WAEC Mathematics 2020 Paper

Concept:A parallelogram and a triangle on the same base and between the same parallel lines have the same perpendicular height.Explanation:In the diagram, parallelogram $PTRS$ and $\triangle SUR$ both stand on base $\overline{SR}$.Since $\overline{PU}$ is parallel to $\overline{SR}$, the perpendicular distance from $U$ to $\overline{SR}$ is the same as the height of parallelogram $PTRS$.Let this height be $h$.Area of $\triangle SUR = \frac{1}{2}\times \overline{SR}\times h$.So $24 = \frac{1}{2}\times 8 \times h$.This gives $24 = 4h$, hence $h = 6$ cm.Now, area of parallelogram $PTRS = \overline{SR}\times h$.Area $= 8\times 6 = 48$ cm$^2$.Answer:B. $48$ cm$^2$