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Questions  >>  Olympiad-Math  >>  Class 10
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Q)

In the adjoining figure P and Q are points on the sides $AB$ and $AC$ respectively of $\Delta ABC$ such that $AP = 3.5 \;cm, PB = 7\;cm, AQ = 3 \;cm, QC = 6\; cm$ and $PQ = 4.5\; cm$. The measure of $BC$ is equal to

$\begin{array}{1 1} 13.5\;cm\\ 15\;cm \\ 12.5 \;cm\\ 9\;cm \end{array} $

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A)
Solution :
In $\Delta ABC$ ,
=> $\large\frac{AQ}{QC}=\frac{AP}{PB}$
=> $\large\frac{3}{6} = \frac{3.5}{7} $
$=> \large\frac{1}{2}=\frac{1}{2}$
Since $\large\frac{AQ}{QC} =\frac{AP}{PB}$
Therefore $\large\frac{AQ}{AC} = \frac{QP}{BC}$
=> $\large\frac{3}{9}=\frac{4.5}{BC}$
$BC= 13.5 \;cm$
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