Question

In given figure, 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, $\angle$B = ${90}^{\circ }$ 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

We know that the lengths of the tangents drawn from an external point to a circle are equal.

DS = DR = 5 cm

$\therefore$ AR = AD – DR = 23 cm – 5 cm = 18 cm

AQ = AR = 18 cm

$\therefore$ QB = AB – AQ = 29 cm – 18 cm = 11 cm

QB = BP = 11 cm

In right $\Delta$PQB, $P{Q}^2=Q{B}^2+B{P}^2=(11cm{)}^2+(11cm{)}^2=2x(11cm{)}^2$

PQ = 11$\sqrt{2}$ cm…..(1)

In right $\Delta$ PQB,

$P{Q}^2=O{Q}^2+O{P}^2=r^2+r^2=2r^2$

PQ = r $\surd 2$ …..(2)

From (1) and (2), we get

r = 11 cm