# Solution to the Circle in a Trapezium Puzzle +-- {.image} [[CircleinaTrapezium.png:pic]] > What fraction of the trapezium is shaded? =-- ## Solution by [[congruent|Congruent Triangles]], [[Angle Between a Radius and Tangent]], [[Area of a Trapezium]], and [[Area of a Circle]] +-- {.image} [[CircleinaTrapeziumLabelled.png:pic]] =-- 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: $$ \frac{1}{2}\Big( (a + b) + (c + d) \Big) \times (2 r) = \frac{15}{4} \times \frac{10}{\pi} = \frac{75}{2\pi} $$ The area of the [[circle]] is given by: $$ \pi \times \left(\frac{10}{2\pi} \right)^2 = \frac{25}{\pi} $$ The fraction of the trapezium that is shaded is therefore $\frac{50}{75} = \frac{2}{3}$.