Question

Find the area of the following regular polygons. Round all answers to nearest 10th

Answer 1

Step 1

Draw the regular polygon to find out how many triangles are in it.

We can, therefore, conclude the regular polygon has 5 triangles in it

Step 2

Find the area of 1 triangle

$\begin{gathered} \text{The angle at the center = 360}^o \\ \text{But since they are 5 triangles, each will have an angle of }\frac{360}{5}=72^o \end{gathered}$

Since the two sides of each triangle are radius, then they share the same angle. Hence,

$\begin{gathered} \text{The angle at the base of each triangle is given by} \\ 180=72+x+x \\ 180-72=2x \\ \frac{2x}{2}=\frac{108}{2} \\ x=54^o \end{gathered}$

We can then find the k and y thus

$\begin{gathered} \sin 54=\frac{k}{6} \\ k=\text{ 6sin54} \\ k=\text{ 4.854101966 units} \\ \cos 54=\frac{y}{6} \\ y=6\cos 54 \\ y=3.526711514\text{ units} \\ 2y=7.053423028 \end{gathered}$

The area of one of the triangles is

$\begin{gathered} A=\frac{1}{2}\times base(2y)\times height(k) \\ A=\frac{1}{2}\times7.053423028\times\text{4.854101966} \\ A=17.11901729\text{ units} \end{gathered}$

The area of the 5 triangles will be

$17.11901729\text{ }\times5=85.59508647unit^2$

Hence the area of a regular polygon to the nearest tenth is given as;

$85.6units^2$