Question

ABCD is a trapezium in which AB || DC and P and Q are the points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, then AD = _________.

Answer 1

ABCD is a trapezium such that AB || CD. P and Q are the points on AD and BC, respectively such that PQ || CD.

Join BD. O is the point of intersection of PQ and BD.

Now,

AB || CD and PQ || CD

∴ AB || PQ || CD

In ∆ADB, OP || AB.

Using basic proportionality theorem, we have

$\frac{\mathrm{DP}}{\mathrm{PA}}=\frac{\mathrm{OD}}{\mathrm{OB}}.....\left(1\right)$

In ∆BCD, OQ || CD.

Using basic proportionality theorem, we have

$$\begin{gathered} \frac{\mathrm{BQ}}{\mathrm{QC}}=\frac{\mathrm{OB}}{\mathrm{OD}} \\ \Rightarrow \frac{\mathrm{QC}}{\mathrm{BQ}}=\frac{\mathrm{OD}}{\mathrm{OB}}.....\left(2\right) \end{gathered}$$

From (1) and (2), we have

$$\begin{gathered} \frac{\mathrm{DP}}{\mathrm{PA}}=\frac{\mathrm{QC}}{\mathrm{BQ}} \\ \Rightarrow \frac{18\mathrm{cm}}{\mathrm{PA}}=\frac{15\mathrm{cm}}{35\mathrm{cm}} \\ \Rightarrow \mathrm{PA}=\frac{18\times 35}{15}=42\mathrm{cm} \end{gathered}$$

∴ AD = AP + PD = 42 cm + 18 cm = 60 cm

ABCD is a trapezium in which AB || DC and P and Q are the points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, then AD = ___60 cm___.