Question

Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 9-inch perimeter A = sq. in.

Answer 1

Answer:

$3.9 in^2$

Step-by-step explanation:

The area of a triangle is given by

$A=\frac{1}{2}bh$

where

b is the base

h is the height

Here we have an equilateral triangle, which has the 3 sides of the same length.

Let's call L the length of one side.

We know that the perimeter of the triangle is

p = 9 in

The perimeter is the sum of the three sides, so:

$p=L+L+L=3L$

Therefore, we find the length of the side:

$L=\frac{p}{3}=\frac{9}{3}=3 in$

Therefore the length is the base of the triangle,

$b=L$

The height can be calculated by considering half triangle: the hypothenuse is equal to L, while one side is equal to half the base (b/2), therefore the height is given by Pythagorean's theorem:

$h=\sqrt{L^2-(\frac{b}{2})^2}=\sqrt{3^2-(\frac{3}{2})^2}=2.6 in$

Therefore, the area of the triangle is:

$A=\frac{1}{2}(3)(2.6)=3.9 in^2$