ABCD is a trapezium in which AB parallel DC a | vedclass

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(A) Given,a trapezium $ABCD$ in which $AB \parallel DC$. $P$ and $Q$ are points on $AD$ and $BC$ respectively such that $PQ \parallel DC$. Thus,$AB \p

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) Given,a trapezium $ABCD$ in which $AB \parallel DC$. $P$ and $Q$ are points on $AD$ and $BC$ respectively such that $PQ \parallel DC$. Thus,$AB \parallel PQ \parallel DC$.Join $BD$. Let $BD$ intersect $PQ$ at point $O$.In $	riangle ABD$,$PO \parallel AB$ (since $PQ \parallel AB$).By the Basic Proportionality Theorem $(BPT)$,we have:$\frac{AP}{PD} = \frac{BO}{OD}$  .......$(i)$In $	riangle BDC$,$OQ \parallel DC$ (since $PQ \parallel DC$).By the Basic Proportionality Theorem $(BPT)$,we have:$\frac{BQ}{QC} = \frac{BO}{OD}$  .......$(ii)$From equations $(i)$ and $(ii)$,we get:$\frac{AP}{PD} = \frac{BQ}{QC}$Substitute the given values $PD = 18 \, cm$,$BQ = 35 \, cm$,and $QC = 15 \, cm$:$\frac{AP}{18} = \frac{35}{15}$$AP = \frac{35 	imes 18}{15}$$AP = \frac{7 	imes 18}{3} = 7 	imes 6 = 42 \, cm$Therefore,$AD = AP + PD = 42 \, cm + 18 \, cm = 60 \, cm$.

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

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