Side-quest 🐉

Problem

Find the area of the shaded part.

At first, I tried to solve this using simple geometry without trigonometry, but eventually gave up.

Solution

Then, brining the big gun:

Using integration:
- First, define a coordinate system:
- Then, write the function describing the blue circle and the red arc.
  - Blue circle: $x^2+(y+5*\sqrt{2})^2=100$
  - Red arc: $x^2+(y)^2=25$
- The coordinate of two crossing point can be found:
  - Point A: $[-\frac{5\sqrt{\frac{7}{2}}}{2},\frac{5}{2\sqrt{2}}]$
  - Point B: $[\frac{5\sqrt{\frac{7}{2}}}{2},\frac{5}{2\sqrt{2}}]$
- Using the $x$ corrdinate of point A and B as lower and upper limit, the integration could be written as:
  $\frac{Area_{total}}{2}=\int^A_B [\sqrt{25-x^2}-(\sqrt{100-x^2}-5\sqrt{2})] dx$
- And the total area:
  $Area_{total}=2*\frac{\sqrt{25}}{2}[\sqrt{7}-8\arcsin{(\frac{\sqrt{\frac{7}{2}}}{4})}+2\arcsin{(\sqrt{7})}]$
  or: $29.2763 cm^2$
Using trigonometry:
- Auxiliary lines can be usefull:
- We know:
  $AC=CB=10$ $AD=DB=5$ $CD=5\sqrt{2}$
- Using Cosin Law, we get the angle of $\angle ACD$ :
  $\angle ACD=\arccos{\frac{AC^2+CD^2-AD^2}{2AC*CD}}$
- Hence the area of sector ACBFA:
  $Area_{ACBFA}=\frac{2*{\angle ACD}}{2 \pi } \pi 10^2$
- The area of the triangle ACBA:
  $Area_{ACBA}=\frac{1}{2} \text{AC}*\text{CB}*\sin (2*\text{$\angle $ACD})$
- Area ABFA can be obtained:
  $Area_{ABFA}=Area_{ACBFA}-Area_{ACBA}$
- Now we perform similar procedure to obtain area of ABEA:
  $\text{$\angle $ADC}=\arccos\left(\frac{\text{AD}^2+\text{DC}^2-\text{AC}^2}{2 \text{AD}*\text{DC}}\right)$ $\text{$\angle $ADB}=2 (\pi -\text{$\angle $ADC})$ $Area_{ADBEA}=\frac{\pi* 5^2 * \text{$\angle $ADB}}{2 \pi }$ $Area_{ADBA}=\frac{1}{2} \text{AD}^2 \sin (\text{$\angle $ADB})$ $Area_{ABEA}=Area_{ADBEA}-Area_{ADBA}$ $Area_{AFBEA}=2*Area_{ABEA}-Area_{ABFA}$
- Result:
  $29.2763 cm^2$

It is intriguing that this problem cannot be simply solved without trigonometry.

I’ve wrote down the whole process in Wolfram Mathematica for download.