In the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle.

Show that the areas of the two shaded regions are equal.

Let the radius of the quarter circle be $r$.

Then $OA = OB = r$

$$\text{Area of the quarter circle: } A = \frac{\pi r^2}{4}$$

$$\text{In right triangle AOB, } AB = r\sqrt{2}$$

Radius of the semicircle on diameter,

$$AB = \frac{AB}{2} = \frac{r\sqrt{2}}{2}$$

Area of the semicircle:

$$\frac{1}{2} \pi \left( \frac{r\sqrt{2}}{2} \right)^2 = \frac{1}{2} \pi \cdot \frac{2r^2}{4} = \frac{\pi r^2}{4}$$

Thus, the area of the semicircle equals the area of the quarter circle. Both figures contain the same unshaded region between chord AB and the arc AB. Removing this common region from equal areas leaves equal remaining areas. Hence, the two shaded regions are equal in area.