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 = (BC)(AD) = (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 =ODČ The area of a circle is given by Area = p·(radius)Č = p·ODČ = p· = Edwin