Question

In trapezium ABCD, side AB || side PQ || side ∆C, AP = 15, PD = 12, QC = 14, Find BQ.

Answer 1

Construction: Join BD intersecting PQ at X.

In △ABD, PX || AB
$\frac{\mathrm{DP}}{\mathrm{PA}}=\frac{\mathrm{DX}}{\mathrm{XB}}...\left(1\right)\left(\mathrm{By}\mathrm{Basic}\mathrm{proportionality}\mathrm{theorem}\right)$
In △BDC, XQ||DC
$\frac{\mathrm{DX}}{\mathrm{XB}}=\frac{\mathrm{CQ}}{\mathrm{QB}}...\left(2\right)\left(\mathrm{By}\mathrm{Basic}\mathrm{proportionality}\mathrm{theorem}\right)$
From (1) and (2), we get
$$\begin{gathered} \frac{\mathrm{DP}}{\mathrm{PA}}=\frac{\mathrm{CQ}}{\mathrm{QB}} \\ \Rightarrow \frac{12}{15}=\frac{14}{\mathrm{QB}} \\ \Rightarrow \mathrm{QB}=17.5 \end{gathered}$$