Question

In the figure given below, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle.

Answer 1

In right triangle BAC,
BC^2= AB^2+ AC^2 [By Pythagoras theorem]
BC = $\sqrt{6^2+8^2}$
BC = $\sqrt{36+64}$
BC = $\sqrt{100}$
Hence, BC = 10cm
Semi-perimeter, s = $\frac{(a+b+c)}{2}$
s = $\frac{(6+8+10)}{2}$ = 12cm
Area of right triangle ABC = $\frac{1}{2}$ x AB x AC
= $\frac{1}{2}$ x 6 x 8 = 24 sq cm
We know that, area of triangle, Δ = r x s
That is, 24 = r x 12
$\therefore$ r = 2cm