# Solution to the [[Three Lengths in A Semi-Circle]] Puzzle +-- {.image} [[ASemiCircle.jpeg:pic]] > What’s the area of this semicircle? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[ThreeLengthsinaSemiCircleLabelled.jpeg:pic]] =-- Consider the diagram as labelled above, in which $F$ is the centre of the semi-circle. Then triangles $D E F$ and $F A B$ are [[right-angled triangles|right-angled]] so [[Pythagoras' theorem]] applies. The line segments $F B$ and $D F$ are both radii of the semi-circle, so have the same length. Let $r$ be the radius of the semi-circle and $x$ the length of the line segment $E F$, so line segment $F A$ has length $6-x$. Then [[Pythagoras' theorem]] says: $$ \begin{aligned} r^2 &= x^2 + 4^2 \\ r^2 &= (6-x)^2 + 2^2 = 6^2 - 12x + x^2 + 2^2 \end{aligned} $$ Subtracting these gives $16 = 40 - 12x$ and so $x = 2$. Then $r^2 = 2^2 + 4^2 = 20$. The area of the semi-circle is therefore $\frac{1}{2} \pi r^2 = 10 \pi$. ## Solution by [[Perpendicular Bisector of a Chord]] and [[Pythagoras' Theorem]] In this version, the diagram is as above but $F$ is taken so that line segment $E F$ has length $2$ and therefore line segment $F A$ has length $4$. Triangles $D E F$ and $F A B$ are therefore [[right-angled triangle|right-angled]] with two lengths the same, and hence are [[congruent]]. This means that $F B$ and $F D$ have the same length, so $F$ lies on the [[perpendicular bisector]] of [[chord]] $D B$. As it also lies on the diameter that goes through $E$ and $A$, it must be the centre of the circle. Therefore $F D$ is a radius of the circle, and [[Pythagoras' theorem]] shows that it has length $\sqrt{4^2 + 2^2} = \sqrt{20}$. The area of the semi-circle is therefore $10\pi$ as before.