
If the frustum of the cone has a height of 6 cm and radius of 5 cm and 9 cm respectively, then its cubical volume will be ________$c{m^3}$.
A. $320\pi $
B. $151\pi $
C. $302\pi $
D. $98\pi $
Answer
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Hint: We are given the height h and both the radius r and R of a frustum and we need to find the volume of the frustum. The volume of the frustum is given by the formula $\dfrac{1}{3}h\left[ {\pi {R^2} + \pi {r^2} + \pi rR} \right]$.
Substituting the given values we get the required volume.
Complete step by step solution:
We are given that the height of the frustum is 6 cm
That is h = 6 cm
We know that a frustum has two radius r and R
Since we are given they are 5 cm and 9 cm
We have r = 5 cm and R = 9 cm
We are asked to find the volume of the frustum
The volume of the frustum is given by the formula
$ \Rightarrow \dfrac{1}{3}h\left[ {\pi {R^2} + \pi {r^2} + \pi rR} \right]$
Lets substitute the given values in this formula
We can have the pi as it is because the given options have a pi in them
$
\Rightarrow Volume = \dfrac{1}{3}\pi \left( 6 \right)\left[ {{{\left( 9 \right)}^2} + {{\left( 5 \right)}^2} + \left( 5 \right)\left( 9 \right)} \right] \\
\Rightarrow Volume = 2\pi \left[ {81 + 25 + 45} \right] \\
\Rightarrow Volume = 2\pi \left[ {151} \right] = 302\pi c{m^3} \\
$
Hence we get the volume of the frustum to be $302\pi c{m^3}$
Therefore the correct answer is option C.
Note :
A conical frustum is a frustum created by slicing the top off a cone (with the cut made parallel to the base). For a right circular cone, let be the slant height and and the base and top radii.
A right circular cone has only one frustum.
Frustum of a right circular cone is that portion of right circular cone included between the base and a section parallel to the base not passing through the vertex.
Substituting the given values we get the required volume.
Complete step by step solution:
We are given that the height of the frustum is 6 cm
That is h = 6 cm
We know that a frustum has two radius r and R
Since we are given they are 5 cm and 9 cm
We have r = 5 cm and R = 9 cm
We are asked to find the volume of the frustum
The volume of the frustum is given by the formula
$ \Rightarrow \dfrac{1}{3}h\left[ {\pi {R^2} + \pi {r^2} + \pi rR} \right]$
Lets substitute the given values in this formula
We can have the pi as it is because the given options have a pi in them
$
\Rightarrow Volume = \dfrac{1}{3}\pi \left( 6 \right)\left[ {{{\left( 9 \right)}^2} + {{\left( 5 \right)}^2} + \left( 5 \right)\left( 9 \right)} \right] \\
\Rightarrow Volume = 2\pi \left[ {81 + 25 + 45} \right] \\
\Rightarrow Volume = 2\pi \left[ {151} \right] = 302\pi c{m^3} \\
$
Hence we get the volume of the frustum to be $302\pi c{m^3}$
Therefore the correct answer is option C.
Note :
A conical frustum is a frustum created by slicing the top off a cone (with the cut made parallel to the base). For a right circular cone, let be the slant height and and the base and top radii.
A right circular cone has only one frustum.
Frustum of a right circular cone is that portion of right circular cone included between the base and a section parallel to the base not passing through the vertex.
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