[[!redirects rectangle across a square solution]] # Solution to the Rectangle Across a Rectangle Puzzle +-- {.image} [[RectangleAcrossaSquare.png:pic]] > Here are two rectangles. $A B=5$, $A E=7$, $A F=1$. What’s the angle? =-- ## Solution by [[Angle at the Circumference is Half the Angle at the Centre]] +-- {.image} [[RectangleAcrossaSquareLabelled.png:pic]] =-- Let $x$ be the length of the diagonal $A C$. Applying [[Pythagoras' theorem]] to the [[right-angled triangle]] $A E C$ shows that $x$ satisfies $x^2 = 7^2 + 1^2 = 50$. Now let $y$ be the length of $B C$, so the height of rectangle $A B C D$. Applying [[Pythagoras' theorem]] to $A B C$ shows that $x^2 = y^2 + 5^2$, so $y^2 = 25$ and $y = 5$. As this is the same length as $A B$, the rectangle $A B C D$ is actually a [[square]]. Now, as in the above diagram, consider the [[circumcircle]] of the square $A B C D$, with the point labelled $O$ at its centre. This also passes through the vertices of the other rectangle as the points $A$ and $C$ are on it. Since the [[angle at the circumference is half the angle at the centre]], angle $A \hat{E} C$ is half of angle $A \hat{O} C$. This latter angle is $90^\circ$, hence $A \hat{E} C = 45^\circ$.