Question

An inverted cone of vertical height 12$12cm$ and the radius of base $9cm$ contains water to a depth of $4cm$. Find the area of the interior surface of the cone not in contact with the water. [Use $\mathrm{\pi }=3.14$]

Answer 1

Step 1: The curved surface area of the conical pot

Height of the cone $h=12cm$

The radius of the cone $r=9cm$

The slant height of the cone:

$$\begin{gathered} \left(l\right)=\sqrt{h^2+r^2} \\ =\sqrt{{\left(12\right)}^2+{\left(9\right)}^2} \\ =\sqrt{144+81} \\ =\sqrt{225} \\ =15cm \end{gathered}$$

The curved surface area of the conical pot will be

$$\begin{gathered} s=\mathrm{\pi rl} \\ =\mathrm{\pi }\times 9\times 15 \\ =135\mathrm{\pi }{\mathrm{cm}}^2 \end{gathered}$$

Step 2: area of the interior surface of the cone not in contact with water

Let $l\text{'}$ the slant height occupied by the water of depth $h\text{'}=4cm$

$$\begin{gathered} \frac{l\text{'}}{l}=\frac{h\text{'}}{h} \\ \Rightarrow l\text{'}=\frac{4}{12}\times 15 \\ \Rightarrow l\text{'}=5cm \end{gathered}$$

The radius of the water cone:

$$\begin{gathered} r\text{'}=\sqrt{l{\text{'}}^2-h{\text{'}}^2} \\ =\sqrt{5^2-4^2} \\ =\sqrt{25-16} \\ =\sqrt{9} \\ =3cm \end{gathered}$$

The curved surface area of the water cone:

$$\begin{gathered} \mathrm{\pi r}\text{'}\mathrm{l}\text{'} \\ =\mathrm{\pi }\times 3\times 5 \\ =15\mathrm{\pi } \end{gathered}$$

The area of the interior surface of the cone not in contact with water will be:

$$\begin{gathered} (135-15)\mathrm{\pi }{\mathrm{cm}}^2 \\ =120\times 3.14{\mathrm{cm}}^2 \\ =376.8{\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the interior surface of the cone not in contact with water will be $376.8c{m}^2$.