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

$</p><p>\large\underline\mathfrak{\red{GIVEN,}} \\ \\ </p><p></p><p>\sf\dashrightarrow \blue{height(H)= 120cm } \\ \\ </p><p></p><p>\sf\dashrightarrow \blue{diameter of roller= 84cm} \\ \\ </p><p></p><p>\sf\therefore \blue{radius= \dfrac{diameter}{2}} \\ \\ </p><p></p><p>\sf\dashrightarrow \blue{ \dfrac{84}{2}} \\ \\ </p><p></p><p>\sf\dashrightarrow \blue{\cancel \dfrac{84}{2}} \\ \\ </p><p> </p><p>\sf\dashrightarrow \blue{radius= 42cm} \\ \\ </p><p></p><p>\large\underline\mathfrak{\purple{TO\:FIND,}} \\ \\ </p><p></p><p>\sf\dashrightarrow \red{AREA\: OF\:PLAYGROUND } \\ \\ </p><p></p><p>\rm{\boxed{\sf{ \circ\:\: C.S.A\: OF\: CYLINDER= 2 \pi rh \:\: \circ}}} \\ \\ </p><p></p><p>\large\underline\mathtt{\purple{SOLUTION,}} \\ \\ </p><p></p><p></p><p></p><p>\purple{\text{AREA COVERED BY ROLLER IN 1 REVOLUTION = PERIMETER OF ROLLER}} \\ \\ </p><p></p><p>\sf\therefore \pink{AREA \:COVERED \:IN\: ONE\: REVOLUTION= 2 \pi r h} \\ \\ </p><p></p><p>\sf\implies \red{ 2 \times \dfrac{22}{7} \times 42 \times 120} \\ \\ </p><p></p><p>\sf\implies \blue{ 2 \times \dfrac{22}{\cancel{7}} \times \cancel{42} \times 120} \\ \\ </p><p></p><p>\sf\implies \red{2 \times 22 \times 6 \times 120} \\ \\ </p><p></p><p>\sf\implies \blue{ 44 \times 72 } \\ \\ </p><p>\sf\implies \pink{ 31680cm^2 } \\ \\ </p><p></p><p>\rm{\boxed{\sf{ \circ\:\: 31680cm^2\:\: \circ}}} \\ \\ </p><p></p><p>\sf\therefore \purple{ THE\:ROLLER\:TAKES\:1000\: REVOLUTIONS TO\:COVER\:AREA\:OF\:THAT\: PARTICULAR\:PALAYGROUND} \\ \\ </p><p></p><p>\sf\therefore \blue{we\: know,\: to\: complete\: one \:revolution\: it \:takes \:31680cm^2 \:area } \\ \\ </p><p></p><p>\sf\therefore \red{then \:area \:of\:rectangle = 1000 \times the \:area\: in\: one\: complete\: revolution} \\ \\ </p><p>\sf\implies \pink{ 1000 \times 31680 } \\ \\ </p><p></p><p>\sf\implies \green{31680000cm^2} \\ \\ </p><p>CONVERSION,</p><p></p><p>\sf\therefore \green{cm^2 \:into\:m^2} \\ \\ </p><p></p><p>\sf\therefore \blue{\dfrac{ 31680000}{ 100 \times 100}} \\ \\ </p><p></p><p>\sf\implies \red{\cancel \dfrac{ 31680000}{ 100 \times 100}} \\ \\ </p><p></p><p>\sf\implies \orange{3168 m^2} \\ \\ </p><p>\rm{\boxed{\sf{ \circ\:\: AREA\:OF\: PLAYGROUND= 3168m^2 \:\: \circ}}} \\ \\ </p><p></p><p>\rm\underline\mathrm{AREA\:OF\:PLAYGROUND\:IS\:3168cm^2}</p><p></p><p>$