the sides BC, AB, AD and CD
at points P, Q, R, S respectively . if AB=29cm, AD =23cm <B=90°and DS=5cm then the radius of the circle is








plz give me the answer now only ​

Answer 1

$\fbox {Given}$

DS = 5 cm

Since DS and DR are tangents from the same external point to the circle, DS = DR = 5 cm

Since AD = 23 cm, AR = AD - DR = 23 - 5 = 18 cm.

Similarly, AR and AQ are the tangents from the same external point to the circle and hence AR = AQ = 18 cm.

Since AB = 29 cm, BQ = AB - AQ = 29 - 18 = 11 cm.

Since CB and AB are the tangents to the circle, ∠OPB and ∠OQB is equal to 90°.

Given that ∠B is 90° and hence ∠POQ is also equal to 90° and hence OQBP is a square.

Since BQ is 11 cm, the side of the square OQBP is 11 cm

 

From the figure it is clear that the side of the square is the radius of the circle and hence radius of the circle is 11 cm.

$\rule {160}{2}\ \sf {\ Be\ Brainly\ \star}$