Question

The sides of a quadrilateral which can be inscribed in a circle are $5,5,12$ and $12$cm.Then the radius of the incircle of the quadrilateral is:

• A. $\frac{15}{17}$cm
• B. $\frac{30}{17}$cm
• C. $\frac{60}{17}$cm
• D. incircle does not exist.

Answer 1

The correct option is C $\frac{60}{17}$cm

Here,$AB=BC=5$cm, $AD=DA=12$cm.
$△ABD=△BDC$ are congruent.
So, $\angle A=\angle C=\frac{{180}^0}{2}={90}^0$
$\therefore$ Area of quadrilateral $ABCD$
$=$Area of $△ABD+$ Area of $△CBD$
$=\frac{1}{2}\times 5\times 12+\frac{1}{2}\times 5\times 12$
$=60$ ${cm}^2$
$\because AB+AD=BC+CD$
So, incircle can be drawn.
Now, if radius of incircle is $r$, then
Area of quadrilateral$=r.s$
$\Rightarrow 60=r\left(\frac{5+5+12+12}{2}\right)=17r$
$\therefore 17r=60$ or $r=\frac{60}{17}$cm