A small circle is placed inside a square just touching the sides. The square is on top of a big circle with its vertices on the circle. Finally the big circle is on top of a trapezium just touching the parallel sides. Using the measurements and clue given, find the area of the trapezium.

Clue: The small circle has a circumference of 12 centimetres.

Here is a solution to "Farhans Major Trapezium" by Abbas Tejani, age 11, from Jack Hunt School, Peterborough. Congratulations Abbas, this is a very neat solution.

The area of the trapezium is half the sum of the parallel sides times the vertical height.

The vertical height of the trapezium is the diameter of the large circle.

From the clue given, the radius of the small circle is $\frac{6}{\pi }$.

Using this value and by Pythagoras theorem, the radius of the large circle is

${\displaystyle \frac{\sqrt{72}}{\pi }}$

.

Therefore, the diameter of the large circle is

${\displaystyle \frac{2\sqrt{72}}{\pi }}$

.

Hence, the area of the trapezium is (2+5) divided by 2 multiplied by the diameter of the large circle which equals 18.907 square centimetres (to three decimal places).

Well done Danniella Yule of the Mount School York for spotting that the diagram as originally drawn was impossible though this did not make any difference to the solution. Good solutions were also sent in by Laura Watson, Rosie Johns, Jenny Gross, Catherine Watton and Kayley Biggs of Davison High School, Worthing; Arti Patel, Ruoyi Sun, Lizzie Levenson and Shelli Welli at the North London Collegiate School (NCLS) Maths Puzzle Club; Alicia Maultby, Nisha Doshi, Suzanne Abbott and Christiane Eaves of the Mount School, York and Hereward Mills, Goldington Middle School, Bedford. Well done all of you!