Question

In Fig., a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P,Q,R and S respectively, If AB = 29 cm, AD = 23 cm, $DB={90}^{\circ }$ and DS = 5 cm, then the radius of the circle (in cm.) is :

• A. 11
• B. 18
• C. 6
• D. 15

Answer 1

The correct option is A

11

Given that AB, BC, CD and AD are tangents to the circle with centre O t Q, P, S and R respectively. AB =
$29cm,AD=23,DS=5cmand\angle B={90}^{\circ }.$
Join PQ
We know that, the lengths of the tangents drawn from an external point to a circle are equal
DS = DR = 5cm
$\therefore AR=AD-DR=23cm-5cm=18cmAQ=AR=18cm$
$\therefore QB=AB-AQ=29cm-18cm=11cmQB=BP=11cm$
In right $\Delta PQB,P{Q}^2=Q{B}^2+B{P}^2(11cm{)}^2+(11cm{)}^2=2\times (11cm{)}^2$
$PQ=11\sqrt{2}cm......\left(1\right)$
In right $\Delta PQB,$
$P{Q}^2=O{Q}^2+O{P}^2+r^2+r^2=2r^2$
$PQ=\sqrt{2r}....\left(2\right)$
From (1) and (2), we get
r = 11 cm
thus, the radius of the circle is 11 cm.
The correct answer is 11 whichis given by option A.