# Two Semi-Circles Diagonally Across a Square +-- {.image} [[TwoSemiCirclesDiagonallyAcrossaSquare.png:pic]] > Two semicircles are placed on the diagonal of a square. What’s the total shaded area? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[TwoSemiCirclesDiagonallyAcrossaSquareLabelled.png:pic]] =-- With the points labelled as above, let $a$ be the diameter of the lower semi-circle and $b$ the diameter of the upper. The total shaded area is then $\frac{1}{8} \pi (a^2 + b^2)$. Triangle $A O E$ is a [[right-angled triangle]]. The full diagonal of the square is then $a + b$, so the length of $A O$ is $\frac{a + b}{2}$. The length of $O E$ is $\frac{a + b}{2} - b = \frac{a - b}{2}$. Applying [[Pythagoras' Theorem]] to triangle $A O E$ yields the equation: $$ \left(\frac{a + b}{2}\right)^2 + \left(\frac{a - b}{2}\right)^2 = 4^2 $$ The left-hand side expands to $$ \frac{1}{4}\left(a^2 + 2 a b + b^2 + a^2 - 2 a b + b^2 \right) = \frac{1}{2}\left( a^2 + b^2 \right) $$ So the shaded area is: $$ \frac{1}{8} \pi (a^2 + b^2) = \frac{1}{4} \pi \frac{a^2 + b^2}{2} = \frac{1}{4} \pi 4^2 = 4 \pi. $$ ## Solution by [[Invariance Principle]] The relative sizes of the semi-circles can be varied. In the first special case, the two are the same size, so the half diagonal of the square is of length $4$ and this is also the diameters of the semi-circles. This gives an area of $2 \times \frac{1}{2} \times \pi \times 2^2 = 4 \pi$. +-- {.image} [[TwoSemiCirclesDiagonallyAcrossaSquareHalf.png:pic]] =-- In the second special case, the lower semi-circle is as large as it can be and the upper as small. This means that the diameter of the lower semi-circle is the diagonal of the square and the side of the square is of length $4$. The diagonal is therefore $4 \sqrt{2}$ and the area of the semi-circle is $\frac{1}{2} \pi (2 \sqrt{2})^2 = 4 \pi$. +-- {.image} [[TwoSemiCirclesDiagonallyAcrossaSquareFull.png:pic]] =--