# Solution to the Triangle Over a Circle Puzzle _Posted on Feb 17, 2021_ +-- {.image} [[TriangleOveraCircle.png:pic]] > The triangle is equilateral. What’s the area of the circle? =-- ## Solution by [[Angles in the Same Segment are Equal]] and [[Equilateral Triangles]] +-- {.image} [[TriangleOveraCircleLabelled.png:pic]] =-- In the above diagram, point $O$ is the centre of the circle and $E$ is the point on the circumference with the property that the lengths of $A E$ and $D E$ are equal, so that triangle $A E D$ is [[isosceles]]. Point $F$ is where the line through $E O$ extends to meet $A D$. By the result that [[angles in the same segment are equal]], angles $A \hat{E} D$ and $A \hat{C} D$ are equal. Since angle $A \hat{C} D$ is the [[interior angle]] of an [[equilateral triangle]], it is $60^\circ$. So triangle $A E C$ is an isosceles triangle with an angle of $60^\circ$ and so is equilateral. The [[height of an equilateral triangle]] is $\frac{\sqrt{3}}{2}$ of its side length, and the [[equilateral triangle|centre of its circumcircle]] is at $\frac{1}{3}$ of its height, so the length of $O E$ is $\frac{2}{3}$ of the height of triangle $A E D$ which is $\frac{\sqrt{3}}{2} \times 9$. This is the radius of the circle, and so the area of that circle is: $$ \pi \left( \frac{2}{3} \times \frac{\sqrt{3}}{2} \times 9\right)^2 = 27 \pi $$