OE


Think of this star as a large equilateral triangle with sides 12 centimeters long, and three additional smaller equilateral triangles (shaded in the figure below) with sides 4 inches long. Using the same 30–60–90 logic you applied in problem 9, you can see that the height of the larger equilateral triangle is

$6 \sqrt{3}$


and the height of the smaller equilateral triangle is $2 \sqrt{3}$


The area of the large triangle is $36 \sqrt{3}$

The small ones are $4 \sqrt{3}$

The sum of both is $48 \sqrt{3}$