Question

A bucket open at the top is in the form of a frustum of a cone with the capacity of $12308.8\text{ }c{{m}^{2}}$. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in making it. (Use $\pi =3.14$)

Answer 1

Hint: This is a simple question based on formula of volume and surface area. We have to use the formulas and substitute the given values in it to get the desired answer. We will use the formulas for the frustum as follows, volume V = $\dfrac{\pi }{3}h\left( {{R}^{2}}+{{r}^{2}}+Rr \right)$ and surface area A = $\pi \left( R+r \right)\sqrt{{{\left( R-r \right)}^{2}}+{{h}^{2}}}$.

Complete step-by-step solution:
We will first draw the figure according to the conditions given in the question.

We have been given in the question that the capacity of the bucket is $12308.8\text{ }c{{m}^{2}}$, and the radii of the top and bottom circular ends are 20 cm and 12 cm. So, let us write down the values provided to us. So, we have,
$\begin{align}
  & V=12308.8\text{ }c{{m}^{2}} \\
 & r=12\text{ }cm \\
 & R=20\text{ }cm \\
\end{align}$
And we have to find the height h and the surface area A. Now, we will us ethe formula of volume to find the height of the bucket. The formula for volume is given by, $\dfrac{\pi }{3}h\left( {{R}^{2}}+{{r}^{2}}+Rr \right)$. So, substituting the values, we get,
$\begin{align}
  & V=\dfrac{\pi }{3}h\left( {{R}^{2}}+{{r}^{2}}+Rr \right) \\
 & \Rightarrow 12308.8=\dfrac{3.14}{3}\times h\left( {{\left( 20 \right)}^{2}}+{{\left( 12 \right)}^{2}}+20\times 12 \right) \\
 & \Rightarrow 12308.8=\dfrac{3.14}{3}\times h\left( 400+144+240 \right) \\
 & \Rightarrow 12308.8=\dfrac{3.14}{3}\times h\left( 784 \right) \\
 & \Rightarrow 12308.8\times 3=\left( 3.14\times 784 \right)h \\
 & \Rightarrow h=\dfrac{12308.8\times 3}{3.14\times 784} \\
 & \Rightarrow h=15\text{ }cm \\
\end{align}$
So, we have obtained the height of the frustum as 15 cm. Now we will find the surface area of the frustum, which is given by the formula, $\pi \left( R+r \right)\sqrt{{{\left( R-r \right)}^{2}}+{{h}^{2}}}$. But it is given in the question that the bucket is open at the top, so the surface of the bucket will be,
$\pi \left( R+r \right)\sqrt{{{\left( R-r \right)}^{2}}+{{h}^{2}}}-\pi {{R}^{2}}$
Now, substituting all the values in the above formula, we will get the surface area as,
$\begin{align}
  & A=\pi \left( R+r \right)\sqrt{{{\left( R-r \right)}^{2}}+{{h}^{2}}}-\pi {{R}^{2}} \\
 & \Rightarrow A=3.14\left( 20+12 \right)\sqrt{{{\left( 20-12 \right)}^{2}}+{{\left( 15 \right)}^{2}}}-3.14\times {{\left( 20 \right)}^{2}} \\
 & \Rightarrow A=3.14\left\{ 32\times \sqrt{{{\left( 8 \right)}^{2}}+{{\left( 15 \right)}^{2}}}-{{\left( 20 \right)}^{2}} \right\} \\
 & \Rightarrow A=3.14\left\{ 32\times 17-{{\left( 20 \right)}^{2}} \right\} \\
 & \Rightarrow A=3.14\left\{ 544-400 \right\} \\
 & \Rightarrow A=3.14\times 44 \\
 & \Rightarrow A=452.16\text{ }c{{m}^{2}} \\
\end{align}$
Therefore, the area of the metal sheet will be $452.16\text{ }c{{m}^{2}}$.

Note: Frustum is a part of a cone if we cut the conical part of any cone, it becomes a frustum. So, in this question, we can use conical volume and area also. Now, let us look at an important concept of the frustum. Let us assume a cone of side H and radius R and cut its conical part from the height of h from the base. Let us assume the radii of the upper part of the frustum as r.

Now consider $\Delta A{{O}_{1}}E$ and $\Delta A{{O}_{2}}C$, these are similar triangles because the angles of both triangles are equal. So, the sides of both triangles will be in the same ratio, $\dfrac{A{{O}_{1}}}{{{O}_{1}}E}=\dfrac{A{{O}_{2}}}{{{O}_{2}}C}\Rightarrow \dfrac{H-h}{r}=\dfrac{H}{R}$ . In this question, we have to remember where a positive sign and where a negative sign is used in the surface area formula, $\pi \left( R+r \right)\sqrt{{{\left( R-r \right)}^{2}}+{{h}^{2}}}$ and don’t interchange them.