Area of a shape defined below

[Guest]

Construct 2 concentric circles O and I where O is larger than I. Then vertically construct a radius of O, from the intersection of I and the radius of O construct 2 lines at an angle of A/2 from the vertical radius which end at the edge of C1.

I've tried looking at this multiple times weeks and months apart. but cannot seem to figure it out.

[shyamjayakannan]

Screenshot 2026-03-12 202303.png (145.02 KiB) Viewed 82 times

The total area can be found as the area of segment FHG + segment JDK + trapezium JKGF. Let us first find the area of segment FHG:

[tex]ar(FHG)=ar(EFHG)-ar(\triangle EFG)=\pi f^2\frac{\theta}{2\pi}-\frac{1}{2}\times2f\sin\theta\times f\cos\theta=f^2\left(\frac{\theta}{2}-f^2\sin\theta\cos\theta\right)[/tex].

[tex]ar(JDK)[/tex] can be similarly found using [tex]\angle IOK[/tex]. Let us now find length IO:

[tex]IO=HO-HI=HO-(EI-EH)=r_2-\left(\frac{f}{\cos\theta}-f\right)=r_2-f\left(\frac{1}{\cos\theta}-1\right)[/tex].

Now, use the sine rule in [tex]\triangle KIO[/tex]: [tex]\displaystyle\frac{KO}{\sin\alpha}=\frac{IO}{\sin{\angle IKO}}\Rightarrow\frac{r_1}{\sin\alpha}=\frac{\displaystyle r_2-f\left(\frac{1}{\cos\theta}-1\right)}{\sin{(\pi-\angle IOK)}}[/tex]

[tex]\Rightarrow\sin{(\pi-\angle IOK)}=\sin{\angle IOK}=\frac{\sin\alpha}{r_1}\left\{r_2-f\left(\frac{1}{\cos\theta}-1\right)\right\}[/tex]

From this, you can find [tex]\angle IOK[/tex], [tex]\alpha=\pi-\frac{A}{2}[/tex] and [tex]\theta=\frac{\pi}{2}-\frac{A}{2}[/tex]. You also have the formula for the area of a sector. Now, you can find lengths [tex]JK=2r_1\sin{\angle IOK}[/tex], [tex]FG=2f\sin\theta[/tex] and the height of the trapezium to calculate its area.

I hope this helps you find the answer.