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Earlier discussion
1 older comment kept for context.
Oluwaseun Adeyemi
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