Searching for a shell
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.