The perimeter of a triangle ABC is 37 cm and the ratio between the lengths of its altitudes be 6 : 5 : 4. Find the lengths of its sides.

Let the sides be x cm, y cm and (37 - x - y) cm. Also, the length of altitudes be 6a, 5a and 4a cm.

By formula,

Area of triangle = $\dfrac{1}{2} \times$ base × height

Area of triangle ABC = $\dfrac{1}{2} \times x \times 6a = \dfrac{1}{2} \times y \times 5a = \dfrac{1}{2} \times (37 - x - y) \times 4a$

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{2} \times x \times 6a = \dfrac{1}{2} \times y \times 5a \\[1em] \Rightarrow 3ax = \dfrac{5ay}{2} \\[1em] \Rightarrow 6ax = 5ay \\[1em] \Rightarrow 6x = 5y \\[1em] \Rightarrow x = \dfrac{5y}{6} \text{ ………(1)}$

Solving L.H.S. of the equation :

$\Rightarrow \dfrac{1}{2} \times y \times 5a = \dfrac{1}{2} \times (37 - x - y) \times 4a \\[1em] \Rightarrow 5ay = 4a(37 - x - y) \\[1em] \Rightarrow 5y = 4(37 - x - y) \\[1em] \Rightarrow 5y = 148 - 4x - 4y \\[1em] \Rightarrow 148 - 4x - 4y - 5y = 0 \\[1em] \Rightarrow 148 - 4x - 9y = 0$

Substituting value of x from equation (1) in above equation, we get :

$\Rightarrow 148 - 4 \times \dfrac{5y}{6} - 9y = 0 \\[1em] \Rightarrow 148 - \dfrac{10y}{3} - 9y = 0 \\[1em] \Rightarrow \dfrac{444 - 10y - 27y}{3} = 0 \\[1em] \Rightarrow 444 - 37y = 0 \\[1em] \Rightarrow 37y = 444 \\[1em] \Rightarrow y = \dfrac{444}{37} = 12.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{5y}{6} \\[1em] \Rightarrow x = \dfrac{5 \times 12}{6} \\[1em] \Rightarrow x = \dfrac{60}{6} = 10.$

Sides : x = 10 cm, y = 12 cm, (37 - x - y) = (37 - 10 - 12) = 15 cm.

Hence, sides of triangle ABC are 10 cm, 12 cm and 15 cm.