Question

The radius and height of a cylinder are in the ratio 1:2. If its volume is 54
$\pi$
, find its curved surface area in terms of n.​

Answer 1

Step-by-step explanation:

Correct Question:-

The radius and height of a cylinder are in the ratio 1:2. If its volume is 54π . Find its curved surface area in terms of π

Solution:-

The radius and height of a cylinder are in the ratio 1:2.

Let the common ratio of radius and height be x

The radius of the cylinder (r) = x

The height of the cylinder (h) = 2x

Volume of the cylinder = πr²h

⇢ πr²h = 54π

⇢ π × (x)² × 2x = 54π

⇢ π × x² × 2x = 54π

⇢ 2x³ = 54

⇢ x³ = $\dfrac{54}{2} = 27$

⇢ x = $\sqrt[3]{27}$

⇢ x = 3

Radius (r) = x = 3 units

Height (h) = 2x = 2 × 3 = 6 units

Curved surface area of cylinder = 2πrh

= 2π × (3) × (6)

= 2π × 18

= 36π

∴ Curved surface area of cylinder = 36π

Answer 2

Given:

• Ratio of Radius and Height of Cylinder = 1:2
• Volume of cylinder = 54π

Find:

• It's curved surface area in terms of π

Solution:

Let, Radius of Cylinder = x unit

Height of Cylinder = 2x unit

we, know that

$\boxed{ \rm Volume \: of \: cylinder = \pi {r}^{2} h}$

where,

• r = x unit
• h = 2x unit
• Volume of cylinder = 54π

So,

$\dashrightarrow \rm Volume \: of \: cylinder = \pi {r}^{2} h$

$\dashrightarrow \rm 54 \pi = \pi \times {x}^{2} \times 2x$

$\dashrightarrow \rm 54 \pi = \pi \times {2x}^{3}$

$\dashrightarrow \rm \dfrac{54 \pi}{ \pi} = {2x}^{3}$

$\dashrightarrow \rm {2x}^{3} = 54$

$\dashrightarrow \rm {x}^{3} = \dfrac{54}{2}$

$\dashrightarrow \rm {x}^{3} = 27$

$\dashrightarrow \rm x = \sqrt[3]{27}$

$\dashrightarrow \rm x = 3$

$\therefore \rm x = 3$

_________________________

Radius = x = 3 unit

Height = 2x = 2×3 = 6 unit

we, know

$\boxed{ \rm C.S.A \: of \: cylinder = 2\pi r h}$

where,

• r = 3 unit
• h = 6 unit

So,

$\dashrightarrow\rm C.S.A \: of \: cylinder = 2\pi r h$

$\dashrightarrow\rm C.S.A \: of \: cylinder = 2\pi \times 3 \times 6$

$\dashrightarrow\rm C.S.A \: of \: cylinder = 2\pi \times 18$

$\dashrightarrow\rm C.S.A \: of \: cylinder = 36\pi \: sq. \: units$

$\therefore\rm C.S.A \: of \: cylinder = 36\pi \: sq. \: unit$