# Find the Radius of the Circle Inside a Right Triangle

# Geometry math problem: When two equal circles are inscribed inside a right triangle, which is tangential to two sides of the triangle then find the radius of the circles

ABC is a right angle triangle, **AB = 12 cm** and **AC = 9 cm**. find the radius of the circle when the radius of the circles are equal

## Solution: Radius of the circle

We can solve this problem in different ways, one of the methods is shown here

This figure can be redrawn as shown below.

Let assume Radius of the circle =* r * and BQ = BR = *x* {BQ & BR are tangent of the circle from point B}

AB = AP + PQ + QB = *r *+ 2*r *+* x *= 3*r* +* x*

⇒ 3*r *+* x *=* *12……………………………………….*eq*(1)

From above figure

Let ∠QBR = *θ * then ∠QOR = 180 – *θ*

## Apply Pythagorean theorem in triangle ABC

BC^{2} = AB^{2} + AC^{2} = 12^{2} + 9^{2} = 144 + 81 = 225

⇒ BC = 15 cm

cos *θ* = AB / BC = 12 / 15 = 4/5

From triangle BQR

QR^{2} = BQ^{2} + BR^{2} – 2 × BQ × BR × cos *θ*

⇒ QR^{2} = *x ^{2} *+

*x*– 2 ×

^{2}*x*×

*x*× 4/5

⇒ QR^{2} =* *2*x ^{2} – *8

*/5*

*x*^{2}⇒ QR^{2} = 2*x ^{2}*/5

From triangle QBR

QR^{2} = OQ^{2} + OR^{2} – 2 × OQ × OR × cos (180-*θ*)

⇒ QR^{2} = *r*^{2} + *r*^{2} +* *2 ×* r *×* r *× 4/5 {cos (180-*θ*) = -cos *θ*}

⇒ QR^{2} = 2*r*^{2} + 8*r*^{2}/5

⇒ QR^{2} = 18*r*^{2}/5

⇒ QR^{2} =* *2*x*^{2}/5 *= *18*r*^{2}/5

⇒ *x*^{2} = 9*r*^{2}

⇒ *x ^{ }*= 3

*r*

From equation 1

3*r *+* x* = 12 *=* 3*r *+ 3*r *=* *6*r*

**⇒** *r *= 2 cm