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/

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.