What is the area of an equilateral triangle who's inscribed circle has a radius equal to 2?

Question

What is the area of an equilateral triangle who's inscribed circle has a radius equal to 2?

Answer 1

Draw and label of diagram.

The base of the triangle is AB.

The center of the circle is O.

Draw a perpendicular from O to AB, label C

ΔOBC is a 30-6-90 triangle.

OC is the short let. mOC = 2, therefore mCB = 2√3

Area of Δ AOB = 4√3.

Three such triangle are in the equilateral triangle.

Are of equilateral triangle equals 12√3.

Answer 2

This problem can be solved in two steps.

 

First:    One side of the inscribed triangle and two radii of the circle drawn to the end points of this side constitute an isosceles triangle with a vertex angle of 120 degree.    The law of cosines can be used on this triangle to get the side, s, of the equilateral triangle because the other two sides are know to have length = 2 (the radius of circle).  

   

   s² = 2² + 2² -  2 x 2 x 2 x cos(120)   .  Since cos(120) = -1/2,  

  

    s² =  12   .

 

Second:    The area of an equilateral triangle  is   A = ( sqrt(3) /4 ) s²  

   So   A = 3  sqrt(3)

 

  The formula   A =  (sqrt(3) /4 ) s²  can be derived in several ways.  The easiest is via Heron's formula.