# Triangle

• Oct 24th 2006, 02:56 PM
Rimas
Triangle
If two equilateral triangles of area A intersect to form a regular hexagon then what is the area of the hexagon?
• Oct 24th 2006, 09:35 PM
Soroban
Hello, Rimas!

An interesting problem . . . Did you make a sketch?

Quote:

If two equilateral triangles of area A intersect to form a regular hexagon,
then what is the area of the hexagon?

Code:

                *               / \           *---*---*---*           \ /o\o/o\ /             * - * - *           / \o/o\o/ \           *---*---*---*               \ /                 *

Each equilateral triangle is comprised of nine smaller triangles.
They overlap in a hexagon comprised of six triangles.

The hexagon has an area which is $\displaystyle \frac{6}{9} = \frac{2}{3}$ of an equilateral triangle.

Area of hexagon: .$\displaystyle \frac{2}{3}A$

• Oct 27th 2006, 08:09 AM
ceasar_19134
Just looking around, but shouldnt the Area of the hexagon only equal 1/2 of the figure?

The hexagon is composed of 6 triangles, while the whole figure is composed of 12.
• Oct 27th 2006, 08:48 AM
CaptainBlack
Quote:

Originally Posted by ceasar_19134
Just looking around, but shouldnt the Area of the hexagon only equal 1/2 of the figure?

The hexagon is composed of 6 triangles, while the whole figure is composed of 12.

The area of the hexagon is given in terms of the area of one of the
equilateral triangles, each of which is 9 of the small triangles compared to
six which comprise the hexagon.

RonL