Question

A circle has a radius of 6 in. Find the area of its circumscribed equilateral triangles.

The circumscribed equilateral triangle will have an area of:

Answer 1

Extend the radius 6" through centre of circle O down to base of triangle ABC.
Then the height of triangle AH = AO+OH where point H is on base BC.
Since the radii AO, BO and CO are also angle-bisectors of angles A, B & C.
We have a right-angle triangle OHC with angle OCH=30 degrees.
OH=6 sin30=3" whereas CH = 6 cos30=5.196".
Thus the height AH=6+3=9" and the base = 2 X 5.196= 10.39"\
Thus the area of triangle= (height x base)/2 --> (9 x 10.39)
=46.8 cm square cm