The area of two similar triangles are $25\ cm^2$ and $36\ cm^2$ respectively. If the altitude of the first triangle is $2.4\ cm$, find the corresponding altitude of the other.

Given:

The area of two similar triangles are $25\ cm^2$ and $36\ cm^2$ respectively.

The altitude of the first triangle is $2.4\ cm$.

To do:

We have to find the corresponding altitude of the other triangle.
 

Solution:

Let the area of the first triangle be $\vartriangle _{1}$ and the area of the second triangle be $\vartriangle _{2}$.

Similarly,

Let the altitude of the first triangle be $h _{1}$ and the altitude of the second triangle be $h _{2}$.

We know that,

The ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

Therefore,

$ \begin{array}{l}
\frac{\vartriangle _{1}}{\vartriangle _{2}} =\left(\frac{h_{1}}{h_{2}}\right)^{2}\\
\\
\frac{25}{36} =\left(\frac{2.4}{h_{2}}\right)^{2}\\
\\
\frac{h_{2}}{2.4} =\sqrt{\frac{36}{25}}\\
\\
h_{2} =\frac{6\times 2.4}{5}\\
\\
h_{2} =\frac{14.4}{5}\\
\\
h_{2} =2.88\ cm
\end{array}$

The altitude of the second triangle is $2.88\ cm$.