The areas of two similar triangles are 48 cm² and 75 cm² respectively. If the altitude of the first triangle is 3.6 cm, find the corresponding altitude of the other.

Let the length of altitude of other triangle be x cm

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding altitudes.

$\therefore \dfrac{\text{Area of first Δ}}{\text{Area of second Δ}} = \Big(\dfrac{\text{Altitude of first Δ}}{\text{Altitude of second Δ}}\Big)^2 \\[1em] \Rightarrow \dfrac{48}{75} = \Big(\dfrac{3.6}{x}\Big)^2 \\[1em] \Rightarrow \dfrac{16}{25} = \Big(\dfrac{3.6}{x}\Big)^2 \\[1em] \Rightarrow \sqrt{\dfrac{16}{25}} = \dfrac{3.6}{x} \\[1em] \Rightarrow \dfrac{4}{5} = \dfrac{3.6}{x} \\[1em] \Rightarrow x = \dfrac{18}{4} \\[1em] \Rightarrow x = 4.5 \text{ cm.}$

Hence, the length of altitude of other triangle = 4.5 cm.