Question

In the given figure, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use $\mathrm{\pi }=\frac{22}{7}\mathrm{and}\sqrt{5}=2.236$). [CBSE 2015]

Answer 1

Now, In ∆OCD and OAB
∠OCD = OAB (Corresponding angles)
∠ODC = OBA (Corresponding angles)
By AA-similarity criterion ∆OCD ∼ ∆OAB
$$\begin{gathered} \therefore \frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{OC}}{\mathrm{OA}} \\ \Rightarrow \frac{\mathrm{CD}}{6}=\frac{4}{12} \\ \Rightarrow \mathrm{CD}=2\mathrm{cm} \end{gathered}$$
The slant height of the bucket is given by
$$\begin{gathered} l=\sqrt{h^2+{\left(R-r\right)}^2} \\ =\sqrt{{\left(8\right)}^2+{\left(6-2\right)}^2} \\ =\sqrt{64+16} \\ =\sqrt{80} \\ =4\sqrt{5}\mathrm{cm} \end{gathered}$$
Surface area of the remaining solid
= Curved surface area of figure + Area of the smaller circle + Area of the larger circle
$$\begin{gathered} =\mathrm{\pi }l\left(R+r\right)+\mathrm{\pi }{R}^2+\mathrm{\pi }{r}^2 \\ =\frac{22}{7}\times 4\sqrt{5}\times \left(6+2\right)+\frac{22}{7}\times 6\times 6+\frac{22}{7}\times 2\times 2 \\ =\frac{22}{7}\times 4\sqrt{5}\times 8+\frac{22}{7}\times 6\times 6+\frac{22}{7}\times 2\times 2 \\ =224.88+113.14+12.57 \\ =350.59{\mathrm{cm}}^2 \end{gathered}$$