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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.
enter image description here

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.]

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One roundabout way would be to prove the Pythagorean theorem geometrically (many standard proofs exist, generally requiring only the formulas for area of squares and triangles that you're given), and use that to compute the area of an equilateral triangle with side 22cm.

It's straight-forward to see that the semi-circle can be divided into 3 such equilateral triangles, plus the 3 $60^o$ circular segments. Thus, you can find the area of one of the circular segments.

The intersection of the semi-circle and quadrant is obviously 1 equilateral triangle plus 2 circular segments. Subtracting that from the formula for the area of a semi-circle gives you the shaded region.

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