lets find dia of arc AEB..
Dia of circle is 20 and triangle ADB will be right isosceles triangle with angle D as 90..
therefore AD or BD will be \(\frac{20}{\sqrt{2}}=10\sqrt{2}\), this is same as RADIUS of the arc ADB and arc AFB

Now area of unshaded portion AEBO..
area of AEBO = area of segment AEBD - area of triangle ABD..
\(pi*(10\sqrt{2})^2*\frac{90}{360}-\frac{1}{2}*10\sqrt{2}*10\sqrt{2}=50pi-100\)..
so the total unshaded portion's area = 2*(50pi-100)=100pi-200

total area of circle = \(pi*10^2=100pi\)

so area of shaded region = 100pi-(100pi-200)=200

D

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Good Question... Kudos to you.. This question tests many Concepts.

For the arc AEB with Centre D, AD = ED = R (Radius of Bigger Circle in for which AEB is an arc).
r = Radius of smaller circle with Centre O.
Now, as AB is the Diameter of Circle with centre O, it will subtend 90 Degree at point C and D.
From Triangle BOD, we have -
\(R^2 = r^2 + r^2\) -----> \(R = \sqrt{2}r.\)

Now, lets find the Area of AEBOA = Area of SECTOR with Centre D - Area of Triangle ADB
Area of AEBFA = \(\frac{90}{360}*pi*R^2 - \frac{1}{2}*2r*r = pi*\frac{r^2}{2} - r^2\)

Now, multiply this area with 2 to get the total area of arc AEBFA = \(pi*r^2 - 2r^2\)
So, area of shaded region = Area of circle with Centre O - Area of unshaded region AEBFA = \(pi*r^2 - (pi*r^2 - 2r^2)\)
Area of shaded region = \(2r^2\) = 200.

Hi

I am able to visualise 2 semi hemispheres here. But unable to split and solve them because of the intersection with each other.

Can someone help?

Thank you.