Question

In Fig. 4, 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.
$(Use\pi =\frac{22}{7})=2.236$

Answer 1

The remaining solid is a frustum of the given cone.
Total surface area of the frustum
$\pi l(r_1+r_2)+\pi r_1^2+\pi r_2^2$
Where
h = Height of the frustum = 12 - 4 = 8 cm
$r_1$ = Larger radius of the frustum = 6 m
$r_2$ = Smaller radius of the frustum
l = Slant height of the frustum

In the given figure, $\Delta ABC\sim \Delta ADE$ by AA similarity criterion.
$\therefore \frac{BC}{DE}=\frac{AB}{AD}$
$\frac{r_2}{6}=\frac{4}{12}$
$\Rightarrow r_2=2cm$
We know
$l=\sqrt{h^2+(r_1-r_2{)}^2}$
$\Rightarrow l=4\sqrt{5}cm$.
$\therefore$ Total surface area of the frustum = $\pi (r_1+r_2)+\pi r_1^2+\pi r_2^2$
$=\pi \times 4\sqrt{5}(6+2)+\pi \times 6^2+\pi \times 2^2$
$=\pi (32\sqrt{5}+40)$
$=350.592c{m}^2$
Hence, the total surface area of the remaining solid is
$350.592c{m}^2$.