Question

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

Answer 1

`A=27 sqrt(3) approx 46.77` inches.

Explanation:

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

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.