[[!redirects five rectangles in a semi-circle solutions]] # Solution to the [[Five Rectangles in a Semi-Circle]] Puzzle +-- {.image} [[FiveRectanglesinaSemiCircle.jpeg:pic]] > What’s the total area of these five congruent rectangles? =-- ## Solution by [[Pythagoras' Theorem]] and the [[Perpendicular Bisector]] of a [[Chord]] +-- {.image} [[FiveRectanglesinaSemiCircleAnnotated.jpeg:pic]] =-- With the points labelled as above, $O$ is the centre of the semi-circle, which has radius $5$. Let the sides of the rectangles be $a$ and $b$, with $a \gt b$. Then [[chord]] $E C$ has length $a + b$, so since $O D$ [[bisects]] this chord, $O A$ has length $\frac{a + b}{2}$. Applying [[Pythagoras' theorem]] to triangle $O A E$ then shows that: $$ 5^2 = \left( \frac{a + b}{2} \right)^2 + (a + b)^2 = \frac{5 (a + b)^2}{4} $$ Hence $a^2 + 2 a b + b^2 = (a + b)^2 = 20$. Then $A G$ has length $a - b$ so $O G$ has length $a - b + \frac{a + b}{2} = \frac{3a - b}{2}$. Applying [[Pythagoras' theorem]] to triangle $O G F$ shows that: $$ 5^2 = \left(\frac{3 a - b}{2}\right)^2 + b^2 = \frac{9 a^2 - 6 a b + 5 b^2}{4} $$ Hence $9a^2 - 6 a b + 5 b^2 = 100$. Subtracting these two equations leads to: $$ 4a^2 - 16 a b = 0 $$ So $a = 4 b$, meaning that the area of one rectangle is $ a b = 4 b^2$, and also: $$ 20 = (a + b)^2 = (5 b)^2 = 25 b^2 $$ Therefore the area of $5$ rectangles is: $$ 5 \times 4 b^2 = 20 b^2 = \frac{400}{25} = 16 $$