What is the approximate area of a regular pentagon with a side lenghtt of 6 feet and a distance from the center to a vertex of 6 feet

A regular pentagon may be divided into 5 congruent triangles.

If the side of the pentagon is 6 feet, each base of the triangles will be 6 feet.

If the distance from the center to a vertex is 6 feet, all the sides of the triangles will 6 feet.

So there are 5 congruent, equilateral triangles each with side equal to 6 feet.

The area one of those triangles is calculateb by:

base * height / 2

height ^2 = (side length) ^2 - (side length /2)^2 = 6^2 - 3^2 = 36 - 9 = 27

=> height = √(27) ≈ 5.2 feet

And, using area = base * height / 2 ,

area of a triangle = 6 feet * 5.2 feet / 2 = 15.6 feet^2

And the area of the entire pentagon equals 5 triangles ≈ 5 * 15.6 feet^2 = 78 feet^2

78 feet^2

Ibelieve y y even though i've only taken chemistry for one year