Math, asked by Anonymous, 4 months ago

solve this!!!
don"t spam ​

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Answers

Answered by hotcupid16
1

\huge\mathtt{\fbox{\orange{★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

Step-by-step explanation:

Given :

Area of rhombus = 120 cm²

Length of diagonal = 8 cm

To find :

Length of another diagonal

According to the question,

\sf{ :  \implies Area \: of \: rhombus =  \dfrac{1}{2}  \times d _{1} \times d_{2}   }

 \\

 \sf  : \implies{ {120 \: cm}^{2} =  \dfrac{1}{2}   \times 8 \: cm \times x}

 \\

 \sf :  \implies{ {120 \: cm}^{2} \times 2 = 8 \: cm \times x }

 \\

 \sf :  \implies{ {240 \: cm}^{2}  = 8 \: cm \times x}

 \\

 \sf  : \implies{ \dfrac{240}{8}  \: cm = x}

 \\

 { \underline{ \boxed{  \sf  \pink{ :  \implies{   \bm3 \bm0 \: c m =x}}}}}

{ \therefore{ \underline{\sf{So \:,the \:  length \:  of \:  other \:  diagonal  \: is \:    3 0 \: cm}}}}

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