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.

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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
  • 1
    $\begingroup$ @Ahmed. OK , here we go. $\endgroup$ – Peter Szilas Apr 8 '17 at 2:33

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 $r$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.