Hello, mathaddict!

I made a sketch . . . and the answer jumped out at me!

An equilateral triangle is inscribed in a regular hexagon.

This triangle's vertices touch the midpoints of three sides of the regular hexagon.

The area of the triangle is 12 cm².

What is the area of the regular hexagon in cm² ?

The diagram looks like this: Code:

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

Divide the hexagon into "unit triangles",

. . and we have this diagram: Code:

*---*---*
/ \ /x\ / \
*---*---*---*
/ \ /x\x/x\ / \
*---*---*---*---*
\ /x\x/x\x/x\ /
*---*---*---*
\ / \ / \ /
*---*---*

We see that the triangle occupies $\displaystyle \frac{9}{24} = \frac{3}{8}$ of the hexagon.

Therefore, the hexagon has: .$\displaystyle \frac{8}{3}\times12 \:=\:32\text{ cm}^2$