Question

A cone of height 24 cm has a curved surface area 550 cm². Find its volume

Answer 1

Given height of the cone, h = 24 cm

Let r cm and l cm be the radius and slant height of the cone respectively.

Curved surface area of the cone = 550 cm2 (Given)

∴ πrl = 550 cm2

Squaring on both sides, we get

r 2 (r 2 + 576) = (175)2 = 30625

∴ r 4 + 576 r 2 – 30625 = 0

⇒ r 4 + 625 r 2 – 49 r 2 – 30625 = 0

⇒ r 2(r 2 + 625) – 49 (r 2 + 625) = 0

⇒ (r 2 + 625) (r 2 – 49) = 0

⇒ r 2 – 49 = 0 ( ∵ r 2 + 625 ≠ 0)

⇒ r 2 = 49

⇒ r = 7 ( ∵ Radius cannot be negative)

Radius of the cone = 7 cm

∴ Volume of the conditions

Thus, the volume of cone is 1232 cm3.

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Answer 2

Answer:

Let r cm be the radius of the base and l cm the slant height. Then,

• l² = r² + 24²

$\implies$ l² = r² + 576

$\implies$ l = $\sf\sqrt{r^2+576}$

Now,

• Curved surface area = 550 cm²

$\implies$ πrl = 550

$\sf\frac{22}{7}×r×$$\sf\sqrt{r^2+576}=550$

r$\sf\sqrt{r^2+576}=550×$$\sf\frac{7}{22}$

$\Rightarrow$ r²$\sf\sqrt{r^2+576}=25×7$

$\Rightarrow$ $\sf{r^2(r^2+576)=(25×7)^2}$

$\Rightarrow$ $\sf{r^4+576r^2-(25^2×7^2)=0}$

$\Rightarrow$ $\sf{r^4+576r^2-(625×49)=0}$

$\Rightarrow$ $\sf{r^4+625r^2-49r^2-625×49=0}$

$\Rightarrow$ $\sf{r^2(r^2-625)-49r^2-625×49=0}$

$\Rightarrow$ $\sf{r^2(r^2+625)-49(r^2+625)=0}$

$\Rightarrow$ $\sf{(r^2+625)(r^2-49)=0}$

$\Rightarrow$ $\sf{r^2-49=0}$

$\Rightarrow$ r = 7

_____________

•°• $\sf{Volume=}$$\sf\frac{1}{3}πr^2h=$

$\sf\frac{1}{3}×$$\sf\frac{22}{7}×7×7×24\:cm^3=1232\:cm^3$