Question

Guys, please help me. Thanks in advance

Find the area of a regular pentagram if one of the sides of the inner pentagon is 3 cm.

Answer 1

hi @Jper area of pentagon can be calculated by more than one way..
try with the help of the tips in this link..
if it is difficult i will help you
http://www.wikihow.com/Find-the-Area-of-a-Pentagon

Answer 2

For some useful info on the shape's properties:

According to the link, we have the following ratios involved:
\[\frac{x}{y}=\frac{y}{x+y}=\frac{y}{z}=\frac{z}{y+z}=\frac{z}{w}\]
You can solve for \(y\):
\[\begin{align*}\frac{3}{y}&=\frac{y}{3+y}\\\\
9+3y&=y^2\\\\
y^2-3y-9&=0\\\\
y&=\frac{3\pm\sqrt{45}}{2}\\\\
y&=\frac{3+\sqrt{45}}{2}&\text{ignore the negative root, }y\text{ must be positive}
\end{align*}\]
From this, you can solve for \(z\) and eventually \(w\):
\[\begin{align*}
\frac{\dfrac{3+\sqrt{45}}{2}}{3+\dfrac{3+\sqrt{45}}{2}}&=\frac{\dfrac{3+\sqrt{45}}{2}}{z}\\\\
\frac{1}{3+\dfrac{3+\sqrt{45}}{2}}&=\frac{1}{z}\\\\
z&=3+\frac{3+\sqrt{45}}{2}\\\\
z&=\frac{9+\sqrt{45}}{2}
\end{align*}\]
and in turn,
\[\begin{align*}
\frac{\dfrac{9+\sqrt{45}}{2}}{\dfrac{3+\sqrt{45}}{2}+\dfrac{9+\sqrt{45}}{2}}&=\frac{\dfrac{9+\sqrt{45}}{2}}{w}\\\\
\frac{1}{6+\sqrt{45}}&=\frac{1}{w}\\\\
w&=6+\sqrt{45}\end{align*}\]