The areas of two similar triangles are 36 cm² and 25 cm². If an altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other triangle.

Let the length of altitude of other △ 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.

$\therefore \dfrac{\text{Area of first △}}{\text{Area of second △}} = \Big(\dfrac{\text{Altitude of first △}}{\text{Altitude of second △}}\Big)^2 \\[1em] \Rightarrow \dfrac{36}{25} = \Big(\dfrac{2.4}{x}\Big)^2 \\[1em] \Rightarrow \dfrac{36}{25} = \dfrac{5.76}{x^2} \\[1em] \Rightarrow x^2 = \dfrac{5.76 \times 25}{36} \\[1em] \Rightarrow x^2 = \dfrac{144}{36} \\[1em] \Rightarrow x^2 = 4 \\[1em] \Rightarrow x^2 - 4 = 0 \\[1em] \Rightarrow (x - 2)(x + 2) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 2 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -2.$

Since length cannot be negative so, x ≠ -2.

Hence, the length of altitude of other triangle = 2 cm.