# How much length of iron sheet of 11 cm wide is required for making an open cylinder 15 cm high and 7 cm as base radius $\left( \pi =\dfrac{22}{7} \right)$?

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**Hint:**We first find the dimensions of the Open cylinder and its surface area. Then we assume the length of the iron sheet which creates the surface for the particular Open cylinder. We solve the equation and find the value of the variable.

**Complete step-by-step solution:**

Open cylinder, shaped like a cylinder, has 15 cm high and 7 cm as base radius.

From the given dimensions of the Open cylinder, we can find its surface area.

We know that the surface area for a cylindrical shape with base radius $r$ and height $h$ is $2\pi rh$.

Therefore, the surface area of the Open cylinder is $2\pi rh=\dfrac{2\times 22\times 7\times 15}{7}=660c{{m}^{2}}$.

Now we know that the Open cylinder’s surface is created by the sheet of 11 cm wide. We assume its length to be $x$ cm.

The area of the sheet is $11\times x=11x\text{ }c{{m}^{2}}$.

We equate them to get $11x=660$ which gives $x=\dfrac{660}{11}=60$.

The length of the iron sheet is 60 cm.

**Note:**In case the cylinder was closed then we would have taken the surface area as $2\pi r\left( r+h \right)$.

We are taking an extra $2\pi {{r}^{2}}$ for two sides of the lid area.

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