Question

In a right angle triangle abc with vertex b being the right angle, the mutually perpendicular sides ab and bc are p cm and q cm long respectively. If the length of the hypotenuse is (p + q 6) cm, the radius of the largest possible circle that can be inscribed in the triangle is

Answer 1

Solution:

In Right Δ a b c

∠ a b c= 90°

a b = p cm

b c= q cm

Using Pythagoras theorem

(ab)²+ (b c)²= (ac)²

→(Hypotenuse)² = p² +q²

→→Hypotenuse=$\sqrt{p^2+q^2}$

Let the largest In circle of right angle triangle a b c has radius r.

OM=ON=r cm

→a M b, a H c and b N c are tangents to the circle.

→→Length of tangents from external point to a circle are equal.

→b M =b N

→b M=  r

→b N =  r

→a M=p -r

→c N = q -r

→a M= a H

→a H= p - r

→c N= c H

→c H = q -r

→a c = $\sqrt{p^2+q^2}$

→p - r + q -r= $\sqrt{p^2+q^2}$

→p +q - $\sqrt{p^2+q^2}$ = 2 r

→r = $r=\frac{p+q-\sqrt{p^2+q^2}}{2}$