Question

A closed cylindrical tank of radius 14m and height 3m is made from a sheet metal. How much sheet of metal is required? \[\]

Answer 1

Hint: The whole of the area of the sheet used will be present in the cylinder. We can say that the area of the sheet used will be equal to the total surface area of the cylindrical tank. We know that the total surface area of the cylinder is given by $A=2\pi rh+2\pi {{r}^{2}}$ . So, we substitute the values in the formulas to get the answer.\[\]

Complete step-by-step answer:
We will start the solution to the above question by drawing the diagram of the cylindrical tank given in the question.\[\]

If we try to understand the question, we will find that the area of the sheet that is asked in the question is actually the total surface area of the cylinder because the sheet metal will made to use the cylindrical tank.\[\]

Now, we know that the total surface area of the cylinder the sum base area of two circular bases $2\pi {{r}^{2}}$and the curved surface area $2\pi rh$ which is equal to $A=2\pi rh+2\pi {{r}^{2}}$ . We are given the question that radius $r$ is 14m and height of the cylinder $h$ is 3m. So we substitute the given values and find the surface area of the cylinder as
\[\begin{align}
  & \text{A}=2\pi rh+2\pi {{r}^{2}} \\
 & =2\pi \times 14\times 3+2\pi \times {{14}^{2}} \\
 & =28\pi \left( 3+14 \right) \\
 & =476\pi \text{ }{{\text{m}}^{2}} \\
\end{align}\]
Therefore, the area of the sheet required is $476\pi \text{ }{{\text{m}}^{2}}$ and if we substitute the value $\pi =\dfrac{22}{7}$ , the answer to the above question is $476\times \dfrac{22}{7}\text{=1496 }{{\text{m}}^{2}}$\[\]

Note: We can also use the other approximation of $\pi =3.14$. We also observe that the cylindrical tank is hollow inside and if we fill water it will occupy the space equal to volume of the cylinder which is given by $V=\pi {{r}^{2}}h$. If the question would have mentioned some wastage of the sheet we would have added that to total surface area.