Question

From the top a conical shaped jaggery of base diameter $10$cm and height $10$cm, a small cone is cut off by slicing parallel to the base, $4$cm from the vertex. Find the surface area and volume of the frustum so obtained.

Answer 1

Diameter of the cone ABO$=10$cm
$\therefore$ radius $r_1=\frac{10}{2}=5$cm
Height of the cone ABO$=h_1=10$cm
Height of the cone $A^{|}{B}^{|}O=h_2=4$cm
First, let us find the other required data.
Radius of cone $A^{|}{B}^{|}O=r_2$
Let the Slant height of cones $ABO$ and $A^{|}{B}^{|}O,$ be $l_1$ and $l_2$ respectively.

$\therefore \frac{r_1}{r_2}=\frac{h_1}{h_2},\frac{5}{r_2}=\frac{10}{4}$

$\therefore r_2=\frac{5\times 4}{10}=2$cm

$l_1^2=h_1^2+r_1^2={10}^2+5^2=100+25=125$

$\therefore l_1=\sqrt{125}=5\sqrt{5}$cm

$l_2^2=h_2^2+r_2^2=4^2+2^2=16+4=20$

$\therefore l_2=\sqrt{20}=2\sqrt{5}$cm

(i) Lateral surface area of cone ABO$=\pi r_1{l}_1=\pi \times 5\times 5\sqrt{5}=25\sqrt{5}\pi$

$\therefore$ LSA of cone ABO$=25\sqrt{5}\pi c{m}^2$

Lateral surface area of cone $A^{|}{B}^{|}O=\pi r_2{l}_2=\pi \times 2\times 2\sqrt{5}=4\sqrt{5}\pi$

$\therefore$ LSA of cone $A^{|}{B}^{|}O=4\sqrt{5}\pi c{m}^2$

$\therefore$ Lateral surface area of frustum$=$ LSA of cone ABO$-$ LSA of come $A^{|}{B}^{|}O$

$=25\sqrt{5}\pi -4\sqrt{5}\pi =21\sqrt{5}\pi =21\times \sqrt{5}\times \frac{22}{7}=66\sqrt{5}c{m}^2$

$\therefore$ LSA of frustum$=66\sqrt{5}$sq. cm. or $147.84c{m}^2$

(ii) Total surface area of cone ABO$=\pi r_1(r_1+l_1)$

$=\pi \times 5\times (5+5\sqrt{5})=\pi \times 5\times 5(1+\sqrt{5})$

$\therefore$ TSA of cone ABO$=25(1+\sqrt{5})\pi c{m}^2$

Lateral surface area of cone $A^{|}{B}^{|}O=\pi r_2{l}_2=\pi \times 2\times 2\sqrt{5}=4\sqrt{5}\pi$

$\therefore$ LSA of cone $A^{|}{B}^{|}O=4\sqrt{5}\pi c{m}^2$

The total surface area of frustum$=$ TSA of cone ABO$-$ LSA of cone $A^{|}{B}^{|}O+\pi r_2^2$

$=25(1+\sqrt{5})\pi -4\sqrt{5}\pi +4\pi$

$=\pi (25+25\sqrt{5}-4\sqrt{5}+4)$

$=\pi (29+21\sqrt{5})$

$=\frac{22}{7}\times 75.95$

$\therefore$ TSA of frustum$=238.70c{m}^2$

(iii)Volume of cone ABO$=\frac{1}{3}\pi r_1^2{h}_1=\frac{1}{3}\times \pi \times 5\times 5\times 10$

$\therefore$ Volume of cone ABO$=\frac{250}{3}\pi c{m}^3$

Volume of cone $A^{|}{B}^{|}O=\frac{1}{3}\pi r_2^2{h}_2=\frac{1}{3}\times \pi \times 2\times 2\times 4$

$\therefore$ Volume of cone $A^{|}{B}^{|}O=\frac{16}{3}\pi c{m}^3$

$\therefore$ Volume of the frustum$=$ Volume of cone ABO$-$ Volume of cone $A^{|}{B}^{|}O$

$=\frac{250}{3}\pi -\frac{16}{3}\pi =\frac{234}{3}\pi$

$\therefore$ Volume of the frustum$=78\pi c{m}^3$ or $245.14c{m}^3$

We can also directly find the lateral surface area, total surface area, and volume of the frustum by using the formulae.