Question

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Find the area of an equilateral triangle (regular 3-gon) with the given measurement. 9-inch perimeter A = sq. in.

Answer 1

9√3/4

or

9/4√3

Explanation:

P=3a=9⇒a=3 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}.A=4a23=4323=493 sq. in.

Answer 2

Aswer: 9 / 4 sqrt 3

Solution:

Perimeter: P=9 inch

Area: A=?

A=(1/4) sqrt(3) L^2

Side of the triangle: L

We need the side "L" of the triangle:

We know: P=3 L

Replacing the perimeter in the equation above:

9 inch=3 L

Solving for L: Dividing both sides of the equation above by 3:

(9 inch) / 3 = 3 L /3

3 inch = L

L = 3 inch

Replacing L=3 inch in the formula of Area:

A=(1/4) sqrt(3) L^2

A=(1/4) sqrt(3) (3 inch)^2

Squaring:

A=(1/4) sqrt(3) (9 inch^2)

Multiplying:

A=(1*9)/4 sqrt(3) inch^2

A=9/4 sqrt(3) inch^2