Question

In Fig. 10.75, 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. $(Use\pi =22/7and\sqrt{5}=2.236)$.

Answer 1

The total area of the remaining frustum $A_f$ is given by

$A_f$ = area of lateral surface of original cone ($L{c}_1$ ) + area of base of original cone ($B{c}_1$) + area of base of base of top part of the cone removed ($B{c}_2$ ) - area of lateral surface of top part of the cone removed ($L{c}_2$)

The area of lateral surface of a cone (L) is given by formula:

$L=\pi rS=\pi r\sqrt{r^2+h^2}$

And the area of base of a cone (B) is given by formula:

$L=\pi r^2$

Where:

r = radius of base of cone

h = height of cone

S = slant height of cone.

Given:

Radius of base of original cone = $r_1$ = 6 cm

Height of original cone = $h_1$ = 12 cm

Radius of base of cone removed =$r_1$ = 6 cm

Height of cone removed =$h_2$ = 12 cm

Radius of the cone removed , $r_2=r_1\times \frac{h_2}{h_1}$

$=6\times \frac{4}{12}=2cm$

Using these symbols and values the area of the frustum Af is calculated as follows>

$A_f=L{c}_1+B{c}_1+B{c}_2-L{c}_2$

$=\pi r_1\sqrt{r_1^2+h_1^2}+\pi r_1^2+\pi r_2^2-\pi r_{2}2\sqrt{r_2^2+h_2^2}$

$=\frac{22}{7}\times 6\times \sqrt{6^2+{12}^2}+\frac{22}{7}\times 6^2+\frac{22}{7}\times 2^2-\frac{22}{7}\times 2\times \sqrt{2^2+4^2}$

$=\frac{22}{7}\times 6\times \sqrt{36+144}+\frac{22}{7}\times 36+\frac{22}{7}\times 4-\frac{22}{7}\times 2\times \sqrt{4+16}$

$=\frac{22}{7}\times 6\times \sqrt{180}+\frac{22}{7}\times 40-\frac{22}{7}\times 2\times \sqrt{20}$

$=\frac{22}{7}[6\times 13.4164+40-2\times 4.472135]$

$=\frac{22}{7}[80.498+40-8.944]=350.598c{m}^2$