What is the area of an equilateral triangle with height of 9 inches?

1 Answer

A=27 sqrt(3) approx 46.77 inches.

Explanation:

In such situations, the first step is to draw a picture.

http://www.mathportal.org/http://www.mathportal.org/

In relation to the notation introduced by the picture, we know that h=9 inches.

Knowing that the triangle is equilateral makes everything easier: the heights are also medians. So the height h is perpendicular to the side AB and it divides it in two halves, which are a/2 long.

Then, the triangle is divided into two congruent right triangles and the Pythagorean Theorem holds for one of these two right triangles: a^2=h^2+(a/2)^2. So 3/4a^2=h^2 i.e. a^2=4/3 h^2. In the end, we get that the side is given by a=[2sqrt(3)]/3 h=[2sqrt(3)]/3 * 9=6 sqrt(3) approx 10.39 inches.

Now the area:
A=(a*h)/2=([2sqrt(3)]/3 h * h)/2=[sqrt(3)]/3 h^2=[sqrt(3)]/3 81=27 sqrt(3) approx 46.77 inches.