Question

In the figure 5.14, $\square$ABCD is a trapezium P and Q are the midpoints of non-parallel sides AB, and DC respectively.

Answer 1

Construction:
Join AC intersecting PQ at O.

We know that a segment joining the midpoint of non-parallel sides of a trapezium is parallel to its parallel sides.

$$\begin{gathered} \mathrm{Consider}∆\mathrm{ABC}. \\ \mathrm{P}\mathrm{is}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{AB}\mathrm{and}\mathrm{PO}\parallel \mathrm{BC}. \\ \mathrm{O}\mathrm{is}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{AC}(\mathrm{By}\mathrm{the}\mathrm{converse}\mathrm{of}\mathrm{midpoint}\mathrm{theorem}) \\ \mathrm{By}\mathrm{midpoint}\mathrm{theorem},\mathrm{we}\mathrm{have}: \\ \mathrm{PO}=\frac{1}{2}\mathrm{BC}...\left(1\right) \\ \mathrm{Now},\mathrm{consider}∆\mathrm{ADC}. \\ \mathrm{Q}\mathrm{and}\mathrm{O}\mathrm{are}\mathrm{the}\mathrm{midpoint}s\mathrm{of}\mathrm{DC}\mathrm{and}\mathrm{AC},\mathrm{respectively},\mathrm{and}\mathrm{OQ}\parallel \mathrm{AD}. \\ \therefore \mathrm{QO}=\frac{1}{2}\mathrm{AD}...\left(2\right) \\ \mathrm{Adding}\mathrm{equation}s\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ \mathrm{PO}+\mathrm{QO}=\frac{1}{2}\mathrm{BC}+\frac{1}{2}\mathrm{AD} \\ \Rightarrow \mathrm{PQ}=\frac{1}{2}\left(\mathrm{BC}+\mathrm{AD}\right) \end{gathered}$$