Three positive numbers are in the ratio $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}.$ Find the numbers if the sum of their squares is 244.

Since, numbers are in ratio $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}.$

Hence, numbers are $\dfrac{x}{2}, \dfrac{x}{3}, \dfrac{x}{4}.$

Given, sum of squares of number = 244.

$\Rightarrow \Big(\dfrac{x}{2}\Big)^2 + \Big(\dfrac{x}{3}\Big)^2 + \Big(\dfrac{x}{4}\Big)^2 \\[1em] \Rightarrow \dfrac{x^2}{4} + \dfrac{x^2}{9} + \dfrac{x^2}{16} = 244 \\[1em] \Rightarrow \dfrac{36x^2 + 16x^2 + 9x^2}{144} = 244 \\[1em] \Rightarrow 61x^2 = 244 \times 144 \\[1em] \Rightarrow x^2 = \dfrac{35136}{61} \\[1em] \Rightarrow x^2 = 576 \\[1em] \Rightarrow x = \sqrt{576} \\[1em] \Rightarrow x = \pm 24.$

Since numbers are positive,

x ≠ -24.

Numbers = $\dfrac{x}{2} = \dfrac{24}{2} = 12, \dfrac{x}{3} = \dfrac{24}{3} = 8, \dfrac{x}{4} = \dfrac{24}{4} = 6.$

Hence, numbers are 6, 8, 12.