Question

☐ PQRS is an isosceles trapezium l(PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm, Distance between two parallel sides is 4 cm, find the area of ☐ PQRS

Answer 1

Draw a perpendicular from Q to line MR. Where it meets the line MR name it point N.
MN = PQ = 7 cm
In $△$PMS,
$$\begin{gathered} {\mathrm{PM}}^2+{\mathrm{SM}}^2={\mathrm{PS}}^2 \\ \Rightarrow 4^2+3^2={\mathrm{PS}}^2 \\ \Rightarrow {\mathrm{PS}}^2=16+9=25 \\ \Rightarrow \mathrm{PS}=5\mathrm{cm} \end{gathered}$$
PQRS is an isosceles trapezium so, PS = QR = 5 cm
PM = QN = 4 cm
So, NR = SM = 3 cm
SR = SM + MN + NR = 3 + 7 + 3 = 13 cm
Area of trapezium PQRS = $\frac{1}{2}\times \left(\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \mathrm{height}$
$$\begin{gathered} =\frac{1}{2}\times \left(7+13\right)\times 4 \\ =40{\mathrm{cm}}^2 \end{gathered}$$