Question

What is the area of a regular pentagon with a radius of 7

Answer 1

What is the area of a regular pentagon with a radius of 7?

Let we have regular pentagon ABCDE. Then if $O$ the centre of the circle then $OA = OB = OC = OD = OE$ and for angles:

$$\angle A O B = \angle B O C = \angle C O D = \angle D O E = \angle E O A$$

Then we have that corresponding triangles are equal:

$$\triangle A O B = \triangle B O C = \triangle C O D = \triangle D O E = \triangle E O A$$

Thus area of pentagon equals to 5 areas of:

$$S (A B C D E) = 5 S (\triangle A O B)$$

Now we must find area of $\triangle AOB$ . We know that $AO = BO = 7$ , and from equation for angles we get that $\angle AOB = \frac{2\pi}{5}$ .

But $S(\triangle AOB) = 0.5 * AO * BO * \sin(\angle AOB) = 0.5 * 7 * 7 * \sin \frac{2\pi}{5} = 24.5 * \sin \frac{2\pi}{5}$ .

And finally

$$\begin{array}{l} S (A B C D E) = 5 S (\triangle A O B) = 5 * 2 4. 5 * \sin \frac {2 \pi}{5} = 1 2 2. 5 * \sin \frac {2 \pi}{5} = \\ = 1 2 2. 5 * \sqrt {\frac {5}{8} + \frac {\sqrt {5}}{8}} \approx 1 1 6. 5 \\ \end{array}$$