Question

What is the area of an equilateral triangle with sides 8 inches long?

Answer 1

We just apply the formula `A= \sqrt{3}/4 s^2` and get `16 sqrt{3}` square inches.

Explanation:

Let's derive the formula. The altitude `h` divides an equilateral triangle of side `s` into two 30,60,90 triangles. The Pythagorean Theorem says

`s^2 = h^2 + (s/2)^2`

`s^2 - s^4/4 = h^2`

`3/4 s^2 = h^2`

`h = \sqrt{3}/2 s`

The area is

`A = 1/2 s h = sqrt{3}/4 s^2 quad quad square`

Let's check this with my favorite theorem, called Archimedes' Theorem, which, like Heron's formula, relates the area of a triangle to its sides:

`16A^2 = 4a^2 b^2 - (c^2-a^2-b^2)^2`

Let's plug in `a=b=c.`

`16A^2 = 4 a^4 - a^4 = 3 a^4`

`A^2 = 3/16 a^4`

`A = \sqrt{3}/4 a^2 quad sqrt`

Answer 2

`A=16sqrt3` `"in".^2`

Explanation:

.

In an equilateral triangle, the altitude divides the base into two equal halves. As shown above, if the measure of the side is `s` then in the right angle triangle that drawing the altitude generates, the small leg is `s/2`.

We can use the Pythagorean formula to get the measure of the long leg (altitude):

`h^2+(s/2)^2=s^2`

`h^2=s^2-s^2/4=(4s^2-s^2)/4=(3s^2)/4`

`h=(ssqrt3)/2`

In our problem, `s=8` in.

`h=(8sqrt3)/2=4sqrt3`

The formula for the area of a triangle is:

`A_("Triangle")=1/2bh` where `b` is the base and `h` is the altitude.

`A=1/2(8)(4sqrt3)=16sqrt3` `"in".^2`