Question

the diameter of a 120 cm long roller is 84 cm It takes 1000 complete revolutions in moving once over to level a playground what is the area of the playground​

Answer 1

$\huge\mathtt{\fbox{\red{Answer✍︎}}}$

$\large\underline\mathfrak{\red{GIVEN,}}$

$\sf\dashrightarrow \blue{height(H)= 120cm }$

$\sf\dashrightarrow \blue{diameter of roller= 84cm}$

$\sf\therefore \blue{radius= \dfrac{diameter}{2}}$

$\sf\dashrightarrow \blue{ \dfrac{84}{2}}$

$\sf\dashrightarrow \blue{\cancel \dfrac{84}{2}}$

$\sf\dashrightarrow \blue{radius= 42cm}$

$\large\underline\mathfrak{\purple{TO\:FIND,}}$

$\sf\dashrightarrow \red{AREA\: OF\:PLAYGROUND }$

FORMULA

$\rm{\boxed{\sf{ \circ\:\: C.S.A\: OF\: CYLINDER= 2 \pi rh \:\: \circ}}}$

$\large\underline\mathtt{\purple{SOLUTION,}}$

© ATQ,

$\purple{\text{AREA COVERED BY ROLLER IN 1 REVOLUTION = PERIMETER OF ROLLER}}$

$\sf\therefore \pink{AREA \:COVERED \:IN\: ONE\: REVOLUTION= 2 \pi r h}$

$\sf\implies \red{ 2 \times \dfrac{22}{7} \times 42 \times 120}$

$\sf\implies \blue{ 2 \times \dfrac{22}{\cancel{7}} \times \cancel{42} \times 120}$

$\sf\implies \red{2 \times 22 \times 6 \times 120}$

$\sf\implies \blue{ 44 \times 72 }$

$\sf\implies \pink{ 31680cm^2 }$

$\rm{\boxed{\sf{ \circ\:\: 31680cm^2\:\: \circ}}}$

$\sf\therefore \purple{ THE\:ROLLER\:TAKES\:1000\: REVOLUTIONS TO\:COVER\:AREA\:OF\:THAT\: PARTICULAR\:PALAYGROUND}$

$\sf\therefore \blue{we\: know,\: to\: complete\: one \:revolution\: it \:takes \:31680cm^2 \:area }$

$\sf\therefore \red{then \:area \:of\:rectangle = 1000 \times the \:area\: in\: one\: complete\: revolution}$

$\sf\implies \pink{ 1000 \times 31680 }$

$\sf\implies \green{31680000cm^2}$

CONVERSION,

$\sf\therefore \green{cm^2 \:into\:m^2}$

$\sf\therefore \blue{\dfrac{ 31680000}{ 100 \times 100}}$

$\sf\implies \red{\cancel \dfrac{ 31680000}{ 100 \times 100}}$

$\sf\implies \orange{3168 m^2}$

$\rm{\boxed{\sf{ \circ\:\: AREA\:OF\: PLAYGROUND= 3168m^2 \:\: \circ}}}$

$\rm\underline\mathrm{AREA\:OF\:PLAYGROUND\:IS\:3168cm^2}$

Answer 2

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✮ Question :

➀ The diameter of a 120 cm long roller is 84 cm .It takes 1000 complete revolutions in moving once over to level a playground what is the area of the playground.

✮ Given:

➡ Number of revolution =1000

➡ diameter =84cm

★solution :

$\\ \implies \sf \: radius \: of \: the \: roller \: = \frac{diameter}{2}$

$\\ \implies \sf \: radius \: of \: the \: roller \: = \frac{84}{2} = \boxed{ \sf{42cm}}$

$\\ \implies \sf \boxed{ \sf \: radius \: of \: the \: roller \: = 42cm}$

➡ Length of the roller =120cm

$\\ \implies \sf \: curved \: surface (cylinderical \: shape)\: area = 2\pi \: rh$

$\\ \implies \sf \: curved \: surface \: area \: = 2 \times \frac{22}{7} \times 42 \times 120$

$\\ \implies \sf \boxed{ \sf{ curved \: surface \: area \: = 31680 {cm}^{2} }}$

➡ Therefore the area covered by the roller in one revolution =31680cm²

➡ Therefore The area covered by the roller in 1000 revolution = $\sf{ 1000 \times 31680 =31680000cm²}$

➡ therefore the area of playground =3168m²

$\\ \implies \sf \: l \: \times b \: = 3168$

$\implies \sf \: l \times 32 = 3168$

$\\ \implies \sf \boxed{\sf{l = 99m}}$

$\large\boxed{\red\bigstar{\sf{Area \: of \: playground =3168m²}}}$

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Additional information :

$\\ \pink\bigstar\large\sf{ Surface \: area \: of \: cube =6s^2}$

$\\ \pink\bigstar\large\sf{ Surface \: area \: of \: cuboid =2(lb+lh+bh)}$

$\\ \pink\bigstar\large\sf{ Surface \: area \: of \: Prism =2B+ph}$

$\\ \pink\bigstar\large\sf{ Surface \: area \: of \: cylinder = 2 \pi r^2+2 \pi rh}$

$\\ \pink\bigstar\large\sf{surface \: area \: of \: sphere =4 \pi r^2}$