English, asked by Anonymous, 5 months ago

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​

Answers

Answered by ItzCaptonMack
20

\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}

Answered by Anonymous
1

\huge\mathtt{\fbox{\red{Answer✍︎}}}Answer✍︎</p><p></p><p>\large\underline\mathfrak{\red{GIVEN,}}GIVEN,</p><p></p><p>\sf\dashrightarrow \blue{height(H)= 120cm }⇢height(H)=120cm</p><p></p><p>\sf\dashrightarrow \blue{diameter of roller= 84cm}⇢diameterofroller=84cm</p><p></p><p>\sf\therefore \blue{radius= \dfrac{diameter}{2}}∴radius=2diameter</p><p></p><p>\sf\dashrightarrow \blue{ \dfrac{84}{2}}⇢284</p><p></p><p>\sf\dashrightarrow \blue{\cancel \dfrac{84}{2}}⇢284</p><p></p><p>\sf\dashrightarrow \blue{radius= 42cm}⇢radius=42cm</p><p></p><p>\large\underline\mathfrak{\purple{TO\:FIND,}}TOFIND,</p><p></p><p>\sf\dashrightarrow \red{AREA\: OF\:PLAYGROUND }⇢AREAOFPLAYGROUND</p><p></p><p>FORMULA</p><p></p><p>\rm{\boxed{\sf{ \circ\:\: C.S.A\: OF\: CYLINDER= 2 \pi rh \:\: \circ}}}∘C.S.AOFCYLINDER=2πrh∘</p><p></p><p>\large\underline\mathtt{\purple{SOLUTION,}}SOLUTION,</p><p></p><p>© ATQ,</p><p></p><p>\purple{\text{AREA COVERED BY ROLLER IN 1 REVOLUTION = PERIMETER OF ROLLER}}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}∴AREACOVEREDINONEREVOLUTION=2πrh</p><p></p><p>\sf\implies \red{ 2 \times \dfrac{22}{7} \times 42 \times 120}⟹2×722×42×120</p><p></p><p>\sf\implies \blue{ 2 \times \dfrac{22}{\cancel{7}} \times \cancel{42} \times 120}⟹2×722×42×120</p><p></p><p>\sf\implies \red{2 \times 22 \times 6 \times 120}⟹2×22×6×120</p><p></p><p>\sf\implies \blue{ 44 \times 72 }⟹44×72</p><p></p><p>\sf\implies \pink{ 31680cm^2 }⟹31680cm2</p><p></p><p>\rm{\boxed{\sf{ \circ\:\: 31680cm^2\:\: \circ}}}∘31680cm2∘</p><p></p><p>\sf\therefore \purple{ THE\:ROLLER\:TAKES\:1000\: REVOLUTIONS TO\:COVER\:AREA\:OF\:THAT\: PARTICULAR\:PALAYGROUND}∴THEROLLERTAKES1000REVOLUTIONSTOCOVERAREAOFTHATPARTICULARPALAYGROUND</p><p></p><p>\sf\therefore \blue{we\: know,\: to\: complete\: one \:revolution\: it \:takes \:31680cm^2 \:area }∴weknow,tocompleteonerevolutionittakes31680cm2area</p><p></p><p>\sf\therefore \red{then \:area \:of\:rectangle = 1000 \times the \:area\: in\: one\: complete\: revolution}∴thenareaofrectangle=1000×theareainonecompleterevolution</p><p></p><p>\sf\implies \pink{ 1000 \times 31680 }⟹1000×31680</p><p></p><p>\sf\implies \green{31680000cm^2}⟹31680000cm2</p><p></p><p>CONVERSION,</p><p></p><p>\sf\therefore \green{cm^2 \:into\:m^2}∴cm2intom2</p><p></p><p>\sf\therefore \blue{\dfrac{ 31680000}{ 100 \times 100}}∴100×10031680000</p><p></p><p>\sf\implies \red{\cancel \dfrac{ 31680000}{ 100 \times 100}}⟹100×10031680000</p><p></p><p>\sf\implies \orange{3168 m^2}⟹3168m2</p><p></p><p>\rm{\boxed{\sf{ \circ\:\: AREA\:OF\: PLAYGROUND= 3168m^2 \:\: \circ}}}∘AREAOFPLAYGROUND=3168m2∘</p><p></p><p>\rm\underline\mathrm{AREA\:OF\:PLAYGROUND\:IS\:3168cm^2}AREAOFPLAYGROUNDIS3168cm2</p><p></p><p>\huge\mathtt{\fbox{\red{Answer✍︎}}}Answer✍︎</p><p></p><p>\large\underline\mathfrak{\red{GIVEN,}}GIVEN,</p><p></p><p>\sf\dashrightarrow \blue{height(H)= 120cm }⇢height(H)=120cm</p><p></p><p>\sf\dashrightarrow \blue{diameter of roller= 84cm}⇢diameterofroller=84cm</p><p></p><p>\sf\therefore \blue{radius= \dfrac{diameter}{2}}∴radius=2diameter</p><p></p><p>\sf\dashrightarrow \blue{ \dfrac{84}{2}}⇢284</p><p></p><p>\sf\dashrightarrow \blue{\cancel \dfrac{84}{2}}⇢284</p><p></p><p>\sf\dashrightarrow \blue{radius= 42cm}⇢radius=42cm</p><p></p><p>\large\underline\mathfrak{\purple{TO\:FIND,}}TOFIND,</p><p></p><p>\sf\dashrightarrow \red{AREA\: OF\:PLAYGROUND }⇢AREAOFPLAYGROUND</p><p></p><p>FORMULA</p><p></p><p>\rm{\boxed{\sf{ \circ\:\: C.S.A\: OF\: CYLINDER= 2 \pi rh \:\: \circ}}}∘C.S.AOFCYLINDER=2πrh∘</p><p></p><p>\large\underline\mathtt{\purple{SOLUTION,}}SOLUTION,</p><p></p><p>© ATQ,</p><p></p><p>\purple{\text{AREA COVERED BY ROLLER IN 1 REVOLUTION = PERIMETER OF ROLLER}}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}∴AREACOVEREDINONEREVOLUTION=2πrh</p><p></p><p>\sf\implies \red{ 2 \times \dfrac{22}{7} \times 42 \times 120}⟹2×722×42×120</p><p></p><p>\sf\implies \blue{ 2 \times \dfrac{22}{\cancel{7}} \times \cancel{42} \times 120}⟹2×722×42×120</p><p></p><p>\sf\implies \red{2 \times 22 \times 6 \times 120}⟹2×22×6×120</p><p></p><p>\sf\implies \blue{ 44 \times 72 }⟹44×72</p><p></p><p>\sf\implies \pink{ 31680cm^2 }⟹31680cm2</p><p></p><p>\rm{\boxed{\sf{ \circ\:\: 31680cm^2\:\: \circ}}}∘31680cm2∘</p><p></p><p>\sf\therefore \purple{ THE\:ROLLER\:TAKES\:1000\: REVOLUTIONS TO\:COVER\:AREA\:OF\:THAT\: PARTICULAR\:PALAYGROUND}∴THEROLLERTAKES1000REVOLUTIONSTOCOVERAREAOFTHATPARTICULARPALAYGROUND</p><p></p><p>\sf\therefore \blue{we\: know,\: to\: complete\: one \:revolution\: it \:takes \:31680cm^2 \:area }∴weknow,tocompleteonerevolutionittakes31680cm2area</p><p></p><p>\sf\therefore \red{then \:area \:of\:rectangle = 1000 \times the \:area\: in\: one\: complete\: revolution}∴thenareaofrectangle=1000×theareainonecompleterevolution</p><p></p><p>\sf\implies \pink{ 1000 \times 31680 }⟹1000×31680</p><p></p><p>\sf\implies \green{31680000cm^2}⟹31680000cm2</p><p></p><p>CONVERSION,</p><p></p><p>\sf\therefore \green{cm^2 \:into\:m^2}∴cm2intom2</p><p></p><p>\sf\therefore \blue{\dfrac{ 31680000}{ 100 \times 100}}∴100×10031680000</p><p></p><p>\sf\implies \red{\cancel \dfrac{ 31680000}{ 100 \times 100}}⟹100×10031680000</p><p></p><p>\sf\implies \orange{3168 m^2}⟹3168m2</p><p></p><p>\rm{\boxed{\sf{ \circ\:\: AREA\:OF\: PLAYGROUND= 3168m^2 \:\: \circ}}}∘AREAOFPLAYGROUND=3168m2∘</p><p></p><p>\rm\underline\mathrm{AREA\:OF\:PLAYGROUND\:IS\:3168cm^2}AREAOFPLAYGROUNDIS3168cm2</p><p></p><p>

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