Bunuel wrote:
If the circle above has a radius of 4, what is the perimeter of the inscribed equilateral triangle?
A. \(6\sqrt{2}\)
B. \(6\sqrt{3}\)
C. \(12\sqrt{2}\)
D. \(12\sqrt{3}\)
E. 24
Attachment:
2018-01-18_1405.png
If we draw a perpendicular from the centre to all the side and also join all the vertices to the centre we will get 6 right angled triangle which is 30-90-60
hypotenuse=radius=4
and side opp 60 degree=\sqrt{3}/2 *hypotenuse=2\sqrt{3}
hence one side is 4\sqrt{3} and perimeter 3* 4\sqrt{3}=12\sqrt{3}