Area of Intersection of Three Circles

The area of the common wedge between the two horizontal semicircles = the area of the wedge between the left horizontal circle and the tilted circle (unfortunately I can't draw a diagram). Now delete the area common to both these regions.

Think: if you rotate back the tilted circle by 20∘, it will superimpose itself on the right horizontal circle.


enter image description here

Let $A$ be the area of the region bounded by arc $QS$, arc $SU$ and line segment $QU$.

The shaded area is equal to $A$ minus the sum of the area of the minor segment with centre angle $60^\circ$ and the area of the sector with centre angle $40^\circ$.

But $A$ minus the sum of the area of the minor segment with centre angle $60^\circ$ is equal to the area of the sector with centre angle $60^\circ$.

So the shaded area is equal to the area of the sector with centre angle $20^\circ$, which is equal to $\displaystyle 72\times\frac{20}{180}=8$.


For variety, we can apply a variation on Cavalieri's principle.

The slanted semicircle is obtained by rotating the other semicircle 20 degrees around $Q$. The variation we need is:

If all of the circular slices of two regions have the same length, then the regions have the same volume

By circular slice, I mean the intersections of the region with circles centered on $Q$.

Because of the rotation, it's easy to see slice of the shaded region has the same length as the corresponding slice of the $20^\circ$ wedge — each region is obtained by rotating a figure $20$ degrees — and thus they have the same area.

For the rigorously inclined, the aforementioned principle can be fairly easily proven via area integrals in polar coordinates.

Tags:

Geometry

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