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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​

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Answer:

\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\: REVOLUTIONSTO\: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 \timesthe \: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}

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