Let T be the centre of the semicircle with diameter QR and let OT produced meet the circumference of the larger semicircle at U .
Semicircular arcs solution

By symmetry, we note that OT is perpendicular to QR . As TR = TO = TQ (radii of the same semicircle), triangles ORT and OQT are both isosceles, right-angled triangles. So QOR is a right angle.

By Pythagoras' Theorem: QR ² = OQ ² + OR ² = 2² + 2² = 8.

So

QR = \sqrt{8}

2 \sqrt{2}

and the radius of semicircle QORis

\sqrt{2}

The area of the shaded region is equal to the area of semicircle QOR plus the area of the quadrant bounded by OQ , OR and arc QUR less the triangle OQR .

So the required area is

\frac{1}{2} π (\sqrt{2})^{2}

\frac{1}{4} π 2^{2}

- (

\frac{1}{2} \times 2 \times 2

) =

π + π - 2

2 π

- 2.