Question

In an equilateral triangle of side 24 cm, a circle is inscribed touching it's sides. find the area of remaining portion of the triangle

Answer 1

Solution:

Given that side of an Equilateral triangle is 24 cm

we know that In-radius of an equilateral triangle =  ( side/2√3) = 24/2√3 = 4√3 cm

Now , we know that area of an equilateral triangle = √3/4 (side)²

                                         = √3/4 ×(s)² = √3×24×24 =  114√3 cm         =144×1.732 = 249.408 cm²

now , we will find the area of in-circle.

we know that area of circle = πr² = 22/7 ×4√3×4√3 = 22×16 ×3/7 = 1056/7         =1150.86 cm²

thus, Area of remaining part = area of triangle - area of inscribed circle

                                                   = 249.408 - 150.86 = 98.54 cm²

Hence the remaining part = 98.54 cm²

Answer 2

Let r is radius of inscribed circle touching sides of given equilateral triangle.

Let ABC is given equilateral triangle, and O is the incentre of triangle.

then, area of equilateral triangle ABC = area of ∆AOB + area of ∆BOC + area of ∆COA.

or, √3/4 × (side)² = 1/2 × r × AB + 1/2 × r × BC + 1/2 × r × CA

or, √3/4 × (24)² = 1/2 × r(AB + BC + CA)

or, √3/4 × 24 × 24 = 1/2 × r (24 + 24 + 24)

or, 144√3 = 36r

or, r = 4√3 = 4√3 cm

so, area of inscribed circle = πr²

= π × (4√3)² = 48π cm²

area of equilateral triangle = √3/4 × 24²

= 144√3 cm²

area of remaining portion of the triangle = 144√3 cm² - 48π cm²

= 98.69 cm²