# Equilateral triangle inscribed in a circle

• Mar 18th 2008, 04:14 AM
struck
Equilateral triangle inscribed in a circle
An equilateral triangle with sides of 8cm length each, is inscribed in a circle. Calculate the diameter of the circle.

I suspect the theorem of intersecting chords may come into play here. However, I am unable to make out how. Btw, I can't use trigonometry here.

Thanks for the help!
• Mar 18th 2008, 04:32 AM
colby2152
Quote:

Originally Posted by struck
Btw, I can't use trigonometry here.

Then I cannot help you. I would use trig to disect the height of the triangle each of the three vertices, and then figure out the distance to the point in the middle.

I'm no pro at Geometry, but I am not sure how the Intersecting Chords Theorem will help you?
• Mar 18th 2008, 06:29 AM
Soroban
Hello, struck!

Quote:

An equilateral triangle with sides of 8cm length each,
is inscribed in a circle. Calculate the diameter of the circle.

Code:

                A               * * *           *    /|\    *         *    / | \    *       *    /  |  \    *             /  |  \ 8       *    /    |    \    *       *  /    O*  r \  *       *  /      |  *  \  *         /      |    * \       B*- - - - + - - - -*C         *      D  4  *           *          *               * * *

Equilateral $\Delta ABC\!:\;\;AB \,=\,BC\,=\,CA\,=\,8$

It is inscribed in circle $O$ with radius $r\!:\;OC \,=\,r$

In right triangle $ADC\!:\;\;AD^2 + 4^2 \:=\:8^2\quad\Rightarrow\quad AD \:=\:4\sqrt{3}$

The medians, altitudes and angle bisectors intersect at $O$,
. . the centroid, which divides the altitude in the ratio 1:2.
Hence: . $OD \:=\:\frac{1}{3}\left(4\sqrt{3}\right) \:=\:\frac{4\sqrt{3}}{3}$

In right triangle $ODC\!:\;r^2 \:=\:CD^2 + OD^2 \;=\;4^2 + \left(\frac{4\sqrt{3}}{3}\right)^2$

[Remember: they asked for the diameter.]

• Mar 18th 2008, 06:40 AM
galactus
The area of a circumscribed circle about an equilateral triangle is given by $\frac{\pi}{3}r^{2}$. Where r is a length of a side of the triangle.

In this case 8. So, we have $\frac{\pi}{3}(64)=\frac{64\pi}{3}$ as the area of the circle.

${\pi}R^{2}=\frac{64\pi}{3}$

$R=\frac{8}{\sqrt{3}}$ is the radius of the circle.

Of course, the diameter is twice that.