ABCD is a parallelogram and O is any point inside it. If ar(ABCD) = $144{\text{cm}}^2$, find ar(△AOB) + ar(△COD) in ${\text{cm}}^2$.

Triangles AOB and COD have bases AB and DC (which are equal and parallel) and their heights from O to AB and from O to DC sum to the full distance between AB and DC. So ar(AOB) + ar(COD) = $\frac{1}{2}$(AB)(total height) = $\frac{1}{2}$ ar(ABCD) = $\frac{1}{2}(144)=72{\text{cm}}^2$.