Notes
two quarter circles in a semi-circle ii solution

Solution to the two quarter circles in a semi-circle ii puzzle

Solution Using Circle Theorems

To find the area of the semi-circle, we need to locate its centre. There are a couple of different ways to do this, but all start with drawing the diagonals of the quarter circles.

Since the larger quarter circle is twice the size of the smaller, so also is the larger triangle, $G F C$ in the diagram, twice the size of the smaller, $F D E$. We can therefore fit two copies of $F D E$ inside $G F C$, as $G H F$ and $C H F$. This shows that the length of $F E$ is the same as the length of $G F$. (This can also be shown using Pythagoras' Theorem together with the area scale factor.)

Angle at centre is twice angle at circumference

Having established that $F E = G F$, triangle $G F E$ is isosceles with angle $G \hat{F} E = 135^\circ$. So angle $F \hat{E} G = 22.5^\circ$ and also $G \hat{E} A = 22.5^\circ$. Then angle $G \hat{B} A = 45^\circ = 2 \times G \hat{E} A$ and hence $B$ is the centre of the circle.

Symmetry

The quadrilateral $G F E B$ is a parallelogram with $G F = F E$ and so is a rhombus. Its diagonals are therefore perpendicular bisectors of each other. Since $G E$ is a chord of the circle, the centre of the circle lies on its perpendicular bisector which is (the extension of) $F B$. As $A E$ is a diameter, the intersection of $A E$ and $F B$ is therefore the centre of the circle, and this is $B$.

Calculating the Area

Once $B$ is established as the centre, calculating the area goes as follows. Since $B$ is the centre, $B G$ is a radius. This is equal in length to $F E$, which is equal to $F G$. The radius of the semi-circle is therefore the same as the radius of the larger quarter circle and so the area of the semi-circle is $8$.