In \(\triangle ABC\), if \(\angle B=90^{\circ}\), BC = 8 cm and AB = 15 cm, then the radius r of the incircle of \(\triangle ABC\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3 cm
The radius of the incircle of any triangle is its area divided by its semi-perimeter. In a right triangle the two sides at the right angle serve as base and height, and the third side comes from Pythagoras’ theorem.
Hypotenuse, by Pythagoras’ theorem: \(AC=\sqrt{15^2+8^2}=\sqrt{289}=17\) cm
Area, half the product of the two legs: \(\displaystyle \frac{1}{2}\times AB\times BC=\frac{1}{2}\times15\times8=60\) cm2
Semi-perimeter, half the sum of the sides: \(\displaystyle s=\frac{15+8+17}{2}=20\) cm
Radius of the incircle: \(\displaystyle r=\frac{\text{Area}}{s}=\frac{60}{20}=3\) cm
Shortcut: In a right triangle, the two tangents from the right-angle corner are each equal to \(r\) (with the radii they form a square), so the two legs minus the hypotenuse leave \(2r\), as the picture shows.
Radius from the legs and the hypotenuse: \(\displaystyle r=\frac{AB+BC-AC}{2}=\frac{15+8-17}{2}=3\) cm
