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

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Given:

In the figure, from the top of a solid cone of height \( 12 \mathrm{~cm} \) and base radius \( 6 \mathrm{~cm} \), a cone of height \( 4 \mathrm{~cm} \) is removed by a plane parallel to the base.

To do:

We have to find the total surface area of the remaining solid.

Solution:

 Total height of the cone $= 12\ cm$

Radius of the base of the cone $= 6\ cm$

Height of the frustum formed $= 12 - 4$

$= 8\ cm$

Let $r$ be the radius of the cone cut out.
This implies,

$\frac{r}{6}=\frac{4}{12}$

$r=\frac{6 \times 4}{12}$

$=2 \mathrm{~cm}$

Let $l$ be the slant height of the whole cone.

Therefore,

$l=\sqrt{r^{2}+h^{2}}$

$=\sqrt{6^{2}+12^{2}}$

$=\sqrt{36+144}$

$=\sqrt{180}$

$=\sqrt{36 \times 5}$

$=6 \sqrt{5} \mathrm{~cm}$

Slant height of the remaining portion of frustum $l_1=6 \sqrt{5}-\frac{6 \sqrt{5}}{3}$

$=6 \sqrt{5}-2 \sqrt{5}$

$=4 \sqrt{5} \mathrm{~cm}$

Surface area of the remaining portion $=\pi(r_{1}+r_{2}) \times l_{1}$

$=\frac{22}{7}(6+2) \times 4 \sqrt{5}$

$=\frac{22}{7} \times 8 \times 4 \sqrt{5}$

$=\frac{704 \sqrt{5}}{7}$

$=\frac{704}{7}(2.236)$

$=224.88 \mathrm{~cm}^{2}$

Area of the base and top $=\pi(6^{2}+2^{2})$

$=\frac{22}{7}(36+4)$

$=40 \times \frac{22}{7}$

$=\frac{880}{7}$

$=125.71 \mathrm{~cm}^{2}$

Total surface area of the remaining solid $=224.88+125.71$

$=350.59 \mathrm{~cm}^{2}$

The total surface area of the remaining solid is $350.59\ cm^2$.