This relies on knowing your special triangles, which include 30-60-90, 45-45-90, and 60-60-60.
If it helps, redraw the triangle so FG is at the bottom. For a 30-60-90, if the short side FG is 4, the hypoteneuse FH is 8. The radius would then be 4, and the area would be pi*r^2 or 16*pi.
If you forget the dimensions of 30-60-90 triangles, imagine doubling this triangle by reflecting FGH along the hypoteneuse FH. Then you would have an equilateral triangle, and each side would be 8.
Carcass wrote:
Attachment:
GRE In the figure above, FG = 4, and FH.png
In the figure above, FG = 4, and FH is a diameter of the circle. What is the area of the circle?
A. \(4 \pi\)
B. \(8 \pi\)
C. \(12 \pi\)
D. \(16 \pi\)
E. \(20 \pi\)