Question

A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of a metal. How much sheet of metal is required?

Answer 1

Hint: We should know that the total surface area of the cylinder whose radius is r and height of the cylinder is equal to l is equal to \[\pi r\left( l+r \right)\]. We know that the radius of a cylindrical tank is equal to 7 m. It is also clear that the height of the cylindrical tank is equal to 3 m. Now by using the above formula, we can find the total surface area of metal. This will give us the sheet of metal required.

Complete step by step answer:
Before solving the question, we should know that the total surface area of the cylinder whose radius is r and the height of the cylinder is equal to l is equal to \[\pi r\left( l+r \right)\].
From the question, it is clear that the radius of the cylinder tank is equal to 7 m and the height of the cylinder tank is equal to 3 m.

Let us assume the radius of the cylinder tank is equal to r.
Then we get
\[r=7.....(1)\]
Now let us assume the height of the cylinder tank is equal to l.
Then we get
\[l=3....(2)\]
We know that the total surface area of the cylinder whose radius is r and the height of the cylinder is equal to l is equal to \[\pi r\left( l+r \right)\].
Let us assume the total surface area of the cylinder is equal to T.S.A.
\[\begin{align}
  & \Rightarrow T.S.A=2\pi (7)(7+3) \\
 & \Rightarrow T.S.A=2(70)\pi \\
 & \Rightarrow T.S.A=(140)\left( \dfrac{22}{7} \right) \\
 & \Rightarrow T.S.A=440{{m}^{2}} \\
\end{align}\]
So, it is clear that the total surface area of the cylinder is equal to \[440c{{m}^{2}}\].
Hence, we can say that the sheet of metal required to make the cylinder is equal to \[440c{{m}^{2}}\].

Note: Students may have a misconception that the lateral surface area of the cylinder gives us the sheet of metal. So, if we calculate the lateral surface area, then
Let us assume the lateral surface area is equal to L.S.A.
\[\Rightarrow L.S.A=\pi rl\]
Here r is radius of cylinder and height of cylinder.
\[\begin{align}
  & \Rightarrow L.S.A=\pi \left( 7 \right)(3) \\
 & \Rightarrow L.S.A=66{{m}^{2}} \\
\end{align}\]
So, we get the sheet of metal required is equal to \[66{{m}^{2}}\] but we know that the sheet of metal required to make the cylinder is equal to \[440c{{m}^{2}}\]. This misconception should be avoided.