Area of Right Triangle with Inscribed Circle

Inradius $r = \frac{\text{diameter}}{2} = 4$. For a right-angled triangle with legs $a = BC = 10$ and $b = AB$, and hypotenuse $c = AC$: $r = \frac{a + b - c}{2} \implies 4 = \frac{10 + b - c}{2} \implies c - b = 2$

Also, by Pythagoras theorem: $c^2 - b^2 = a^2 = 100 \implies (c - b)(c + b) = 100$ Since $c - b = 2$, we get $c + b = 50$. Adding $c - b = 2$ and $c + b = 50$ gives $2b = 48 \implies b = 24$.

Area of right triangle $\Delta ABC = \frac{1}{2} \times a \times b = \frac{1}{2} \times 10 \times 24 = 120\text{ square inches}$.