Question

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about 8 cm. Find the volume and the curved surface of the cone so formed

Answer 1

The cone so formed has h= 8cm as the height and base radius as r= 6cm and slant height as l=10cm

 volume =1/3h

            =1/3*3.14*6*6*8

            =301.44cubic cm

curved surface area=rl

                               =3.14*6*10

                               =188.4 square cm

Answer 2

Given:

A right triangle with sides 6 cm, 8 cm, and 10 cm is revolved about 8 cm.

To find:

The volume and the curved surface area of the cone.

Solution:

The volume and the curved surface of the cone formed by revolving a right triangle with sides 6 cm, 8 cm, and 10 cm about 8 cm are 301.7cm³ and 188.57cm² respectively.

We can solve the above mathematical question using the following mathematical approach.

The sides of the right triangle are the height, radius, and slant height of the formed cone respectively.

⇒ Height of the cone (h) = 8 cm

⇒ Radius of the cone (r) = 6 cm

⇒ Slant height of the cone (l) = 10 cm

We know that, the volume of the cone:

$\begin{aligned} V &=\frac{1}{3} \pi r^{2} h \\ &=\frac{1}{3} \times \frac{22}{7} \times(6 \mathrm{~cm})^{2} \times 8 \mathrm{~cm} \end{aligned}$

$=\frac{2112}{7} = 301.7 cm^3.$

Curved surface area of the cone:

$\begin{aligned} C S A &=\pi r l \\ &=\frac{22}{7} \times 6 \mathrm{~cm} \times 10 \mathrm{~cm} \\ &=188.57 \mathrm{~cm}^{2} \end{aligned}$

Therefore, the volume and the curved surface area of the cone are 301.7cm³ and 188.57cm² respectively.