Radian Measure

Hello,

Let H be the intersection point of AB and O1O2.

- First compute the measure (in radians) of $\alpha=\widehat{AO_1H}$

Consider the triangle $AO_1H$ , which is a right angle triangle.
We know that $O_1H=8 \text{ cm}$
Hence $\cos \alpha=\frac{8}{10}=\frac 45$

So $\alpha=\arccos \left(\frac 45 \right) \approx 0.6435$ (radian)

- Now compute the area $\mathcal{A}_1$ of the arc of circle $AO_1B$
Note that $\widehat{AO_1B}=2 \alpha$

I'm too rusty to remember if there is a formula for it, but here is the reasoning.
The whole circle has a center angle of $2 \pi$ radians.
The arc of circle we consider has a center angle of $2 \alpha$ radians.
So basically, we're considering $\frac{2 \alpha}{2 \pi}=\frac{\alpha}{\pi}$ th of the circle.

It can be used for the area : $\mathcal{A}_1=\frac \alpha \pi \cdot \underbrace{(\pi \cdot 10^2)}_{\text{total area of the circle}}=\alpha \cdot 10^2 \approx 64.35 \text{ cm}^2$

- You now have to substract the area of the triangle $AO_1B$ to get half of the shaded area.

Note that $AO_1B$ has twice the area of $AO_1H$

So let's consider the triangle $AO_1H$ .
It's a right angle triangle.
By the Pythagorean theorem, $AO_1^2=AH^2+O_1H^2$
That is $10^2=8^2+AH^2$
Hence $AH=6 \text{ cm}$

So the area of this triangle is $\frac 12 \cdot 6 \cdot 8=24 \text{ cm}^2$
Twice this area = area of triangle $AO_1B$ = 48 cm².

So the area of half the shaded area is $\approx 64.35-48=16.35 \text{ cm}^2$

Hence the shaded area is $\approx 32.70 \text{ cm}^2$

I hope it's clear enough (Whew)