Notes
two concentric circles solution

Solution to the Two Concentric Circles Puzzle

Two Concentric Circles

These two circles have the same centre. What’s the shaded area?

Solution by Pythagoras Theorem

Two Concentric Circles by Pythagoras

Let us write RR for the radius of the outer circle, so on the diagram above we have R=OAR = O A, and rr for the radius of the inner circle, so r=OCr = O C. In the diagram, EE is the midpoint of the chord BCB C and let us write hh for the distance OEO E. Note that OEO E meets the line ABCDA B C D at a right-angle.

Since OEAO E A is a right-angled triangle, Pythagoras' Theorem says that

R 2=12.5 2+h 2 R^2 = 12.5^2 + h^2

Since OECO E C is a right-angled triangle, Pythagoras’ Theorem also says that

r 2=2.5 2+h 2 r^2 = 2.5^2 + h^2

Eliminating h 2h^2 from these gives

R 2r 2=12.5 22.5 2 R^2 - r^2 = 12.5^2 - 2.5^2

We can directly calculate that, or we can use the difference of two squares to write it as (12.5+2.5)(12.52.5)=15×10=150(12.5 + 2.5)(12.5 - 2.5) = 15 \times 10 = 150.

The area we want to calculate is the area of the large circle minus the area of the smaller one. This is:

πR 2πr 2=π(R 2r 2)=150π \pi R^2 - \pi r^2 = \pi (R^2 - r^2) = 150 \pi

Solution by Intersecting Chords Theorem

Two Concentric Circles by Intersecting Chords

In this diagram, the line FGF G is chosen to be the diameter that passes through CC.

The intersecting chords theorem says that in the above diagram then:

FC×CG=AC×CD=15×10=150 F C \times C G = A C \times C D = 15 \times 10 = 150

With RR and rr as above, then FC=RrF C = R - r and CG=R+rC G = R + r so then

FC×CG=(Rr)(R+r)=R 2r 2 F C \times C G = (R - r) (R + r) = R^2 - r^2

Hence R 2r 2=150R^2 - r^2 = 150 and so the orange area can be calculated as:

πR 2πr 2=π(R 2r 2)=150π \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = 150 \pi