The volume of a right circular cone is 9856 cm³ and the area of its base is 616 cm². Find

(i) the slant height of the cone.

(ii) total surface area of the cone.

Given, area of base = 616 cm².

Area of base = πr².

∴ πr² = 616

$\Rightarrow \dfrac{22}{7} \times r^2 = 616 \\[1em] \Rightarrow r^2 = \dfrac{616 \times 7}{22} \\[1em] \Rightarrow r^2 = \dfrac{4312}{22} \\[1em] \Rightarrow r^2 = 196 \\[1em] \Rightarrow r = \sqrt{196} = 14 \text{ cm}.$

Volume of cone = $\dfrac{1}{3}πr^2h$

$\therefore \dfrac{1}{3}πr^2h = 9856 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times (14)^2 \times h = 9856 \\[1em] \Rightarrow h = \dfrac{9856 \times 3 \times 7}{22 \times 14 \times 14} \\[1em] \Rightarrow h = \dfrac{206976}{4312} \\[1em] \Rightarrow h = 48 \text{ cm}.$

(i) We know that,

    l² = r² + h²
⇒ l² = 14² + 48²
⇒ l² = 196 + 2304
⇒ l² = 2500
⇒ l = $\sqrt{2500}$ = 50 cm.

Hence, the slant height of cone = 50 cm.

(ii) Total surface area of cone = πr(l + r).

Putting values in equation we get,

Total surface area of cone = $\dfrac{22}{7} \times 14 \times (50 + 14)$ = 22 × 2 × 64 = 2816 cm².

Hence, the total surface area of cone = 2816 cm².