o is the center of the regular octagon below. find its area. round to the nearest tenth if necessary.
o is the center of the regular octagon below. find its area. round to the nearest tenth if necessary.
Answer
Explanation:
Step1: Divide octagon into triangles
A regular octagon can be divided into 8 isosceles triangles by connecting the center to each vertex. The central - angle of each triangle is $\theta=\frac{360^{\circ}}{8}=45^{\circ}$.
Step2: Find area of one triangle
The two equal - sides of each isosceles triangle have length $r = 6$. The area of a triangle with two sides $a$ and $b$ and included - angle $\theta$ is $A_{\triangle}=\frac{1}{2}ab\sin\theta$. Here, $a = b=r = 6$ and $\theta = 45^{\circ}$, so $A_{\triangle}=\frac{1}{2}\times6\times6\times\sin45^{\circ}=\frac{1}{2}\times36\times\frac{\sqrt{2}}{2}=9\sqrt{2}$.
Step3: Find area of octagon
The area of the octagon $A$ is the sum of the areas of the 8 isosceles triangles. So $A = 8\times A_{\triangle}=8\times9\sqrt{2}=72\sqrt{2}\approx101.8$.
Answer:
$101.8$