Question

a regular pentagon with center c is shown below. what is the area of the pentagon? round your answer to the nearest tenth. square inches

Explanation:

Step1: Calculate the central angle

The central angle of a regular pentagon is $\frac{360^{\circ}}{5} = 72^{\circ}$.

Step2: Divide the pentagon into 5 isosceles triangles

Each triangle has a vertex angle of $72^{\circ}$ and two equal sides of length $r = 6$ inches.

Step3: Find the area of one triangle

Using the formula $A=\frac{1}{2}r^{2}\sin\theta$ (where $\theta$ is the included angle between the two sides of length $r$). Here, $r = 6$ and $\theta=72^{\circ}$.
So, $A_{triangle}=\frac{1}{2}\times6^{2}\times\sin(72^{\circ})=\frac{1}{2}\times36\times\sin(72^{\circ}) = 18\times\sin(72^{\circ})$.
Since $\sin(72^{\circ})\approx0.9511$, then $A_{triangle}\approx18\times0.9511 = 17.1198$.

Step4: Calculate the area of the pentagon

The area of the pentagon $A = 5\times A_{triangle}$.
$A=5\times17.1198=85.599\approx85.6$.

Answer:

$85.6$