# Solution to the [[Triangle and Semi-Circle in a Circle]] Puzzle +-- {.image} [[TriangleandSemiCircleinaCircle.png:pic]] > The semicircle and equilateral triangle have the same base. What fraction of the circle is shaded? =-- ## Solution by Lengths in an [[Equilateral Triangle]] and in a [[Circle]] +-- {.image} [[TriangleandSemiCircleinaCircleAnnotated.png:pic]] =-- Consider the diagram as labelled above, in which $O$ is the centre of the circle and $E$ is the centre of the semi-circle. As triangle $A B C$ is an [[equilateral triangle]], angle $E \hat{A} D$ is $60^\circ$, and sides $A E$ and $D E$ are both radii of the semi-circle so are the same length, hence triangle $A E D$ is also an equilateral triangle. By a similar argument, so is triangle $B E F$, and then so must be triangle $D E F$. This also establishes $D$ as the [[midpoint]] of $A C$ since $A D$ is the same length as $A E$, and $A C$ is the same length as $A B$. So triangle $C D F$ is also equilateral, and the four smaller triangles are all [[congruent]]. The missing [[segment]] $D F$ which is unshaded in triangle $C D F$ matches the shaded segment $F B$. Thus the top and right shaded parts together form a complete equilateral triangle. This equilateral triangle is congruent to triangle $A E D$, so the total shaded area is equivalent to the [[sector]] $A E D$. Therefore the area of the shaded regions is one sixth of the area of circle with diameter $A B$. Using the relationship between the side length and height of an [[equilateral triangle]], the radius of the semi-circle, $E B$, has length $\frac{\sqrt{3}}{2}$ times that of $O B$, which is the radius of the outer circle. The area of the circle with radius $E B$ is therefore $\frac{3}{4}$ths of the area of the outer circle, and so the shaded regions have area $\frac{1}{6} \times \frac{3}{4} = \frac{1}{8}$th of the area of the full circle.