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

Question 29 Mensuration Exercise 17.2

The height and the radius of the base of a right circular cone is half the corresponding height and radius of another bigger cone.

Find: the ratio of their lateral surface areas.



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.

Let r be the radius of bigger cone. Then the radius of smaller cone is r/2.

Let h be the height of bigger cone. Then the height of smaller cone is h/2.

slant height of bigger cone = √(h2+r2)

slant height of smaller cone = √((h/2)2+(r/2)2) = √(h2/4+r2/4) = ½ √(h2+r2)

Curved surface area of bigger cone, s1 = π rl

=π r√(h2+r2)

Curved surface area of smaller cone, s2 =π rl

= π ×(r/2)× ½ √(h2+r2)

= ¼ π r√(h2+r2)

ratio of curved surface area of smaller cone to bigger cone, s2/s1 = ¼ π r√(h2+r2) ÷ π r√(h2+r2)

= ¼ r√(h2+r2) ×1/(r√(h2+r2))

= 1/4

Hence the ratio of curved surface area of smaller cone to bigger cone is 1:4 M L Aggarwal - Understanding ICSE Mathematics - Class 10 chapter Mensuration Question 29 Solution image

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