3420.51 – Area of a Regular Pentagon

Let a regular pentagon have side length 10. What is its area?

Solution

Referring to the diagram,

h5=tan⁡(54°)h=5tan⁡(54)≈6.88area of △ABC=1210h≈34.41area of pentagon =5×area of △ABC≈172.05\begin{aligned}\dfrac{h}{5} &= \tan(54°) \\h &= 5 \tan(54) \approx 6.88 \\\text{area of } \triangle ABC &= \dfrac{1}{2} 10h \approx 34.41 \\\text{area of pentagon } &= 5 \times \text{area of } \triangle ABC \approx 172.05\end{aligned}