Question

8. In a right triangle ABC, right angled at B, BC
= 5 cm and AB = 12 cm. The cirlce is
touching the sides of triangle ABC. Find the
radius of the circle.
​

Answer 1

Answer:

Step-by-step explanation:

Consider ABC be the right angled triangle such that ∠A = 90° and AB = 5cm, AC = 12 cm.  

And O be the centre and r be the radius of the incircle.

AB, BC and CA are tangents to the circle at P, N and M.

∴ OP = ON = OM =  r  (radius of the circle)

Area of ΔABC = ½ × 5 × 12 = 30 cm^2

By Pythagoras theorem,

 BC^2  = AC^2  + AB^2

 ⇒ BC^2  = 12^2  + 5^2

⇒ BC2  = 169

⇒ BC = 13 cm

Area of ∆ABC = Area ∆OAB + Area ∆OBC + Area ∆OCA

30 = 1 2 r × AB + 1 2 r × BC + 1 2 r × CA  

30 = 1 2 r(AB+BC+CA)

⇒ r = 2 × 30 (AB+BC+CA)  

⇒ r = 60 5+13+12  

⇒ r = 60/30 = 2 cm.