Let a regular pentagon have side 10. What is its area? Solution Referring to the diagram, h5=tan(54°)⇝h=5tan(54)≈6.88,area of △ABC=1210h≈34.41,area of pentagon =5×area of △ABC≈172.05. \begin{aligned} \frac{h}{5} &=& \tan(54°) \leadsto \\ h &=& 5 \tan(54) \approx 6.88, \\ \text{area of } \triangle ABC &=& \frac{1}{2} 10h \approx 34.41, \\ \text{area of pentagon } &=& 5 \times \text{area of } \triangle ABC \approx 172.05. \end{aligned} 5hharea of △ABCarea of pentagon ====tan(54°)⇝5tan(54)≈6.88,2110h≈34.41,5×area of △ABC≈172.05.
