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In the figure below, $ΔACB\ ∼\ ΔAPQ$. If $BC\ =\ 10\ cm$, $PQ\ =\ 5\ cm$, $BA\ =\ 6.5\ cm$, $AP\ =\ 2.8\ cm$, find $CA$ and $AQ$. Also, find the $area\ (ΔACB)\ :\ area\ (ΔAPQ)$.
"
Given:
In the given figure $ΔACB\ ∼\ ΔAPQ$.
$BC\ =\ 10\ cm$, $PQ\ =\ 5\ cm$, $BA\ =\ 6.5\ cm$ and $AP\ =\ 2.8\ cm$.
To do:
We have to find $CA$, $AQ$ and $area\ (ΔACB)\ :\ area\ (ΔAPQ)$.
Solution:
$ΔACB ∼ ΔAPQ$ (given)
Therefore,
$\frac{AB}{AQ} = \frac{BC}{PQ} = \frac{AC}{AP}$ (Corresponding parts of similar triangles)
$\frac{AB}{AQ} = \frac{BC}{PQ}$
$\frac{6.5}{AQ} = \frac{10}{5}$
$AQ=\frac{6.5}{2}$
$AQ = 3.25\ cm$
Similarly,
$\frac{BC}{PQ} = \frac{CA}{AP}$
$\frac{10}{5}=\frac{CA}{2.8}$
$CA=2\times2.8$
$CA = 5.6\ cm$
We know that,
The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.
Therefore,
$ \begin{array}{l}
\frac{ar\vartriangle ACQ}{ar\vartriangle APQ} =\left(\frac{BC}{PQ}\right)^{2}\\
\\
\frac{ar\vartriangle ACQ}{ar\vartriangle APQ} =\left(\frac{10}{5}\right)^{2}\\
\\
\frac{ar\vartriangle ACQ}{ar\vartriangle APQ} =\left(\frac{2}{1}\right)^{2}\\
\\
\frac{ar\vartriangle ACQ}{ar\vartriangle APQ} =\frac{4}{1}
\end{array}$
Therefore, $area\ (ΔACB)\ :\ area\ (ΔAPQ)$ is $4:1$.