# 10th Class Mathematics Mensuration

Mensuration

Category : 10th Class

Mensuration

We are familiar with some of the basic solids like cuboid, cone, cylinder and sphere. In this chapter we will discuss about how to find the surface area and volume of these figures. In our daily life, we come across number of solids made up of combinations of two or more of the basic solids.

Surface Area of Solids

We may get the solids which may be combinations of cylinder and cone or cylinder and hemisphere or cone and hemisphere and so on. In such cases we find the surface area of each part separately and add them to get the surface area of entire solid.

Cylinder

If ‘r’ is the radius and "h" is the height of a cylinder, then

Curved surface area of the cylinder $=2\pi rh$

Total surface area of the cylinder $=2\pi r(r+h)$

Cone

If ‘r’ be the radius and 'h1 be the height of a cone, then

Curved surface area of the cone$=\pi rl$

Total surface area of the cone $=\pi r(r+l)$

Where, l is the slant height of the cone and is given by

$l=\sqrt{{{r}^{2}}+{{h}^{2}}}$

Sphere

If ‘r’ be the radius of a sphere, then Surface area of the sphere $=4\pi {{r}^{2}}$

Hemisphere

If ‘r’ be the radius of a hemisphere, then

Curved surface area of the hemisphere $=2\pi {{r}^{2}}$

Total Surface area of the hemisphere$=3\pi {{r}^{2}}$

Volume of Solids

The volume of the combined figures is obtained by finding the volume of each part separately and then adding them together.

Cylinder

If ‘r’ be the radius and "h" be the height of a cylinder, then

Volume of the cylinder$=\pi {{r}^{2}}h$

Cone

If ‘r’ be the radius and 'h' be the height of a cone, then

Volume of the cone$=\frac{1}{3}\pi {{r}^{2}}h$

Sphere

If ‘r’ be the radius of a sphere, then

Volume of a the sphere$=\frac{4}{3}\pi {{r}^{3}}$

Hemisphere

If ‘r’ be the radius of a hemisphere, then

Volume of the hemisphere $=\frac{2}{3}\pi {{r}^{3}}$

• Example:

A toy is in the form of a cone of radius 77 cm and height 36 cm. Find the area of the cardboard required to make the toy.

(a) 18720$c{{m}^{2}}$                       (b) 20570$c{{m}^{2}}$

(c) 21426$c{{m}^{2}}$                       (d) 22480$c{{m}^{2}}$

(e) None of these

Explanation: Area of the cardboard required

= curved surface area of the toy$=\pi rl$

Here, $l=\sqrt{{{r}^{2}}+{{h}^{2}}}=\sqrt{{{77}^{2}}+{{36}^{2}}}=85$

$\therefore$Curve surface area

$=\frac{22}{7}\times 77\times 85=22\times 11\times 85=20570c{{m}^{2}}$

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