find thr volume of the iron usrd to make a hollow cylinder of hright 9cm and whose internal and external radius are 21cm and 28cm respectively​

sorry I don't know the answer to

Volume of iron used is 58982cm

Step-by-step explanation:

For hollow cylinder,

\bold{Height=9cm}Height=9cm

\bold{Internal \ radius=21cm}Internal radius=21cm

\bold{Externa \ radius=28cm}Externa radius=28cm

volume \: of \: cylinder = \pi \times {r}^{2} \times h \:volumeofcylinder=π×r

2

×h

\bold{Volume \ of \ cylinder \ with \ hollow \ part}Volume of cylinder with hollow part

\bold{=\frac{22}{7}×28^2×9}=

7

22

×28

2

×9

\bold{=22×4×28×9}=22×4×28×9

\bold{=71456cm^3}=71456cm

3

\bold{Volume \ of \ hollow \ part}Volume of hollow part

\bold{=\frac{22}{7}×21^2×9}=

7

22

×21

2

×9

\bold{=22×3×21×9}=22×3×21×9

\bold{=12474cm^3}=12474cm

3

\bold{Volume \ of \ iron \ used=}Volume of iron used=

\bold{Volume \ of \ cylinder \ with \ hollow \ part}Volume of cylinder with hollow part

\bold{- Volume \ of \ hollow \ part}−Volume of hollow part

{=71456-12474}=71456−12474

{=58982cm^3}=58982cm

3

\bold\red{Volume \ of \ iron \ used \ is \ 58982cm^3}Volume of iron used is 58982cm

3

is this part of a question or something

no idea

step-by-step explanation: