The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm³, find its slant height. (Take π = 3.14)

Let radius of cone (r) = 5a and height (h) = 12a.

Volume of cone (V) = $\dfrac{1}{3}πr^2h$

Given, V = 2512 cm³.

$\therefore \dfrac{1}{3}πr^2h = 2512 \\[1em] \Rightarrow \dfrac{1}{3} \times 3.14 \times (5a)^2 \times 12a = 2512 \\[1em] \Rightarrow \dfrac{3.14 \times 25a^2 \times 12a}{3} = 2512 \\[1em] \Rightarrow \dfrac{942a^3}{3} = 2512 \\[1em] \Rightarrow 314a^3 = 2512 \\[1em] \Rightarrow a^3 = \dfrac{2512}{314} \\[1em] \Rightarrow a^3 = 8 \\[1em] \Rightarrow a^3 = (2)^3 \\[1em] \Rightarrow a = 2 \text{ cm}.$

r = 5a = 5 × 2 = 10 cm.

h = 12a = 12 × 2 = 24 cm.

$\text{l} = \sqrt{r^2 + h^2} \\[1em] = \sqrt{10^2 + 24^2} \\[1em] = \sqrt{100 + 576} \\[1em] = \sqrt{676} \\[1em] = 26 \text{ cm}.$

Hence, the slant height of cone is 26 cm.