In the figure (2) given below, the perimeter of ∆ABC is 37 cm. If the lengths of the altitudes AM, BN and CL are 5x, 6x, and 4x respectively, calculate the lengths of the sides of ∆ABC.

Let us consider BC = p, CA = q.

From figure,

Perimeter of ∆ABC = AB + BC + CA

⇒ 37 = AB + p + q

⇒ AB = 37 – (p + q)

Area of ∆ABC = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × AM

= $\dfrac{1}{2}$ × p × $5x$

∴ Area of ∆ABC = $\dfrac{1}{2}$ × p × $5x$ ………(i)

Also,

Area of ∆ABC = $\dfrac{1}{2}$ × CA × BN

= $\dfrac{1}{2}$ × q × $6x$

∴ Area of ∆ABC = $\dfrac{1}{2}$ × q × $6x$ ………(ii)

Also,

Area of ∆ABC = $\dfrac{1}{2}$ × AB × CL

= $\dfrac{1}{2}$ × (37 - p - q) × $4x$

∴ Area of ∆ABC = $\dfrac{1}{2}$ × (37 - p - q) × $4x$ ………(iii)

From equation (i) and (ii) we get,

$\Rightarrow \dfrac{1}{2} \times p \times 5x = \dfrac{1}{2} \times q \times 6x \\[1em] \Rightarrow p = q \times \dfrac{6x}{5x} \\[1em] \Rightarrow p = \dfrac{6q}{5}.$

From equation (ii) and (iii) we get,

$\Rightarrow \dfrac{1}{2} \times q \times 6x = \dfrac{1}{2} \times (37 - p - q) \times 4x \\[1em] \Rightarrow q = (37 - p - q) \times \dfrac{4x}{6x} \\[1em] \Rightarrow q = \Big(37 - \dfrac{6q}{5} - q\Big) \times \dfrac{2}{3} …..\text{ (From (i))} \\[1em] \Rightarrow \dfrac{3q}{2} = \dfrac{185 - 6q - 5q}{5} \\[1em] \Rightarrow 3q \times 5 = 2(185 - 11q) \\[1em] \Rightarrow 15q = 370 - 22q \\[1em] \Rightarrow 37q = 370 \\[1em] \Rightarrow q = \dfrac{370}{37} \\[1em] \Rightarrow q = 10.$

⇒ CA = q = 10 cm.

⇒ BC = p = $\dfrac{6q}{5} = \dfrac{6}{5} \times 10 = 12$ cm.

⇒ AB = (37 - p - q) = (37 - 12 - 10) = 15 cm.

Hence, AB = 15 cm, BC = 12 cm and CA = 10 cm.