Question

a regular decagon has a radius of 8 cm. what is the approximate area of the decagon? recall that a decagon is a polygon with 10 sides. 188 cm² 198 cm² 304 cm² 375 cm²

Explanation:

Step1: Find central angle

A decagon has 10 sides. The central angle $\theta=\frac{360^{\circ}}{n}$, where $n = 10$. So $\theta=\frac{360^{\circ}}{10}=36^{\circ}$.

Step2: Consider one - isosceles triangle

The area of an isosceles triangle formed by two radii and a side of the decagon with two equal sides of length $r = 8$ cm. The area of a triangle is $A_{\triangle}=\frac{1}{2}r^{2}\sin\theta$. Substituting $r = 8$ cm and $\theta = 36^{\circ}$, we get $A_{\triangle}=\frac{1}{2}\times8^{2}\times\sin36^{\circ}=\frac{1}{2}\times64\times\sin36^{\circ}= 32\times\sin36^{\circ}$.

Step3: Find area of decagon

The area of the decagon $A$ is composed of 10 such congruent isosceles triangles. So $A = 10\times A_{\triangle}=10\times32\times\sin36^{\circ}$. Since $\sin36^{\circ}\approx0.5878$, then $A=10\times32\times0.5878 = 320\times0.5878=188.096\approx188$ $cm^{2}$.

Answer:

$188$ $cm^{2}$