Math, 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 MysticalRainbow
0

Answer:

Radius of the roller = diameter ÷ 2 = 84 ÷ 2 = 42 cm

Length of the roller = 120 cm

Curved Surface Area of the Roller (Cylindrical shape) = begin mathsize 16px style 2 πrh space space equals space 2 space cross times 22 over 7 cross times 42 space cross times space 120 space equals space 31680 space cm squared end style

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

Therefore the area covered by the roller in 1000 revolutions = 31680 × 1000 = 31680000 cm2

Therefore the area of the play ground = 31680000 cm2 = 3168 m2

→ l × b = 3168

→ l × 32 = 3168

→ l = 99 m

Answered by ItzCaptonMack
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}

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