Question

The areas of two similar triangles are $16{\text{cm}}^2$ and $36{\text{cm}}^2$ respectively. If the altitude of the first triangle is $3\text{cm}$, then the corresponding altitude of the other triangle is:

• A. $4\text{cm}$
• B. $6.5\text{cm}$
• C. $4.5\text{cm}$
• D. $6\text{cm}$

Answer 1

The correct option is C $4.5\text{cm}$

Let $A_1$ and $A_2$ be the areas of the similar triangles and $s_1$ and $s_2$ be the altitudes of these triangles respectively.

Then, using the property that the ratio of the areas of similar triangles is equals to square of the ratio of the corresponding sides or altitudes.

i.e. $\frac{A_1}{A_2}=\frac{s_1^2}{s_2^2}$

Substituting the values of $A_1,A_2$ and $s_1$ in the above equation:

$\frac{16}{36}=\frac{{\left(3\right)}^2}{s_2^2}$

$s_2^2=\frac{36\times 9}{16}$

$s_2=\frac{6\times 3}{4}$

$=4.5\text{cm}$