$ \mathrm{ABCD} $ is a trapezium in which $ \mathrm{AB} \| \mathrm{DC} $ and $ \mathrm{P} $ and $ \mathrm{Q} $ are points on $ \mathrm{AD} $ and $ B C $, respectively such that $ P Q \| D C $. If $ P D=18 \mathrm{~cm}, B Q=35 \mathrm{~cm} $ and $ \mathrm{QC}=15 \mathrm{~cm} $, find $ \mathrm{AD} $.

Given: 

\( \mathrm{ABCD} \) is a trapezium in which \( \mathrm{AB} \| \mathrm{DC} \) and \( \mathrm{P} \) and \( \mathrm{Q} \) are points on \( \mathrm{AD} \) and \( B C \), respectively such that \( P Q \| D C \).

\( P D=18 \mathrm{~cm}, B Q=35 \mathrm{~cm} \) and \( \mathrm{QC}=15 \mathrm{~cm} \)

To do: 

We have to find $AD$.

Solution:

Join $BD$

In $\vartriangle ABD$,

$PO \| AB$        [Since $AB \| CD \| PQ$]

Therefore, by basic proportionality theorem,

$\Rightarrow \frac{DP}{AP}=\frac{DO}{OB}$..........(i)

In $\vartriangle BDC$,

$OQ \| DC$        [Since $AB \| CD \| PQ$]

Therefore, by basic proportionality theorem,

$\frac{BQ}{QC}=\frac{OB}{OD}$

$\Rightarrow \frac{QC}{BQ}=\frac{OD}{OB}$..............(ii)

From (i) and (ii), we get,

$\frac{DP}{AP}=\frac{QC}{BQ}$

$\Rightarrow \frac{18}{AP}=\frac{15}{35}$

$\Rightarrow AP=\frac{18\times 35}{15}$

$\Rightarrow AP=42$

Therefore,

$AD=AP+DP$

$=42+18$

$=60\ cm$

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Updated on: 10-Oct-2022

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