Circle gets cozy in triangle

We know that incenter is the point of intersection of the angular bisectors. So, $\text{D}$ is the incenter of $\triangle{\text{ABC}}$ Now, since $\text{AFDE}$ is a rectangle, $\angle{\text{AEB}} = {90}^{\circ}$ . So, $\text{DE}$ is the inradius.

Using pythagoras' theorem:-
$\begin{aligned} {\text{AD}}^2 & = {\text{AE}}^2 + {\text{DE}}^2\\ \text{DE} & = \sqrt{{10}^2 - 8^2}\\ \text{DE} & = \sqrt{36}\\ \text{DE} & = 6 \end{aligned}$

So, area of incircle $= \pi r^2\\ = \dfrac{22}{7} × 6^2\\ = 113.142\\ \approx \color{#69047E}{\boxed{113 {\text{cm}}^{2}}}$

Since the angle bisectors meet at pint D, then by definition, D is the inradius if the incircle. The line De is a radius and length of sqrt(100 - 64) by Pythagoras. Therefore the area is 36*pi , or approximately 113. Ed Gray