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

Here, $AB=BC=5\mathrm{cm},CD=DA=12\mathrm{cm}$

$\therefore \mathrm{△}\mathrm{ABD}=\mathrm{△}\mathrm{BDC}$ are congruent 

So, $\mathrm{\angle }\mathrm{A}=\mathrm{\angle }\mathrm{C}=\frac{{180}^{\circ }}{2}={90}^{\circ }$

$\therefore$Area of quadrilateral ABCD = Area of $\mathrm{△}\mathrm{ABD}+\text{ Area of }\mathrm{\Delta CBD}$

$=\frac{1}{2}\cdot 5\cdot 12+\frac{1}{2}\cdot 5\cdot 12=60{\mathrm{cm}}^2$

$\because AB+CD=BC+AD.$ So, incircle can be drawn Now, if radius of incircle is r, then Area of quadrilateral = $r.s$

$\Rightarrow 60=r\left(\frac{5+5+12+12}{2}\right)=17r\Rightarrow \therefore r=\frac{60}{17}\mathrm{cm}$