Equilateral Triangle

• Nov 20th 2008, 05:53 AM
magentarita
Equilateral Triangle
Find the perimeter of the equilateral triangle inscribed in a circle of radius 20.0 inches
• Nov 20th 2008, 06:31 AM
Soroban
Hello, magentarita!

Quote:

Find the perimeter of the equilateral triangle inscribed in a circle of radius 20 inches
Code:

                A               * * *           *    /|\    *         *    / | \    *       *    /  |  \    *             / 20|  \  D       *    /    |    *    *       *  /    *    \  *       *  /      O      \  *         /              \     B *- - - - - - - - -* C         *              *           *          *               * * *

The triangle is $\displaystyle ABC$
The circle has center $\displaystyle O$ and radius $\displaystyle OA = 20$
Draw $\displaystyle OD \perp AC.$

Since $\displaystyle \Delta ABC$ is equilateral, $\displaystyle \angle A \:=\:\angle B \:=\:\angle C \:=\:60^o$
. . Then: $\displaystyle \angle OAD = 30^o$

In right triangle $\displaystyle ADO\!:\;\;\cos30^o \:=\:\frac{AD}{20} \quad\Rightarrow\quad AD \:=\:20\cos30^o \:\approx\:17.32$ inches

The side of the triangle is: .$\displaystyle 2\times 17.32 \:=\:34.64$ inches

Therefore, the perimeter is: .$\displaystyle 3 \times 34.64 \:=\:103.92$ inches.

• Nov 20th 2008, 02:13 PM
magentarita
wonderfully done.........
Quote:

Originally Posted by Soroban
Hello, magentarita!

Code:

                A               * * *           *    /|\    *         *    / | \    *       *    /  |  \    *             / 20|  \  D       *    /    |    *    *       *  /    *    \  *       *  /      O      \  *         /              \     B *- - - - - - - - -* C         *              *           *          *               * * *
The triangle is $\displaystyle ABC$
The circle has center $\displaystyle O$ and radius $\displaystyle OA = 20$
Draw $\displaystyle OD \perp AC.$

Since $\displaystyle \Delta ABC$ is equilateral, $\displaystyle \angle A \:=\:\angle B \:=\:\angle C \:=\:60^o$
. . Then: $\displaystyle \angle OAD = 30^o$

In right triangle $\displaystyle ADO\!:\;\;\cos30^o \:=\:\frac{AD}{20} \quad\Rightarrow\quad AD \:=\:20\cos30^o \:\approx\:17.32$ inches

The side of the triangle is: .$\displaystyle 2\times 17.32 \:=\:34.64$ inches

Therefore, the perimeter is: .$\displaystyle 3 \times 34.64 \:=\:103.92$ inches.

It's amazing how you used trig to answer a geometry question.