# ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.2 in Chapter 17 - Mensuration

Question 25 Mensuration Exercise 17.2

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. Find the volume and the curved surface of the cone so formed. (Take π = 3.14)

Solution:

The shape of a cone is produced by stacking several triangles and spinning them around an axis. It has a total surface area and a curved surface area because it has a flat base.

The triangle is rotated about the side 8cm.

So the height of the resulting cone, h = 8 cm

Radius, r = 6 cm

Slant height, l =10 cm

Volume of the cone, V = (1/3)r2h

V = (1/3)×3.14×62×8

V = (1/3)×3.14×36×8

V = 3.14×12×8

V = 301.44 cm3

Hence the volume of the cone is 301.44 cm3.

Curved surface area of the cone = π rl

= 3.14×6×10

= 188.4

Hence the curved surface area of the cone is 188.4 cm2.

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