Two circles of equal radius (R) intersect as shown below. Assuming more points are uniformly distributed in an area with dimensions D*D, where D = 4*R. What is the probability that a point will be within the shaded area ?
How can I go about solving a proof for my answer ?
On
Let $C$ and $D$ be the intersection points. Let $\theta$ be the angle $CAD$. The area of the diamond $ACBD$ is $\frac{1}{2}R^2\sin \theta + \frac{1}{2}R^2\sin \theta = R^2\sin \theta$. Twice the area of the sector $CAD$ of the circle is $R^2 \theta$. So the area shaded is $R^2(\theta - \sin \theta)$
Probability of being within the shaded area = (shaded area)/(total area).
Total area = $16R^2$.
Shaded area - bit harder to work out!
Copied from @user3491648's answer:
Let $C$ and $D$ be the intersection points. Let $\theta$ be the angle $CAD$. The area of the diamond $ACBD$ is $\frac{1}{2}R^2\sin \theta + \frac{1}{2}R^2\sin \theta = R^2\sin \theta$. Twice the area of the sector $CAD$ of the circle is $R^2 \theta$. So the area shaded is $R^2(\theta - \sin \theta)$.
So probability required is $\frac{(\theta - \sin \theta)}{16}$. (credit to @toby's comment).
The additional information we need is that $\cos(\theta)=\frac {d/2} R$.