There are three dimensions that can be measured, length, width, and height, for any object that you can see or touch. There are certain dimensions of our home that we live in. The rectangular display screen/Monitor you’re looking at has the width and breadth of its own length. For every three-dimensional geometrical structure, surface area and volume are measured.

The area or zone covered by the object’s surface is the surface area of any given object. Whereas the quantity of space available in an object is volume.

**Surface Area**

The area occupied by a three-dimensional object by its outer surface is called the surface area. It is measured in square units. The area is of two types:

- Total surface area
- Curved surface area/Lateral surface area

**Total surface area**

The area including the base(s) and the curved portion corresponds to the overall surface area. It is the amount of the area enclosed by the object’s surface. If the form has a curved base and surface, so the sum of the two regions would be the total area.

**Curved surface area/Lateral surface area**

Curved surface area, except its center, corresponds to the area of only the curved portion of the shape (s). For shapes such as a cone, it is often called the lateral surface area.

**Volume**

The Volume is the amount of space in a certain 3D object. The total amount of space, that an object or substance occupies is called volume. It is measured in cubic units.

### Formulae of Surface Area and Volume

The table given contains the Total Surface Area, Curved Surface Area/Lateral Surface Area, and Volume of various shapes.

Name of the shape | Curved Surface Area | Total Surface Area | Volume |
---|---|---|---|

Cuboid | 2h(l + b) | 2(lb + bh + hl) | l * b * h |

Cube | 4a | 6a | a |

Cylinder | 2πrh | 2πr(r + h) | πr |

Sphere | 4πr | 4πr | 4/3π r |

Cone | πrl | πr(r + l) | 1/3π r |

Hemisphere | 2πr | 3πr | 2/3π r |

### Examples

**Example 1: 2 cubes each of volume 512 cm ^{3 }are joined end to end. Find the surface area of the resulting cuboid?**

**Solution:**

Given,

The Volume (V) of each cube is = 512 cm

^{3}we can now implies that a

^{3}= 512 cm^{3}∴ The side of the cube, i.e. a = 8 cm

Now, the breadth and length of the resulting cuboid will be 8 cm each while its height will be 16 cm.

So, the surface area of the cuboid (TSA) = 2(lb + bh + lh)

Now, by putting the values, we get,

= 2(8 × 16 + 8 × 8 + 16 × 8) cm^{2}

= (2 × 320) = 640 cm^{2}

Hence, TSA of the cuboid = 640 cm^{2}

**Example 2: We have a cylindrical candle, 14 cm in diameter and of length 2cm.** **It is melted to form a cuboid candle of dimensions 7 cm × 11 cm×1 cm. How many Cuboidal candles can be obtained?**

**Solution:**

Dimensions of the cylindrical Candle:

Radius of cylindrical candle = 14/2 cm = 7 cm

Height/Thichkness=2 cm

Volume of one cylindrical candle = πr

^{2}h = π x 7 x 7 x (2) cm^{3 }= 308 cm^{3}.Volume of cuboid candle = 7 x 11 x 1 = 77 cm

^{3}Hence, number of Cuboidal candles = Volume of cuboid candle/Volume of one cylindrical candle = 308/77 = 4

Hence we can get 4 Cuboidal shaped candles.

**Example 3: A woman wants to build a spherical toy ball of clay whose radius is equal to the radius of the bangle she wears. Given that the bangle is circular in shape, she also wants that the area of the bangle is equal to the volume of the sphere. Find out the radius of the bangle she is wearing?**

**Solution:**

Let r be the radius of the bangle as well as the sphere,

We have been given that the volume of the sphere is equal to the area of the bangle:

Hence,

πr

^{2}= 4/3 πr^{3}⇒ r = 3/4

Hence the radius of the bangle is 3/4 units.

**Example 4: It is given that the slant height of a right circular cone is 25 cm and its height is 24 cm. Find the curved Surface area of the cone?**

**Solution:**

The formula for the curved surface area of the cone is πrl. Where r is the radius of the cone and l is the slant height of the cone.

Here the cone is the Right Circular Cone.

So the radius of the cone would be :

=>

=> r = 7 cm.

Now calculating the curved surface are:

Required Area = (22/7) * 7 * 25 = 550 cm

^{2}

Hence the curved surface area of the cone is 550 cm^{2}.