# Solution to the Rectangle, Circle, and Triangle Puzzle +-- {.image} [[RectangleCircleandTriangle.png:pic]] > A rectangle, a circle, and a $3{-}4{-}5$ triangle. What fraction of the rectangle is shaded? =-- ## Solution by [[Properties of Chords]], the [[Intersecting Chords Theorem]], and [[Similar Triangles]] +-- {.image} [[RectangleCircleandTriangleLabelled.png:pic]] =-- With the points labelled as above, triangle $A B E$ has a [[right-angle]] at $B$ so since the [[angle in a semi-circle]] is $90^\circ$, $A E$ is a [[diameter]] of the circle. Then as $B D$ is a [[chord]] of the circle and $A E$ is [[perpendicular]] to this chord and passes through the centre of the circle, it must pass through the [[midpoint]] of $B D$. Therefore $A D$ has the same length as $A B$, which is $4$. Applying the [[intersecting chords theorem]] to the chords $B C$ and $A C$ shows that: $$ \begin{aligned} B C \times E C &= A C \times D C \\ 3 \times E C &= 5 \times 1 \end{aligned} $$ So the length of $E C$ is $\frac{5}{3}$ meaning that the length of $B E$ is $\frac{4}{3}$. Therefore $B A$ is $3$ times the length of $B E$. Triangles $B F E$ and $B A F$ are both [[similar]] to triangle $B A E$, so the length of $B F$ is $4$ times that of $F E$, and of $A F$ is $4$ times of $B F$. Putting these together, $B F$ is $\frac{4}{17}$ of the length of $A E$. As the rectangle and triangle have the same "height", the fraction of the rectangle that is shaded is given by comparing half the length of $B D$ to the length of $G H$. Half the length of $B D$ is the length of $B F$, and the length of $B D$ is a diameter of the circle, which has the same length as $A E$. Therefore $\frac{4}{17}$ths of the rectangle is shaded.