# Thread: Area of five-sided coin

1. ## Area of five-sided coin

stuck on another problem from Abbotts "Teach Yourself" Trigonometry:

A new five-sided coin is to be made in the shape of figure /image attached/

Тhe point А on the cirсumferenсe of the coin is the сentre of the
arc CD, which has а radius of 2 сm. Similarly B is the сentre of
the arс DЕ, and so on. Find the area of one faсe of the сoin.

hope you can help me, thanks in advance!

2. Originally Posted by dmi3
stuck on another problem from Abbotts "Teach Yourself" Trigonometry:

A new five-sided coin is to be made in the shape of figure /image attached/

Тhe point А on the cirсumferenсe of the coin is the сentre of the
arc CD, which has а radius of 2 сm. Similarly B is the сentre of
the arс DЕ, and so on. Find the area of one faсe of the сoin.

hope you can help me, thanks in advance!
1. Draw a sketch (see attachment)

2. Draw a pentagon ABCDE. Determine the interior angles.

3. The area of the coin consists of
triangle
• ACD: $a_{ACD} = \frac12 \cdot 2 \cdot 2 \cdot \sin(36^\circ)$
• 2 triangles ABC: $a_{ABC} = \frac12 \cdot 2 \cdot h = \tan(36^\circ)$
• 5 segments: area of sector - area of triangle ACD: $a_{segm}= \frac{36^\circ}{360^\circ} \cdot \pi \cdot 2^2 - \frac12 \cdot 2 \cdot 2 \cdot \sin(36^\circ)$

4. I've got $a_{coin} \approx 3.034\ cm^2$

3. 3.034 is correct, thanks!