Impossible C2 Circles Proof Question?

Been on this question for about 40 minutes
Can't understand it even with the ms, can someone please explain it?
Thanks
Hasan

right okay here's what you have to do....replicate the diagram first and draw in the other circles, then draw two circle sectors where the arcs of each are OC and OBThen, draw a hexagon inside the centre circle (the first one you drew)Split the hexagon into six equilateral triangles (every angle is 60 degrees) and you should see that two of the triangles fall exactly into those sectors, the sides are length r (one side of each will line up with the radius already drawn in).use Area = 1/2 absin(c) to work out the area of one of these triangles and you should get Area = 1/2 r x r sin(30) = 1/2 r^2 x (3^1/2)/2Look at the diagram again and you'll see that because of the hexagon you can tell that the two circle segments are a sixth of the circle each as each of them fits inside one of the six triangles (within the hexagon) therefore the area of ONE of the sectors is (pi x r^r)/6 If you do the area of the segment - the are of the triangle you get this small area where the arc is and if you look at the triangle that the shaded region is inside you'll see that TWO of these areas (the one you just found) cross into that triangle and the rest is the shaded region that you want.Therefore, double the area of that you just found (should be ((r)^2(3)^1/2)/4 Then do the area of one of the triangles made by the hexagon minus this area (which you just doubled) and then simplify by taking 1/6 and r^2 outside the bracket and multiplying the value with (3)^1/2 by 2 and you're sorted

Been on this question for about 40 minutes
Can't understand it even with the ms, can someone please explain it?
Thanks
Hasan

Don't know how good your integration is but assume you can do some as it's C2, don't worry if you don't follow this entirely.

We know the equation of the central circle. we also know that the unshaded area in the right half of the circle is four times the area between the circumference and the line where y=r/2.

We can then solve to find the intersection point between the top and central circles to get the intersection point $(\frac{\sqrt{3}r}{2}, \frac{r}{2})$

knowing the above we can set up an integral to find this area.

$\displaystyle A = \int^{\frac{\sqrt{3}r}{2}}_0( \sqrt{r^2-x^2}-\frac{r}{2})dx$

This integral requires a substitution to solve

$x = rsin\theta$ giving $dx=rcos\theta d\theta$

our upper and lower limits become $\frac{\pi}{3}$ and 0 respectively.

Now our integral becomes

$\displaystyle A = \int^{\frac{\pi}{3}}_0 r^2cos^2\theta -\frac{r^2cos\theta}{2} d\theta$

solving this integral yields

$\displaystyle A = \frac{r^2}{2}[\theta + \frac{sin2\theta}{2} - sin\theta]^{\frac{\pi}{3}}_0 = \frac{r^2}{2}[\frac{\pi}{3} - \frac{\sqrt{3}}{4}]$

to solve the from earlier we recall that A that we have calculated is only a quarter of the unshaded area on that side so we can multiply it by 4 and subtract from the area of a semicircle and it gives us the solution.

Again, been a while since I did A-level so not sure if that is a C2 level solution...

Thank you so much for the help
I followed your reasoning and ended up with this?