# How do you find the area of the shaded area of this circle?

I need to find the area of the shaded area. The triangle is equilateral. So far, I have found the area of the triangle to be $\sqrt 3$, but I cannot figure out how to find the radius of the circle in order to find the area of the circle. Any advice would be appreciated.

• Is the triangle equilateral? Apr 7 '17 at 3:06
• Yes it is equilateral. Apr 7 '17 at 3:10
• Do you know how to find the areas of circles and triangles? Oops, sorry, I see that you do. My bad. Try drawing in a radius and looking for relationships. Hint: Choose your radius wisely.
– Arby
Apr 7 '17 at 3:11
• Lookup circular segment, and think what all that comes down to for an equilateral triangle.
– dxiv
Apr 7 '17 at 3:13

Let's label Ahmed's drawing: Triangle $ABC$, lower left $A$, then counterclockwise $B$, and $C$ (top). Let the center of the circle be $M$. Extend $CM$ to intersect $AB$ in $D$. Note length $AD$ $=$ length $DB$ $=1$, $MD$ being the perpendicular bisector of $AB$. Triangle $ADM$ is a right angled triangle. Angle $MAD = 30°$.

$$\cos (30°) = \frac{1}{r}$$

$$r = \frac{1}{\cos (30°)}$$

Using $\cos (30°) = \frac{\sqrt{3}}{2}$ we get $r$.

We're looking for the radius right? So let's draw them..

Ok. Now we have an isosceles triangle $30-30-120$. If $r$ is the radius then law of sines tells us,
$$\frac{2}{\sin 120}=\frac{r}{\sin 30}$$
So $r=2 \frac{\sin 30}{\sin 120}=2\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$. I think you can take it from here.
• You're welcome. That's about right. The exact answer is $\frac{4}{3}\pi-\sqrt{3}$ feet squared. @JustinLam Apr 7 '17 at 3:44