Question

The area of the incircle of an equilateral triangle of side 42 cm is

Options

• \[22\sqrt{3} c m^2\]

•  231 cm²

• 462 cm²

• 924 cm²

Answer 1

Let ABC be the equilateral triangle such that AB = BC = CA = 42 cm. Also, let O be the centre and r be the radius of its incircle.

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

`∴ OM=ON=OP=r`

Area of ΔABC = Area (ΔOAB) + Area (ΔOBC) + Area (ΔOCA) 

`⇒ sqrt3/4(42)^2=1/2xxrxxAB+1/2xxrxxBC+1/2xxrxx CA`

`⇒ sqrt3/4xx42xx42=1/2 r(AB+BC+CA)`

`⇒ 441sqrt3=1/2xxrxx(42+42+42)`

`⇒441sqrt3=1/2xxrxx(126)`

`⇒441sqrt3=63r`

`⇒r=(441sqrt3)/63`

`⇒ r=7sqrt3 cm`

Area of the circle `= pir^2=22/7 (7sqrt3)^2=22/7xx147=462 cm^2`