Question

The diameter of a roller is 98cm and its length is

130cm.It takes 450 complete revolutions to move once

over to level a playground. Find the area of the

playground in m2

.​

Answer 1

Answer:

Explanation:

Given :

• Diameter of a roller (d) = 98 cm
• Length (h) = 130 cm
• Number of revolutions = 450

To Find :

• The area of the playground in m².

Solution :

We know,

• Diameter = r × 2

=> 98 = r × 2

=> r = 98/2

=> r = 49 cm

Converting units,

Radius (r) = 49 cm => 0.49 m

Length (h) = 130 cm => 1.3 m

We need to find curved surface area of cylinder,

• CSA = 2πrh

=> CSA = 2 × 3.14 × 0.49 × 1.3

=> CSA = 6.28 × 0.637

=> CSA = 4.00036 m²

Now,

• Area of playground = CSA × Number of revolutions

=> Area of playground = 4.00036 × 450

=> Area of playground = 1800.162 m²

Hence :

The area playground is 1800.162 m².

Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{diameter \: of \: roller \: = 98 \: cm} \\ &\sf{length \: of \: roller \: = 130 \: cm}\\ &\sf{number \: of \: revolutions \: = 450} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{area \: of \: playground \: in \: {m}^{2} }  \end{cases}\end{gathered}\end{gathered}$

Formula Used:-

${{ \boxed{{\bold\green{Curved \: Surface \: Area_{(Cylinder)}\: = \:2\pi rh)}}}}}$

$\large\underline\purple{\bold{Solution :- }}$

Given Dimensions of roller

• Diameter of roller = 98 cm
• Therefore, radius of roller, r = 49 cm
• Height of roller, h = 130 cm

☆ Area covered in 1 revolution is equals to Curved Surface Area of roller.

$\tt \:   \longrightarrow \:CSA_{cylinder} = 2\pi \: rh$

$\tt \:   \longrightarrow \:CSA_{cylinder} = 2 \times \dfrac{22}{7} \times 49 \times 130$

$\tt \:   \longrightarrow \:CSA_{cylinder} = 40040 \: {cm}^{2}$

$\tt \:   \longrightarrow \:CSA_{cylinder} = 40040 \times \dfrac{1}{100} \times \dfrac{1}{100} {m}^{2}$

$\tt \:   \longrightarrow \:CSA_{cylinder} = 4.004 \: {m}^{2}$

$\tt\implies \:area \: covered \: in \: 1 \: revolution \: = 4.004 \: {m}^{2}$

$\tt\implies \:area \: covered \: in \: 450 \: revolutions \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 450 \times 4.004 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1801.8 \: {m}^{2}$

$\tt\implies \:area \: of \: playground \: = 1801.8 \: {m}^{?}$