Notes
circle in a trapezium solution

Solution to the Circle in a Trapezium Puzzle

Solution by Congruent Triangles, Angle Between a Radius and Tangent, Area of a Trapezium, and Area of a Circle

In the above diagram, the point labelled $O$ is the centre of the circle and the points $B$, $D$, $F$, and $H$ are where the sides of the trapezium meet the circle.

Consider triangles $A O H$ and $A O B$. These are both right-angled triangles since $A \hat{H} O$ and $O \hat{B} A$ are both the angle between a radius and tangent. They share the side $A O$ as their hypotenuse, and the sides $O H$ and $O B$ are the same length as they are radii of the same circle. Therefore triangles $A O H$ and $A O B$ are congruent. So the lengths of $A H$ and $A B$ are the same. A similar argument holds at each other vertex of the trapezium. So writing $a$, $b$, $c$, and $d$ for the lengths of $A B$, $B C$, $E F$, and $F G$ respectively, the perimeter of the trapezium is $2a + 2 b + 2 c + 2 d$ and this is equal to $15$.

Let $r$ be the radius of the circle, then from considering the circumference of the circle, $2 \pi r = 10$.

The area of the trapezium is given by:

The area of the circle is given by:

The fraction of the trapezium that is shaded is therefore $\frac{50}{75} = \frac{2}{3}$.