Subtopic: Inscribed circle in right-angled triangle
Question Statement:
For a right-angled triangle with perpendicular sides $a$ and $b$ and hypotenuse $c$, the inradius $r$ of the inscribed circle is given by:
Options:
(A) $r = \frac{a + b - c}{2}$
(B) $r = \frac{a + b + c}{2}$
(C) $r = \frac{a b}{c}$
(D) $r = \sqrt{a b}$
Answer Key:
A
Step-by-Step Solution & Explanation
In any triangle, inradius $r = \frac{\text{Area}}{s}$, where semi-perimeter $s = \frac{a + b + c}{2}$. For a right triangle, $\text{Area} = \frac{a b}{2}$. Since $c^2 = a^2 + b^2$, $(a + b)^2 - c^2 = 2 a b = 4 \text{Area}$. Thus:
$r = \frac{a b}{a + b + c} = \frac{a + b - c}{2}$.