Question

a cone of radius 12cm and height
20cms is cut parallel to the base from the top 3cms down and removed out. the remaining part becomes the frustom of a cone.find the ratio between the frustom of cone and the cone which is removed out​

Answer 1

Volume of the frustum will be 2970 cm^3cm

3

Step-by-step explanation:

\textrm{Volume of cone is given as}V=\frac{1}{3} \pi r^2hVolume of cone is given asV=

3

1

πr

2

h

\textrm{Now the volume of cone }=\frac{1}{3} \times\frac{22}{7} \times12^{2} \times20Now the volume of cone =

3

1

×

7

22

×12

2

×20

= 3017.14cm^3=3017.14cm

3

\textrm{Now the height of smaller cone will be calculated as}Now the height of smaller cone will be calculated as

\begin{lgathered}\frac{20}{12} =\frac{x}{3} \\\end{lgathered}

12

20

=

3

x

\textrm{after solving this we can get x} =5cmafter solving this we can get x=5cm

\textrm {So the volume of smaller cone will be}=\frac{1}{3} \times\frac{22}{7} \times3^{2} \times5So the volume of smaller cone will be=

3

1

×

7

22

×3

2

×5

=47.14 cm^3=47.14cm

3

\begin{lgathered}\textrm {Now volume of frustum of cone is given by} \\=\textrm {Volume of bigger cone -Volume of smaller cone}\\=3017.14 - 47.14\\=\bold{2970 cm^3}\end{lgathered}

Now volume of frustum of cone is given by

=Volume of bigger cone -Volume of smaller cone

=3017.14−47.14

=2970cm