a regular decagon has a radius of 8 cm. what is the approximate area of the decagon? recall that a decagon…

Explanation:

Step1: Find central angle

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

Step2: Divide decagon into isosceles triangles

A regular decagon can be divided into 10 isosceles triangles with two - sides equal to the radius $r = 8$ cm.

Step3: Find area of one isosceles triangle

The area of a triangle is $A_{\triangle}=\frac{1}{2}r^{2}\sin\theta$. Substituting $r = 8$ cm and $\theta = 36^{\circ}$, we have $A_{\triangle}=\frac{1}{2}\times8^{2}\times\sin36^{\circ}=\frac{1}{2}\times64\times\sin36^{\circ}= 32\times\sin36^{\circ}$. Since $\sin36^{\circ}\approx0.5878$, $A_{\triangle}\approx32\times0.5878 = 18.8096$ $cm^{2}$.

Step4: Find area of decagon

The area of the decagon $A = 10\times A_{\triangle}$. So $A=10\times18.8096\approx188$ $cm^{2}$.

Answer:

$188$ $cm^{2}$