Question -
Answer -
In right ∆ABC, ∠B = 90°
BC = 6 cm, AB = 8 cm
Let r be the radius of incircle whose centre is O and touches the sides A B, BC and CA at P, Q and R respectively
AP and AR are the tangents to the circle AP = AR
Similarly CR = CQ and BQ = BP
OP and OQ are radii of the circle
OP ⊥ AB and OQ ⊥ BC and ∠B = 90° (given)
BPOQ is a square
BP = BQ = r
AR = AP = AB – BD = 8 – r
and CR = CQ = BC – BQ = 6 – r
But AC² = AB² + BC² (Pythagoras Theorem)
= (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
=> AR + CR = 10
=> 8 – r + 6 – r = 10
=> 14 – 2r = 10
=> 2r = 14 – 10 = 4
=> r = 2
Radius of the incircle = 2 cm