Question

Please help me thank you!!!

Answer 1

There are two acceptable ways of solving this. I'll show both (just in case one of them doesn't align with the information from your textbook).

Option 1 (fast and easy):

The following is the equation for the area of a regular pentagon with side length s.

$A= \dfrac{s^2}{4} \times \sqrt{5(5+2\sqrt{5})}$

Plug s=12 into the equation.

$A= \dfrac{12^2}{4} \times \sqrt{5(5+2\sqrt{5})}$

$\approx 247.7$

The third choice should be your answer.

Option 2 (much more complicated):

The area of a regular polygon can be solved with the following formula.

$A=\dfrac{\text{perimeter} \times \text{apothem}}{2}$

Let's solve for the apothem.

A pentagon can be divided into 5 congruent isosceles triangles. The vertex angle will be 360/5, or 72 degrees.

An isosceles triangle can be divided into two right triangles. One of the angles of the right triangle will be 72/2, or 36 degrees.

The leg opposite of the 36 degree angle will be 12/2, or 6 inches long.

$\text{apothem}=\dfrac{6}{\tan{36}}$

$\approx 8.26$

The apothem is about 8.26 inches.
The perimeter is 12*5, or 60 inches.

$A=\dfrac{\text{perimeter} \times \text{apothem}}{2}$

$=\dfrac{60 \times 8.26}{2}$

$\approx 247.8$

(it seems that we're off by 0.1 due to rounding errors)

The answer is the third choice. Hope this helps! :)