$\triangle A B C \sim \triangle D E F$. If the areas of $\triangle A B C$ and $\triangle D E F$ are $100 \mathrm{~cm}^2$ and $81 \mathrm{~cm}^2$, respectively and the altitude of $\triangle D E F$ is $6.3 \mathrm{~cm}$, then the corresponding altitude of $\triangle A B C$ is:
Answer & explanation
Correct answer: option 3
Let the altitude of \(\Delta \)ABC be x
So,
\(\frac{100}{81}\) = \( {x }^{2 } \)/\( {6.3}^{2 } \)
= \(\frac{√100}{√81}\) = \(\frac{x}{6.3}\)
= \(\frac{10}{9}\) = \(\frac{x}{6.3}\)
= 9 × x = 10 × 6.3
= x = 7 cm
Therefore, the altitude of \(\Delta \)ABC is 7 cm.