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

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

Answers (1)

Answer : [AD=60cm]

Given:- ABCD is a trapezium in which AB || DC, P and Q are points on AD and BC. Such that PQ || DC.

PD = 18 cm, BQ = 35, QC = 15 cm

To prove:- Find AD

Proof:-

Construction:- Join BD

In \Delta ABD, PO || AB

And we know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then according to basic proportionality theorem the other two sides are divided in the same ratio.

\therefore \frac{DP}{AP}=\frac{DO}{OB}

     \Rightarrow \frac{DP}{AP}=\frac{OD}{OB}\; \; \; \; \; \; \; \; \; .....(1)

In \Delta BDC, OQ || DC

Similarly by using basic proportionality theorem.

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

\Rightarrow \frac{QC}{BQ}=\frac{OD}{OB}\; \; \; \; \; \; \; \; \; .....(2)

from equation (1) and (2) we get

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

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

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

AP=42 \; cm

\\AD = AP + PD\\ 42 + 18 = 60 cm\\ \therefore AD = 60 cm

Posted by

infoexpert23

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