WAEC Mathematics 2014 Paper

Concept:When side $QR$ is produced to $M$ so that $QR = RM$, the extra triangle formed has the same area as half the parallelogram. Compare the quadrilateral areas by splitting them into triangles.Explanation:Let the area of parallelogram $PQRS$ be $2x$.The diagonal $PR$ divides parallelogram $PQRS$ into two equal triangles.So, $[\triangle PRS] = x$ and $[\triangle PQR] = x$.Since $Q$, $R$, $M$ are collinear and $QR = RM$, triangles $\triangle QRS$ and $\triangle MRS$ have equal bases on the same line and the same height from $S$.Hence, $[\triangle MRS] = [\triangle QRS] = x$.Now, area of $PQMS$ = area of parallelogram $PQRS$ + area of $\triangle MRS$.So, $[PQMS] = 2x + x = 3x$.Also, area of $PRMS$ = area of $\triangle PRS$ + area of $\triangle MRS$.So, $[PRMS] = x + x = 2x$.Therefore, required fraction $= \frac{[PRMS]}{[PQMS]} = \frac{2x}{3x} = \frac{2}{3}$.Answer:$\frac{2}{3}$Correct option: C.