Question

How do I solve? So far no one could be of help. It’s asking for the area of this regular polygon.

Answer 1

We can notice the triangle on the square, hypotenuse is 8m and the base is half of the lengt of the square we will name it x

then

we can use pythagoras to solve

$a^2+b^2=h^2$

where a and b aer sides of the triangle and h the hypotenuse

replacing

$\begin{gathered} x^2+x^2=8^2 \\ 2x^2=64 \\ x^2=\frac{64}{2} \\ \\ x=\sqrt[]{32}=4\sqrt[]{2} \end{gathered}$

the length of x is half of the side of the square then the side of the square is

$4\sqrt[]{2}\times2=8\sqrt[]{2}$

each side of the square is 8v2 meters

Area

we use formula of the area

$\begin{gathered} A=l\times l \\ A=8\sqrt[]{2}\times8\sqrt[]{2} \\ \\ A=128 \end{gathered}$

area of the square is 128 square meters

Perimeter

we use formula of the perimeter

$\begin{gathered} P=4l \\ P=4\times8\sqrt[]{2} \\ P=32\sqrt[]{2} \end{gathered}$

perimeter of the square is 32v2 meters