Question

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Find the area of an equilateral triangle (regular 3-gon) with the given measurement.

9-inch perimeter

A = sq. in.

Answer 1

An equilateral triangle has all sides congruent, denote them to be a inches. If the perimeter of this triangle is 9 inches, then

$P=3a=9\Rightarrow a=3\ \text{inches}.$

The area of the equialteral triangle is

$A=\dfrac{a^2\sqrt{3}}{4}=\dfrac{3^2\sqrt{3}}{4}=\dfrac{9\sqrt{3}}{4}\ \text{sq. in}.$

Answer: $\dfrac{9\sqrt{3}}{4}$ sq. in.

Answer 2

The area of the equilateral triangle whose perimeter is 9 inches is 3.897 inches².

What is an equilateral triangle?

An equilateral triangle is a type of triangle whose length of all sides is equal.

$\text{Perimeter of the triangle} = 3a$

$\text{Area of equilateral triangle} = \dfrac{\sqrt3}{4}a^2\\\\\text{where a is one of the sides of the triangle}$

The perimeter of the triangle is the sum of three sides now since the length of the three sides of the triangle is equal. Therefore,

The perimeter of the triangle = 3a

where a is one of the sides of the triangle.

Now, if find the three sides of the triangle,

$\text{Perimeter of the triangle} = 3a\\9=3a\\a=3$

We know that the area of an equilateral triangle is given as,

$\text{Area of equilateral triangle} = \dfrac{\sqrt3}{4}a^2$

                                          $= \dfrac{\sqrt3}{4}\times (3)^2\\\\= \dfrac{\sqrt3}{4}\times 9\\\\= 3.897\rm\ in^2$

Hence, the area of the equilateral triangle whose perimeter is 9 inches is 3.897 inches².

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