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If $\vartriangle ABC \sim\vartriangle DEF$, $AB = 4\ cm,\ DE = 6\ cm,\ EF = 9\ cm$ and $FD = 12\ cm$, find the perimeter of $ABC$.
Given: $\vartriangle ABC ~ \vartriangle DEF$,\ $AB = 4 cm,\ DE = 6 cm,\ EF = 9 cm$ and $FD = 12\ cm$
To do: To find the perimeter of ABC.
Solution:
As per the dimensions give in the questions,
Now, we have to find out the perimeter of $\vartriangle ABC$
Let $\vartriangle ABC\sim\vartriangle DEF$
So, $\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}$
Consider, $\frac{AB}{DE}=\frac{AC}{DE}$
$\frac{4}{6}=\frac{AC}{12}$
By cross multiplication we get,
$\Rightarrow AC=\frac{( 4\times 12)}{6}$
$\Rightarrow AC=\frac{48}{6}$
$\Rightarrow AC=8\ cm$
Then, consider $\frac{AB}{DE}=\frac{BC}{EF}$
$\Rightarrow \frac{4}{6}=\frac{BC}{9}$
$\Rightarrow BC=\frac{( 4\times 9)}{6}$
$\Rightarrow BC=\frac{36}{6}$
$\Rightarrow BC=6\ cm$
Therefore, the perimeter of $\vartriangle ABC=AB+BC+AC$
$=4+6+8$
$=18\ cm$.
Updated on: 10-Oct-2022
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