Side-quest 🐉



Find the area of the shaded part.

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


Then, brining the big gun:

  1. Using integration:

    • First, define a coordinate system:

    Alt text

    • 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:


      or: \(29.2763 cm^2\)

  2. Using trigonometry:

    • Auxiliary lines can be usefull:

    Alt text

    • 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:

    • 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.

Author | Chengcheng Xiao

Currently a PhD student at Imperial College London. Predicting electron behaviour since 2016.