Circle in a trapezium

Since it is a trapezium, $\angle A=\angle B, \angle D= \angle C, AD=BC$ . It follows that $DF=FC=4$ and $AE=EB=9$ . Tangents to a circle drawn from an external point are equal, therefore, $BH=EB=9$ and $FC=CH=4$ . Thus, $BC=9+4=13$ . It follows that $AD=BC=13$ . From my figure, $FE=DG=diameter~of~the~circle$ .

Consider $\triangle DGA:$ Since $GE=DF=4$ , $AG=9-4=5$

Applying pythagorean theorem on $\triangle DGA$ , we get

$DG=\sqrt{13^2-5^2}=\sqrt{144}=12$

Hence, the radius is $\dfrac{DG}{2}=\dfrac{12}{2}=$ $\boxed {6}$

According to Pitot's Theorem we conclude black sides are equal to $13$ , we can clearly see that the circle diameter is the height of the trapezium so we apply Pythagorean Theorem

$25 + h^2 = 169$

$\therefore$

$h=12$

Radius is equal to $6$