Question

In the adjoining figure quadrilateral ABCD is circumscribed. If the radius of the incircle with centre O is 10 cm and AD perpendicular to DC .

Answer 1

Step-by-step explanation:

Your question incomplete.

Answer 2

Step-by-step explanation:

$\huge{Answer}$

˂A = ˂OPA =˂OSA = 90˚

Hence, ˂SOP=90˚

Also, AP=AS

Hence, OSAP is a square

AP=AS=10cm

CR=CQ=27cm

BQ=BC-CQ=38-27=11cm

BP=BQ=11 cm

X=AB=AP+BP=10+11=21 cm

Please refer the attachment for the diagram..