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circle with triangle inscribed in it

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.

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  • $\begingroup$ Is the triangle equilateral? $\endgroup$ – Ahmed S. Attaalla Apr 7 '17 at 3:06
  • $\begingroup$ Yes it is equilateral. $\endgroup$ – Justin Lam Apr 7 '17 at 3:10
  • $\begingroup$ 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. $\endgroup$ – Arby Apr 7 '17 at 3:11
  • $\begingroup$ Lookup circular segment, and think what all that comes down to for an equilateral triangle. $\endgroup$ – dxiv Apr 7 '17 at 3:13
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We're looking for the radius right? So let's draw them..

enter image description here

Please excuse the drawing.

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.

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  • $\begingroup$ Thanks for the help! I got about 1.15 feet squared as the radius. Now I can do the rest. $\endgroup$ – Justin Lam Apr 7 '17 at 3:22
  • $\begingroup$ My final answer is about 2.5 feet squared. Is this correct? $\endgroup$ – Justin Lam Apr 7 '17 at 3:29
  • $\begingroup$ You're welcome. That's about right. The exact answer is $\frac{4}{3}\pi-\sqrt{3}$ feet squared. @JustinLam $\endgroup$ – Ahmed S. Attaalla Apr 7 '17 at 3:44
  • $\begingroup$ Ahmed's drawing: cos(30°) = 1/r. Hence r = 1/cos(30°). Using cos(30°) = (1/2)×(√3) we get r. $\endgroup$ – Peter Szilas Apr 7 '17 at 6:04
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    $\begingroup$ @Ahmed. OK , here we go. $\endgroup$ – Peter Szilas Apr 8 '17 at 2:33
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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 $r$.

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