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- Surface Areas and Volumes - Exercise 13.1
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- Surface Areas and Volumes - Exercise 13.4
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- Surface Areas and Volumes - Exercise 13.6
- Surface Areas and Volumes - Exercise 13.7
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- Surface Areas and Volumes - Exercise 13.9
NCERT Solutions Class 9 Mathematics Solutions for Surface Areas and Volumes - Exercise 13.2 in Chapter 13 - Surface Areas and Volumes
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The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to
move once over to level a playground. Find the area of the playground in \mathrm{m}^{2} ? (Assume π=22/7)
Answer:
A roller is shaped like a cylinder.
Let h be the height of the roller and r be the radius.
h = Length of roller = 120 cm
Radius of the circular end of roller = r = (84/2) cm = 42 cm
Now, CSA of roller = 2 πrh
= 2 x 22/7 x 42 x 120
=31680 \mathrm{cm}^{2}
Area of field = 500 × CSA of roller
\begin{aligned} &=(500 \times 31680) \mathrm{cm}^{2}\\ &=15840000 \mathrm{cm}^{2}\\ &=1584 \mathrm{m}^{2}\\ &\text { Therefore, area of playground is } 1584 \mathrm{m}^{2} \text { . Answer! } \end{aligned}
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