I noticed this question in a Math Problem-Set book.
These were the only formulas allowed:
$1. Area(quadrant) = \frac{1}{4}\pi r^2$
$2. Area(square) = (side)^2$
$3.Area(semicircle) = \frac{1}{2} \pi r^2$
$4. Area(\Delta) = \frac{1}{2}(base)(height)$
Any straight-line constructions can be done.
In the figure, AO(22 cm) is the radius of a quadrant. AO is also the radius of a semicircle, and so is OB. We need the area of the dark region. I just want to know the logic. Calculations are not required.
[I got as far as constructing a square with AO as the bottom side and a right-angled isosceles triangle with AO as the base, and then using it to find a part of the left half of the semicircle. Finding the area of the part to the right of the constructed square is of course, a piece of cake.]