Question

Referring to the figure, find the perimeter of the regular polygon

shown. [Note: do not approximate the value of any radicals; thus, if you
obtain, for example, 45 times the square root of 3, simply type 45SQR(3)
as your final answer.]



Referring to the Fig. in Question #1, find the area of the regular polygon
shown. [Note: do not approximate the value of any radicals; thus, if you
obtain, for example, 45 times the square root of 3, simply type 45SQR(3)
as your final answer.]

Answer 1

Perimeter = 18√3 = 18SQR(3)

Area = 27√3 = 27SQR(3)

Explanation:

The sides of a regular polygon are congruent.

Therefore, as the triangle is a regular polygon, it is an equilateral triangle since its sides (and interior angles) are congruent.

The interior angles of a triangle sum to 180°. Therefore, each interior angle of the equilateral triangle is 60°.

Therefore, the right triangle formed inside the equilateral triangle by the perpendicular bisectors is a 30-60-90 triangle with hypotenuse of 6 units.

The ratio of the side lengths of a 30-60-90 triangle is b : b√3 : 2b, where:

• b = shortest side opposite the 30° angle.
• b√3 = side opposite the 60° angle.
• 2b = longest side (hypotenuse) is opposite the right angle.

As the hypotenuse of the 30-60-90 right triangle is 6 units:

`2b = 6 \implies b=3`

Therefore, the length of the longest leg of the right triangle is 3√3.

As this is half the length of one side of the equilateral triangle:

`\implies \sf Perimeter = 6 * 3 √(3)=18√(3)\; units`

Heron's Formula allows us to find the area of a triangle in terms of its side lengths.

Heron's Formula

`\sf Area = √(s(s-a)(s-b)(s-c))`

where:

• a, b and c are the side lengths of the triangle.
• s is half the perimeter.

Given:

• Each side length = 6√3 units
• Half perimeter = 9√3 units

Substitute these values into the Heron's formula to calculate the area of the equilateral triangle:

`\begin{aligned}\implies \sf Area &= \sf \sqrt{9√(3)(9√(3)-6√(3))^3}\\ &= \sf \sqrt{9√(3)(3√(3))^3}\\ &= \sf \sqrt{9√(3)(81√(3))}\\&=\sf √(2187)\\&=\sf √(729 \cdot 3)\\&=\sf √(729)√(3)\\&=\sf 27√(3) \;\;units^2\end{aligned}`