Question

Aregular hexagon with a perimeter of 48 yd. Find the area of the polygon and round to the nearest tenth.

Answer 1

GivenL

A) The hexagon having the perimeter 48 yd.

As hexagon is six sided polygon and its perimeter is given by,

`\begin{gathered} P=6* a \\ a\text{ is the measure of each side.} \\ 48=6* a \\ a=8 \end{gathered}`

The area is calculated with the formula,

`\begin{gathered} A=6*\frac{\sqrt[]{3}}{4}* a^2 \\ A=6*\frac{\sqrt[]{3}}{4}*8^2 \\ A=166.3yd^2 \end{gathered}`

B) the side length of pentagon is 6 ft . its area is calculated as,

`\begin{gathered} A=(1)/(4)\sqrt[]{5(5+2\sqrt[]{5})}s^2 \\ A=(1)/(4)\sqrt[]{5(5+2\sqrt[]{5})}(6)^2 \\ =61.9ft^2 \end{gathered}`