Question

A cylindrical bucket 32cm high and with radius of base 18cm 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 24cm then find the radius and slant height of the heap

Answer 1

Answer: Its very simple

Step-by-step explanation: First we need to find the volume of the cylindrical bucket which can be found by πr^2h. Then keep this volume equal to the volume of conical heap formed. i.e πr1^2h= 1/3πr2^2H.

volume of cylindrical bucket= volume of conical heap formed. Just simplify it.

Hope it helps!!!!!

Answer 2

GIVEN :-

cylinder :

• hieght ( h1 ) = 32 cm

• radius ( r1 ) = 18 cm

cone :

• hieght ( 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,

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