1. Equilateral Triangle

Find the perimeter of the equilateral triangle inscribed in a circle of radius 20.0 inches

2. Hello, magentarita!

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.

3. wonderfully done.........

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.