Question

In the following figure, a circle is inscribed in the quadrilateral ABCD. If BC = 38 cm. QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.

• A. 34 cm
• B. 24 cm
• C. 13 cm
• D. 14 cm

Answer 1

The correct option is D

14 cm

A circle is inscribed in a quadrilateral ABCD.
Given, BC = 38 cm, QB = 27 cm, DC= 25 cm and AD $\perp$ BC.

Construction: Join OP and OS.

Here, OP and OS are the radii of the circle
By Theorem- The tangent at any point of a circle is perpendicular to the radius through the point of contact.
We have, OP $\perp$ AD and OS $\perp$ CD

$\Rightarrow OP\parallel DSandOS\parallel DP$
[Since all angles are 90 degrees in the quadrilateral OPDS]
$\Rightarrow$ OPDS is a parallelogram.

Also, OP = OS (Radii of the circle) and DP = DS (Tangents from an external point to a circle are equal in length.)
$\therefore$ OP = OS = DP = DS
$\therefore$ OPDS is a square
[A parallelogram with all sides equal and all angles equal is a square]

Let length of radius of the circle = $r$ then DP = DS = $r$

We have, DC = CS + DS
$\Rightarrow$ CS = DC - DS
$\Rightarrow$ CS = 25 - $r$

BQ = BR = 27 cm (Tangents from external point to the circle are equal in length)

BC = CR + BR
$\therefore$ CR = BC - BR = 38 - 27 = 11 cm

We have, CR = CS (Tangents from external points are equal in length)

$\therefore 11=25-r$
$\Rightarrow r=25-11=14$

$\therefore$ Radius of the circle = 14 cm