Notes
three lengths in a semi-circle solution

Solution to the Three Lengths in A Semi-Circle Puzzle

Solution by Pythagoras' Theorem

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 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:

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 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.