# What is the probability that a point randomly selected inside the rectangle will also be inside both semicircles?

I tried to solve this problem by finding the area of the rectangle and the area shared by both semicircle. My difficult is how to find the area of the created by the two semicircles. Is my approach wrong?

Two semicircles are inscribed in a rectangle as shown so their diameters are opposite sides of the rectangle. What is the probability that a point randomly selected inside the rectangle will also be inside both semicircles? Express your answer as a decimal rounded to the nearest hundredth.

• Did you try integrating to find the area created by the two semi-circles? – wesssg Mar 1 '17 at 4:56
• This is much more a geometry question than probability, I'd suggest tagging it as such. – David Mar 1 '17 at 4:58
• Yes, you want the ratio of the area of the region between the semicircles to the area of the rectangle. Assume the semicirlces have radius 2. Draw a vertical line between the centers, and a horizontal line between the points of intersection of the two semicircles. Also draw the line segments from the semicircle centers to the points of intersection. You now have 4 congruent triangles. Compute the side lengths of one of the triangles. That should get you started. – quasi Mar 1 '17 at 5:02
• Of course, if you have Calculus knowledge, you can get equations, solve for the intersection points, and then integrate. Are you taking Calculus? But the geometric approach is easier. – quasi Mar 1 '17 at 5:04
• a middle school student – math Mar 1 '17 at 5:10

This is my approach in an example. Take one of the circles to be $x^2+y^2=1$, and the other to be $x^2+(y-1)^2=1$. We are interested in the area where the two circles overlap. First you need to find their intersection points by setting the equations equal. Then, you can evaluate the integral: $\int Top circle -Bottomcircle$ $dx$. This will give you the area. I'll end here since you seemed to have trouble with this bit. NOTE: These equations are just used as an example of how to calculate such an area
Integration is completely unnecessary. The area common to both semicircles can be found by observing that two equilateral triangles of side length equal to the common radius $r$, can be inscribed in this region. The area of one such triangle is therefore $$\frac{r}{2} \cdot \frac{r \sqrt{3}}{2} = \frac{\sqrt{3}}{4} r^2.$$ Because these triangles are equilateral, the circular arcs of the region each subtend an angle of $2\pi/3$ radians, and the area of the sector of one circle is simply $\pi r^2/3$. Thus the total area is $$\frac{2\pi}{3} r^2 - \frac{\sqrt{3}}{4} r^2.$$ The area of the rectangle is trivially $2r^2$, thus the desired probability is....