Question
$\triangle\text{ABC}\sim\triangle\text{DEF}$ and their areas are respectively $100\ cm^2 $ and $49\ cm^2.$ If the altitude of $\triangle\text{ABC}$ is $5\ cm$, the corresponding altitude of $\triangle\text{DEF}.$
✓
Answer
It is given that $\triangle\text{ABC}\sim\triangle\text{DEF}.$
Therefore, the ratio of the areas of these triangles will be equal to the ratio of squares of their corresponding sides.
Also, the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.
Let the altitude of $\triangle\text{ABC}$ be $AP,$ drawn from $A$ to $BC$ to meet $BC$ at $P$ and the altitude of $\triangle\text{DEF}$ be $DQ,$ drawn from $D$ to meet EF at $Q$
Then,
$\frac{\text{ar}(\triangle\text{ACB})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AP}^2}{\text{DQ}^2}$
$\Rightarrow\frac{100}{49}=\frac{\text{5}^2}{\text{DQ}^2}$
$\Rightarrow\frac{100}{49}=\frac{\text{25}}{\text{DQ}^2}$
$\Rightarrow\text{DQ}^2=\frac{49\times25}{100}$
$\Rightarrow\text{DQ}=\sqrt{\frac{49\times25}{100}}$
$\Rightarrow\text{DQ}=3.5\text{cm}$
Hence, the altitude of $\triangle\text{DEF}$ is $3.5\ cm$
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