# Pentagon

How do you find the area of a pentagon without using trig!?

Solution

The diagram at the end of the page is useful!

The process of calculating the area of the pentagon isn't too bad once we've established the length of the diagonal, $d$. Effectively, we have three isosceles triangles $ABC$, $ACD$ and $ADE$ the combined area of which is the area of the pentagon

Establishing the length of $d$ is effectively the tricky bit without trig and we do it by considering similar triangles, to establish the lengths $x$ & $y$ since $d = 2y + x$

Let us label the intersection of $AC$ and $BE$, $M$ and the intersection of $AD$ and $BE$, $N$.

We can see immediately a pair of similar triangles, $AMN$ is similar to $ACD$. This gives us $$\frac{1}{x} = \frac{2y+x}{y}$$

The tricky thing to spot (to my mind at least) is that $ABN$ is an isosceles triangle, which gives us $$1 = x+y$$

These two equations now let us determine the values of $x,\;y$ and hence $d$

From the last equation, we have $y = 1-x$, substitution of which yields

$$\frac{1}{x} = \frac{2(1-x) + x}{1-x}$$ This simplifies to $$\frac{1}{x} = \frac{2-x}{1-x} \Rightarrow 1-x = x(2-x)$$ this in turn yields the quadratic $x^2 -3x + 1 = 0$. Solving the quadratic gives two values of $x$, $$x = \frac{3 \pm \sqrt{5}}{2}$$ From the diagram, it is clear that $x\lt 1$, so, $x = \frac{3 - \sqrt{5}}{2}$. This gives $y = \frac{-1+\sqrt{5}}{2}$.

The length of the diagonal is $$d = x + 2y = \frac{3 - \sqrt{5}}{2} + 2\frac{-1+\sqrt{5}}{2} = \frac{1+\sqrt{5}}{2}$$

With the diagonal established, the rest of the question is now pretty easy. If we take the diagonal to be the base of triangle $ABC$, we can then establish its height, $h_{ABC}$ through Pythagoras. $$h_{ABC} = \sqrt{1^2 - \left(\frac{d}{2}\right)^2}$$ The area of ABC is then $$A_{ABC} = \frac{1}{2}bh_{ABC} = \frac{1}{2}d\sqrt{1^2 - \left(\frac{d}{2}\right)^2}$$ We note that $A_{ADE} = A_{ABC}$. Now we need the area of $A_{ACD}$. If we say CD is the base, then the height is given by $$h_{ACD} = \sqrt{d^2 - \left(\frac{1}{2}\right)^2}$$ The area is therefore $$A_{ACD} = \frac{1}{2}bh_{ACD} = \frac{1}{2}\sqrt{d^2 - \left(\frac{1}{2}\right)^2}$$ From these results, the pentagons area is $$A = ABC + ACD + ADE = 2\left(\frac{1}{2}d\sqrt{1^2 - \left(\frac{d}{2}\right)^2}\right) + \frac{1}{2}\sqrt{d^2 - \left(\frac{1}{2}\right)^2}$$ It is left to the reader to establish the final simplification of this expression to its usual form of $$A = \frac{1}{4}\sqrt{25 + 10\sqrt{5}}$$