A regular polygon has all sides and all angles equal, so it can be cut from its center into congruent isosceles triangles. The height of each triangle, from the center to the middle of a side, is the polygon's apothem. With the special right triangles of 45-45-90 and 30-60-90 Triangles you can often find it exactly.
Why it works
A regular $n$-gon with side $s$ splits into $n$ triangles, each with base $s$ and height $a$ (the apothem). Their total area is $n \cdot \frac{1}{2} s a = \frac{1}{2} a (ns) = \frac{1}{2} a P$, where $P$ is the perimeter. The same idea explains the circle: cut it into very thin slices, and the "apothem" becomes the radius and the perimeter becomes the circumference, so $A = \frac{1}{2} r \cdot 2\pi r = \pi r^2$.
Key idea
For a regular polygon, $\text{Area} = \frac{1}{2} \cdot \text{apothem} \cdot \text{perimeter} = \frac{1}{2}aP$.
Worked example: With the apothem given
A regular pentagon has side $10$ cm and apothem $6.9$ cm. Find its area.
- Perimeter: $5 \cdot 10 = 50$ cm.
- $A = \frac{1}{2} \cdot 6.9 \cdot 50 = 172.5$ cm$^2$.
Answer: $172.5$ cm$^2$
Worked example: A regular hexagon, exactly
Find the area of a regular hexagon with side $6$.
- The six triangles are equilateral with side 6. The apothem is the height of one of them: the longer leg of a 30-60-90 triangle with shorter leg $3$, so $a = 3\sqrt{3}$.
- Perimeter $= 36$. $A = \frac{1}{2} \cdot 3\sqrt{3} \cdot 36 = 54\sqrt{3} \approx 93.5$.
Answer: $54\sqrt{3}$
Worked example: An equilateral triangle
Find the area of an equilateral triangle with side $10$.
- Its height splits it into two 30-60-90 triangles with shorter leg $5$, so the height is $5\sqrt{3}$.
- $A = \frac{1}{2} \cdot 10 \cdot 5\sqrt{3} = 25\sqrt{3}$. (In general, $\frac{\sqrt{3}}{4}s^2$.)
Answer: $25\sqrt{3} \approx 43.3$
Common mistake
Using the radius (center to a vertex) where the apothem (center to the middle of a side) belongs. In a hexagon with side 6, the radius is 6 but the apothem is $3\sqrt{3} \approx 5.2$. The apothem always meets the side at a right angle.