# How do you find the area of an equilateral triangle that has a perimeter of 30 inches?

Dec 1, 2015

$25 \sqrt{3}$ ${\text{in}}^{2}$

#### Explanation:

We can see that if we split an equilateral triangle in half, we are left with two congruent right triangles. Thus, one of the legs of one of the right triangles is $\frac{1}{2} s$, and the hypotenuse is $s$. We can use the Pythagorean Theorem or the properties of 30˚-60˚-90˚ triangles to determine that the height of the triangle is $\frac{\sqrt{3}}{2} s$.

If we want to determine the area of the entire triangle, we know that $A = \frac{1}{2} b h$. We also know that the base is $s$ and the height is $\frac{\sqrt{3}}{2} s$, so we can plug those in to the area equation to see the following for an equilateral triangle:

$A = \frac{1}{2} b h \implies \frac{1}{2} \left(s\right) \left(\frac{\sqrt{3}}{2} s\right) = \frac{{s}^{2} \sqrt{3}}{4}$

In your case, the perimeter of the triangle is $30$, so the length of each side is $10$, since all $3$ sides are congruent.

We can plug in $10$ for $s$:

$\frac{{10}^{2} \sqrt{3}}{4} = \frac{100 \sqrt{3}}{4} = 25 \sqrt{3}$ ${\text{in}}^{2}$