# Solution to the [[Three Squares and a Rectangle]] Puzzle +-- {.image} [[ThreeSquaresandaRectangle.jpeg:pic]] > Three squares and a rectangle. What’s the shaded area? =-- ## Solution by [[Similar Triangles]] and Lengths in a [[Square]] +-- {.image} [[ThreeSquaresandaRectangleAnnotated.jpeg:pic]] =-- Consider the diagram as labelled above. In this diagram, triangles $A B C$ and $E D C$ are [[right-angled triangle|right angled]] and angle $A \hat{C} B$ is equal to angle $D \hat{C} E$ since each together with angle $A \hat{C} E$ make a right-angle. Therefore, triangles $A B C$ and $E D C$ are [[similar]]. This means that the ratios $A C : B C$ and $E C : D C$ are equal. So the lengths satisfy: $$ \frac{A C}{B C} = \frac{E C}{D C} $$ Multiplying up gives: $$ A C \times D C = E C \times B C $$ Then $E C$ is the [[diagonal]] of a [[square]] of side length $3$, and $B C$ of $4$, so $E C$ has length $3 \sqrt{2}$ while $B C$ has length $4 \sqrt{2}$. Hence $$ A C \times D C = E C \times B C = 3 \sqrt{2} \times 4 \sqrt{2} = 24 $$ ## Solution by [[Invariance Principle]] The tilt of the shaded rectangle is not specified, allowing a range of angles with two special cases. +-- {.image} [[ThreeSquaresAndARectangleInvarianceA.png:pic]] =-- In this configuration, the small square is shrunken to a point meaning that the sides of the rectangle are the diagonals of the squares. So the area is $4\sqrt{2} \times 3\sqrt{2} = 24$. +-- {.image} [[ThreeSquaresAndARectangleInvarianceB.png:pic]] =-- In this configuration, the small square is the same size as the square of side length $4$, meaning that the shaded rectangle has sides $3$ and $8$ so area $24$.