# Two Semi-Circles Inside a Semi-Circle +-- {.image} [[TwoSemiCirclesInsideaSemiCircle.png:pic]] > The two shaded semicircles are the same size. What fraction of the larger semicircle do they cover? =-- ## Solution by [[Angle Between a Radius and Tangent]], [[Pythagoras' Theorem]], and [[Properties of Chords]] +-- {.image} [[TwoSemiCirclesInsideaSemiCircleLabelled.png:pic]] =-- In the above diagram, point $O$ is the centre of the larger semi-circle while points $C$ and $B$ are the centres of the shaded ones. Let $r$ be the radius of the smaller semi-circles and $R$ the radius of the largest one. Then $r$ is the length of $B D$ and of $C A$, while $B C$ has length $2 r$. Both $O D$ and $O E$ have length $R$. As $A B$ is tangent to the left-hand semi-circle, angle $B \hat{A} C$ is $90^\circ$ since it is the [[angle between a radius and tangent]]. Therefore, triangle $C A B$ is [[right-angled triangle|right-angled]] and applying [[Pythagoras' theorem]] to it shows that $A B$ has length $\sqrt{3} r$. Then $A D$ has length $(1 + \sqrt{3}) r$ and so $O A$ has length $(1 + \sqrt{3})r - R$. Applying [[Pythagoras' theorem]] to triangle $C A O$ shows that the length of $O C$ is given by: $$ \sqrt{ ((1 + \sqrt{3})r - R)^2 + r^2} $$ Since $E F$ is a [[chord]] of the outer semi-circle, and $C$ is its [[midpoint]], the line $C O$ is [[perpendicular]] to $E F$ and so triangle $E C O$ is [[right-angled triangle|right-angled]]. Applying [[Pythagoras' theorem]] to this yields the equation: $$ \begin{aligned} R^2 &= ((1 + \sqrt{3})r - R)^2 + r^2 + r^2 \\ &= 6 r^2 + 2 \sqrt{3} r^2 - 2(1 + \sqrt{3})r R + R^2 \\ 2(1 + \sqrt{3})r R &= 6 r^2 + 2\sqrt{3} r^2 \\ R &= \frac{3 + \sqrt{3}}{1 + \sqrt{3}} r \\ &= \sqrt{3} r \end{aligned} $$ The area of the larger semi-circle is therefore $\frac{1}{2} \pi R^2 = \frac{3}{2} \pi r^2$. The shaded area is $\pi r^2$. Therefore the fraction that is shaded is $\frac{2}{3}$.