Question

In the given figure, $△ABC$ is right-angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of $△ABC$. [Take $\pi$ = 3.14 ]

Answer 1

Sol: ABC is a right angled triangle at right angled at A.
BC = 10 cm, AB = 6 cm.

Let ‘O’ be the centre and r be the radius of the incircle.
AB, BC and CA are tangents to the circle at P, M and N.
∴ IP = IM = IN = r (radius of the circle)

In right ΔBAC,

BC^2 = AB^2+ AC^2 [Pythagoras theorem,]
⇒ 10^2 = 6^2+ AC^2
⇒ AC^2 = 100 - 36
⇒ AC^2 = 64
⇒ AC = √64 = 8 cm

Area of ∆ ABC = 1/2 × base × height

=1/2×AC×AB
= 1/2 × 8 × 6 = 24 cm^2

Area of ∆ABC = Area ∆IAB + Area ∆IBC + Area ∆ICA
⇒ 24 = 1/2 r (AB) + 1/2 r (BC) + 1/2 r (CA)
24= 1/2 r (AB + BC + CA)
24= 1/2 r (6 + 8 + 10)
24= 12 r = r = 2

Area of the circle = πr^2 = 3.14 x 2 x 2 = 12.56 cm2

Area of shaded region = Area of ∆ABC - Area of circle = 24 -12.56 = 11.44 cm^2. Hence, the area of the shaded region is 11.44 cm^2.