In trapezium ABCD, AB is parallel to DC and AB is twice DC. The diagonals AC and BD meet at O, so triangle AOB is similar to triangle COD.
If area(COD) = $15{\text{cm}}^2$, find area(AOB) in ${\text{cm}}^2$.

AB : DC = 2 : 1, so area(AOB) : area(COD) = $2^2:1^2=4:1$. Hence area(AOB) = $15\times 4=60{\text{cm}}^2$. Using the side ratio 2 instead of its square gives $30{\text{cm}}^2$, inverting the ratio gives $7.5{\text{cm}}^2$, and $15{\text{cm}}^2$ assumes the diagonals cut the trapezium into four equal areas, which they do not.