# Solution to the [[Two Squares and a Quarter Circle]] Puzzle +-- {.image} [[TwoSquaresandaQuarterCircle.jpeg:pic]] > The area of the red square is 12. What’s the total yellow area? =-- ## Solution by [[Pythagoras' Theorem]] and the [[Area of a Square]] +-- {.image} [[TwoSquaresandaQuarterCircleLabelled.jpeg:pic]] =-- Consider the diagram as labelled above. The red square has area $12$, so its side length is $\sqrt{12} = 2\sqrt{3}$. Let $r$ be the radius of the quarter circle, so $r$ is the length of line segment $A E$. The length of line segment $A C$ can then be found using [[Pythagoras' theorem]]: $$ \sqrt{r^2 - (2\sqrt{3})^2} = \sqrt{r^2 - 12} $$ The area of the outer square is $r^2$, so the yellow area is: $$ r^2 - (\sqrt{r^2 - 12})^2 = r^2 - r^2 + 12 = 12 $$ ## Solution by [[Invariance Principle]] and Lengths in a [[Square]] +-- {.image} [[TwoSquaresandaQuarterCircleInvariance.jpeg:pic]] =-- The radius of the quarter circle is not fixed, so can be varied. At one extreme, the white area shrinks to non-existence. The radius of the quarter circle is the diagonal of the red square, and so has length $\sqrt{2}$ times the side length of the [[square]]. The outer square therefore has double the area of the red square, and so the area of the yellow region is the same as that of the red square, namely $12$.