# Solution to the Two Concentric Circles Puzzle +-- {.image} [[TwoConcentricCircles.png:pic]] > These two circles have the same centre. What’s the shaded area? =-- ## Solution by Pythagoras Theorem +-- {.image} [[TwoConcetricCirclesPythagoras.png:pic]] =-- Let us write $R$ for the radius of the outer circle, so on the diagram above we have $R = O A$, and $r$ for the radius of the inner circle, so $r = O C$. In the diagram, $E$ is the [[midpoint]] of the [[chord]] $B C$ and let us write $h$ for the distance $O E$. Note that $O E$ meets the line $A B C D$ at a [[right-angle]]. Since $O E A$ is a [[right-angled triangle]], [[Pythagoras theorem|Pythagoras' Theorem]] says that $$ R^2 = 12.5^2 + h^2 $$ Since $O E C$ is a right-angled triangle, Pythagoras' Theorem also says that $$ r^2 = 2.5^2 + h^2 $$ Eliminating $h^2$ from these gives $$ R^2 - r^2 = 12.5^2 - 2.5^2 $$ We can directly calculate that, or we can use the [[difference of two squares]] to write it as $(12.5 + 2.5)(12.5 - 2.5) = 15 \times 10 = 150$. The area we want to calculate is the area of the large circle minus the area of the smaller one. This is: $$ \pi R^2 - \pi r^2 = \pi (R^2 - r^2) = 150 \pi $$ ## Solution by Intersecting Chords Theorem +-- {.image} [[TwoConcetricCirclesChords.png:pic]] =-- In this diagram, the line $F G$ is chosen to be the diameter that passes through $C$. The [[intersecting chords theorem]] says that in the above diagram then: $$ F C \times C G = A C \times C D = 15 \times 10 = 150 $$ With $R$ and $r$ as above, then $F C = R - r$ and $C G = R + r$ so then $$ F C \times C G = (R - r) (R + r) = R^2 - r^2 $$ Hence $R^2 - r^2 = 150$ and so the orange area can be calculated as: $$ \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = 150 \pi $$