question_answer

ABCD is a quadrilateral such that \[\angle \mathbf{D}=\mathbf{9}{{\mathbf{0}}^{{}^\circ }}\]. A circle C(O, r) touches the sides AB, BC, CD and DA at P, Q R and S respectively. If \[\mathbf{BC}=\mathbf{38}\]cm. \[\mathbf{CD}=\mathbf{25}\]cm and \[\mathbf{BP}=\mathbf{27}\]cm then radius 'r, is equal to

A)  14 cm 

B)  11 cm         

C)  12 cm                         

D)  10 cm

Correct Answer: A

Solution :

(a): \[\angle ORD=\angle OSD={{90}^{{}^\circ }}\] \[\therefore \]ORDS is a square Since tangents from an exterior point to a circle are equal in length. \[\therefore \] \[BP=BQ:CQ=CR\text{ }and\text{ }DR=DS.\] \[BQ=27\] \[\Rightarrow \] \[BC-CQ=27\] \[\Rightarrow \] \[38-CQ=27\] \[\Rightarrow \] \[CQ=11\text{ }cm\] \[\Rightarrow \] \[CR=11\text{ }cm\] \[\Rightarrow \] \[CD-DR=11\text{ }cm\] \[\Rightarrow \] \[25-DR=11\text{ }cm\] \[\Rightarrow \] \[DR=14cm\] \[\Rightarrow \] \[OR=DR=14cm\]