In trapezium ABCD the side AB is parallel to the side DC and the diagonals meet at O. The area of triangle AOB is $16{\text{cm}}^2$ and the area of triangle COD is $25{\text{cm}}^2$. What is the area of triangle BOC?

AB ∥ DC gives ∠OAB = ∠OCD and ∠OBA = ∠ODC, so triangles AOB and COD are similar. Their areas are 16 : 25, so their sides — and with them AO : OC — are in the ratio 4 : 5. Triangles AOB and BOC share the apex B and have bases AO and OC on the same straight line, so their areas are also in the ratio AO : OC = 4 : 5, giving area BOC = $16\times 5/4=20{\text{cm}}^2$. It is the geometric mean, √$(16\times 25)=20$, not the average: $22.5{\text{cm}}^2$ is that trap, and $41{\text{cm}}^2$ is simply the sum.