ABCD is a trapezium in which AB is parallel to DC. Its diagonals AC and BD cut each other at O. If AO = 4 cm, OB = 6 cm and OC = 10 cm, find OD (in cm).

Because $AB$ is parallel to $DC$, the diagonals of the trapezium are cut proportionally: $\frac{AO}{OC}=\frac{OB}{OD}$. So $\frac{4}{10}=\frac{6}{OD}$, giving $OD=\frac{6\times 10}{4}=15$ cm. The distractor 2.4 comes from writing the proportion upside down $(OD=\frac{4\times 6}{10})$; 6 comes from assuming the diagonals bisect each other as in a parallelogram, so $OD=OB$; 10 comes from matching $O$ with the wrong pair and copying $OC$.