From the figure, PS = QR = YT = 7cm Area of parallelogram PQRS = 56cm 56 = base x height, where base = 7 7 x h = 56cm, h = \(\frac{56}{7}\) = 8cm Area of trapezium \(\frac{1}{2}\) (sum of two sides)x height where two sides are QT and PS but QT = QR + RY + YT = 7 +9 + 7 = 23cm Area of trapezium PQTS = \(\frac{1}{2}\)(23 + 7) x 8 \(\frac{1}{2}\) x 30 x 8 = 120cmsq