SOLUTION: what is the area of the circle inscribed inside an equilateral triangle whose area is 4 square root of 3 square centimeters?

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Question 919088: what is the area of the circle inscribed inside an equilateral triangle whose area is 4 square root of 3 square centimeters?
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!

Draw the altitude AD, which divides the base BC into two equal parts,
BD = DC.



Since TABC is equilateral, AB = BC = 2·BD. 

By the Pythagorean theorem on right triangle BAD,

BDČ+ADČ=ABČ
BDČ+ADČ=(2·BD)Č
BDČ+ADČ=4·BDČ
    ADČ=4·BDČ-BDČ
    ADČ=3·BDČ
     AD=√3·BD



The Area of TABC = 1%2F2(BC)(AD)
                 = 1%2F2(2·BD)(√3·BD)
                 = 4√3·BDČ

We are told that the area of TABC is 4·√3
So we have the equation 

      4√3 = √3·BDČ

Divide both sides by √3

        4 = BDČ
        2 = BD

Draw AO, where O is the center of the circle.



OB bisects < ABC which is 60°, so < OBD is 30° and  BOD = 60°,
so TBOD is a 30°-60°-90° triangle and so OB = 2·OD

By the Pythagorean theorem applied to TBOD,

BDČ+ODČ=OBČ
 2Č+ODČ=(2·OD)Č
  4+ODČ=4·ODČ
  4+ODČ=4·ODČ
      4=3·ODČ
      4%2F3=ODČ
      
The area of a circle is given by

    Area = p·(radius)Č
         = p·ODČ               
         = p·4%2F3
         = 4pi%2F3

Edwin