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A metallic solid cylinder has one end surmounted by a cone of the same radius and a hemisphere, which is also of same radius at that of cone.

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Question

A metallic solid cylinder has one end surmounted by a cone of the same radius and a hemisphere, which is also of same radius at that of cone. The conical portion, the cylindercal portion and the hemispherical portion are separated as shown below:

If the radius of the given solid is r cm and the height of the cone = the height of the cylinder = h cm, find expression for the the total surface area of the three parts obtained.

The image displays a composite solid figure decomposed into its individual component shapes consisting of a right circular cone, a right circular cylinder, and a hemisphere.

Sum
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Solution

When the three portions are separated, each becomes a complete solid. So, we add the entire surface area of the cone, the cylinder and the hemisphere.

For the cone:

Radius = r cm and height = h cm.

Slant height (l) = `sqrt (h^2+r^2​)` cm.

Total surface area of cone = πrl + πr2 = `sqrt (h^2+r^2​)`​ + πr2.

For the cylinder:

Radius = r cm and height = h cm.

Total surface area of cylinder = 2πrh + 2πr2

For the hemisphere:

Radius = r cm.

Total surface area of hemisphere = 2πr2 + πr2 = 3πr2

∴ Total surface area of the three parts

= `(πr sqrt (h^2+r^2​) ​+ πr^2) + (2πrh + 2πr^2) + 3πr^2`

= `πr sqrt (h^2+r^2) ​+ 2πrh + 6πr^2`

= `πr sqrt (h^2+r^2)+2h+6r)`

Hence, total surface area of the three parts = `πr sqrt (h^2+r^2)+2h+6r)`.

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Chapter 20: Cylinder, Cone and Sphere - TEST YOURSELF [Page 326]

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Selina Concise Mathematics [English] Class 10 ICSE
Chapter 20 Cylinder, Cone and Sphere
TEST YOURSELF | Q 19. | Page 326
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