In Delta ABC,A-P-B,A-Q-C and overline{PQ} par | vedclass

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(B) Given that in $\Delta ABC$,$\overline{PQ} \parallel \overline{BC}$.According to the Basic Proportionality Theorem (Thales Theorem),if a line is dr

Vedclass · Accepted Answer

$7.5$

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(B) Given that in $\Delta ABC$,$\overline{PQ} \parallel \overline{BC}$.According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,then the other two sides are divided in the same ratio.Therefore,$\frac{AP}{PB} = \frac{AQ}{QC} = \frac{3}{4}$.Let $AQ = 3x$ and $QC = 4x$.We know that $AC = AQ + QC$.Substituting the values,$17.5 = 3x + 4x$.$17.5 = 7x$.$x = \frac{17.5}{7} = 2.5$.Now,$AQ = 3x = 3 	imes 2.5 = 7.5$.

In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. If $\frac{AP}{PB} = \frac{3}{4}$ and $AC = 17.5$,find $AQ$.

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