Math, asked by nehu215, 7 months ago

\huge\sf\underline{Question}A cylindrical bucket 32 cm high and 18 cm of radius of the base is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm. Find the radius and slant height of the heap.

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Answered by Anonymous
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Answered by Anonymous
2

GIVEN :-

cylinder :

  • hieght ( h1 ) = 32 cm

  • radius ( r1 ) = 18 cm

cone :

  • height ( h2 ) = 24 cm

TO FIND :-

  • radius ( r2 ) and slant hieght ( s ) of heap ( cone )

SOLUTION :-

since in cylinder bucket is emptied to make a colonial heap

volume of cylinder = volume of cone

now volume of cylinder :-

hieght ( h1 ) = 32 cm

radius ( r1 ) = 18 cm

we know the formula that volume of cylinder :-

 \implies  \boxed {\rm{volume \:  = \pi {r}^{2} h}}

\implies \rm{\pi ({18})^{2}(32) }

\implies  \rm{ \bold{ volume = \pi  \times 18 \times 18 \times 32 }}

now volume of cylinder :-

height ( h2 ) = 24 cm

let radius be r cm

and slant hieght be l cm

we know the formula of volume of cone :-

 \implies  \boxed {\rm{volume \:  =  \dfrac{1}{3} \pi {r}^{2} h}}

\implies \rm{  \dfrac{1}{3} \pi {r}^{2} (24)}

\implies \rm{ \bold{ volume \: of \: cone =  8\pi {r}^{2} }}

now volume of cylinder = volume of cone hence ,

\implies \rm{ \pi \times 18 \times 18 \times 32=  8\pi {r}^{2} }

\implies \rm{   {r}^{2} =  \dfrac{  18 \times 18 \times 32}{8} }

\implies \rm{   {r}^{2} =    18 \times 18 \times 4}

\implies \rm{   {r}^{2} =     {36}^{2} }

\implies \rm{  \bold{  {r} =     {36 \: cm}} }

now we have to find slant hieght :-

  \implies \boxed{ \rm{{l}^{2}  =  {h}^{2}  +  {r}^{2}}}

 \implies  \rm{{l}^{2}  =  {24}^{2}  +  {36}^{2}}

 \implies  \rm{{l}^{2}  =  1872}

 \implies  \rm{l  =   \sqrt{1872} }

\implies  \rm{ \bold{l  = 12  \sqrt{13} \: cm }}

HENCE,

\implies  \boxed{ \boxed{ \rm{l  = 12  \sqrt{13} \: cm \: \:  and  \:  \: \: r \:  = 36cm }} }

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