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10-10-2013, 11:35 AM
In the figure below the triangle DGF has an area of 12 and DG=GF, what is the area of the shaded region?


Figure not drawn to scale

(A) 6\sqrt{3})\frac{\pi}{3}-12

(B) 4\sqrt{3})\frac{\pi}{3}-12

(C) 10\sqrt{3})\frac{\pi}{3}-12

(D) 8\sqrt{3})\frac{\pi}{3}-12


10-10-2013, 11:37 AM
The triangle DGF is equilateral since DG=DF as radii and we also know that DG=GF.
Therefore, each angle of the triangle DGF measures 60^o.
We’ ll use two formulas for areas.

The first one gives us the area of triangle in terms of any two sides a,b and their included angle \hat{C}

A_{T}=\frac{1}{2}ab Sin\hat{C} =\frac{1}{2}r^2 Sin(\frac{\pi}{3}) =12\Rightarrow

\frac{1}{2}r^2 \frac{\sqrt{3}}{2}=12 \Rightarrow r^2 =\frac{48}{\sqrt{3}}\Rightarrow r^2 =\frac{48\sqrt{3}}{3}\Rightarrow r^2 =16\sqrt{3}

and the other gives us the area of a sector in terms of radius r and the central angle \theta (in radians!!) of the sector


So, the area of the shaded region A_{R} is given by the following expression

A_{R}= A_{S} - A_{T}= \frac{1}{2}r^2\frac{\pi}{3}-12 =\frac{1}{2}(16\sqrt{3})\frac{\pi}{3}-12=8\sqrt{3})\frac{\pi}{3}-12

Therefore the correct answer is (D).

Hope these help!

10-10-2013, 11:40 AM
Thanks miranda!!