On the how to find Pentagon Area page, explained was how to find the area of a Pentagon with the values of side length and apothem.
But there is also a formula we can use that requires only a side length in order to establish the area of a Pentagon.
As well as a formula that can be used to find a Pentagons area when we know the length of a radius in a Pentagon.
This page will show how to find a regular Pentagon area from side length alone.
Pentagon Area from Side Length & Apothem
The how to find Pentagon area page showed and explained how we can find the area of a Pentagon when we know a side length and the apothem length.
The area of a Pentagon formula shown above come from a Pentagon having 5 interior triangles.
So the area of one of these triangles multiplied by 5, results in a value for whole Pentagon area.
Now if we can find a way to write the apothem a in terms of the side length s.
We can obtain a formula for a Pentagon area where we only need to know the side length s.
If we look at one of the Pentagon interior triangles a bit closer, we can see in more detail how we can write things out.
Writing a Formula with Side Length only
One of the 5 interior triangles in a Pentagon can be split further into 2 right angle triangles.
The image below shows the properties of one of these right angle triangles, which we can use as a starting point to obtain a formula for Pentagon area from side length only.
The area of a triangle is given by. \frac{1}{2} × base × height
For one of the 5 interior triangles, the base is s, and the height is a.
So we have. \frac{1}{2} × s × a
Multiplied by 5 gives the whole Pentagon area. 5 × \frac{1}{2} × s × a = \frac{5}{2} × s × a
Using the Right Angle Triangle:
Let’s focus on the smaller right angle triangle, where the height is a.
Looking at the above angle of 36°, we can write the tan ratio.
tan (36\degree) \space=\space {\LARGE{\frac{\frac{s}{{\scriptsize{2}}}}{a}}}
The right side of the equals sign can be flipped by writing the left side under a numerator of 1, and we can proceed from there.
\frac{1}{tan (36\degree)} \space{\scriptsize{=}}\space \frac{a}{\frac{{\scriptsize{s}}}{2}} ( × \frac{s}{2} ) both sides
{\Large{\frac{s}{2 tan (36\degree)}}} \space\space=\space\space a
Now with an expression for a in terms of s.
We can put the Pentagon area formula together.
5 × \frac{1}{2} × s × a
= 5 × \frac{1}{2} × s × \frac{s}{2 tan (36\degree)} = 5 × \frac{1}{2} × \frac{s^2}{2 tan (36\degree)}
= 5 × \frac{s^2}{4 tan (36\degree)} = \frac{5s^2}{4 tan (36\degree)}
Example
(1.1)
Establish the area of the following Pentagon to 2 decimal places.
Solution
Area = \frac{5s^2}{4 tan (36\degree)} , s = 9
\frac{5(9)^2}{4 tan (36\degree)} = \frac{5 \space \times \space 81}{4 tan (36\degree)}
= \frac{405}{4 tan (36\degree)} = 139.3586...
The area of the Pentagon to 2 decimal places is, 139.36cm2.
