Question

AB=diameter of the circle O. OC=radius. Arc AXB is an arc of the circle with centre C. Prove areas of the shaded region are =

Answer 1

1.
Draw a circle with center C And radius CA, as shown in the attached picture.

Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|=$\sqrt{2}R$.

these are all shown in the picture.

2.

Area of triangle ABC is 1/2 * 2R * R= R^2

3.

Let the area between arc BXA and chord AB be Y. (the yellow region).

and let G be the shaded region between arcs AB and AXB.

G=1/2(Area circle with center O)-Y
   =$\frac{1}{2} \pi R^{2}-Y$

To find Y:

Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.

So Area of sector ACB = $\frac{1}{4} \pi (\sqrt{2} R)^{2}=\frac{1}{4} \pi*2 R^{2} =\frac{1}{2} \pi R^{2}$

Y =area of sector ABC-Area(triangle ABC)=$\frac{1}{2} \pi R^{2}- \frac{1}{2}*2R*R=\frac{1}{2} \pi R^{2}- R^{2}$


4. 

Finally,

$G=\frac{1}{2} \pi R^{2}-Y=\frac{1}{2} \pi R^{2}-(\frac{1}{2} \pi R^{2}- R^{2})=R^{2}$

This proves that the 2 shaded regions have equal area.