Question 14.

We know that inradius

$r = \dfrac{\text{Perpendicular} + \text{Base} - \text{Hypotenuse}}{2}$

So,

$4 = \dfrac{p + 10 - h}{2}$

$h - p = 2$ or $h = p + 2$

Now,

$p^2 + 100 = h^2$

$p^2 + 100 = (p + 2)^2$

$p^2 + 100 = p^2 + 4p + 4$

$4p = 96$

$p = 24$

Hence, area

$= \dfrac{1}{2} \times 10 \times 24 = 120$