**Surface Areas and Volumes – Cube, Cuboid, and Cylinder | Class 10 Maths**

In this chapter, we will learn about the CBSE Board **Class 10** topic **Surface Areas and Volumes** of solids like Cube, Cuboid, Cylinder, and Sphere from Chapter-13. Our Maths notes are self-satisfactory and easy to understand.

**Introduction**

In our day to day life, we come across various forms of solids. For Example – An ice cream cone is a combination of cone and semi-sphere. In this article, we will discuss the calculations regarding finding the surface area and volume of the different solids.

**Units of measurements**

To solve the problems first we need to know the interrelationships between different units of measurement.

**Length**

1 Centimetre (cm) = 10 millimetre (mm)

1 Decimetre (dm) = 10 Centimetres (cm)

1 Metre (M) = 10 decimetre= 100 Centimetres (cm) = 1000 millimetre (mm)

1 Decametre (dam) = 10 Metre (m) = 1000 Centimetres (cm)

1 Hectometre (hm) = 10 Decametre (dam) = 100 Metre (m)

1 kilometre (km) = 1000 Metre (m)= 100 1 Decametre (dam) = 10 Hectometre (hm)

1 myriametre= 10 kilometre (km)

**Area**

1 cm^{2 } = 1 cm x 1 cm = 10 mm x 10 mm= 100mm^{2 }

1 dm^{2 } = 1 dm x 1 dm= 10 cm x 10 cm= 100cm^{2 }

1 m ^{2 } = 1 m x 1m = 10 dm x10 dm= 100dm^{2 }

1 dm ^{2 } = 1 dm x 1dm = 10 m x 10 m= 100m^{2 }

1 hm^{2 } (hectare) = 1 hm x1 hm= 100 m x 100 m= 10000 = 100m^{2 }

1 km^{2 } = 1km x 1 km= 10 hm x10 hm= 100hm^{2 } (hectare)

**Volume **

1 cm ^{3 }= 1 ml =1 cm x 1cm x 1 cm = 10 mm x 10 mm x 10 mm= 1000m^{3}

1 litre= 1000 ml =1000cm^{3}

1m^{3} = 1 ml =1 m x 1m x 1 m = 100 cm x 100 cm x 100 cm = 106 cm^{3}= 1000 litre = 1 kilo litre

1 dm ^{3}= 1000 cm ^{3}

1 m ^{3}= 1000 dm ^{3}

1 km ^{3}= 109 m ^{3}

**CUBOID**

A cuboid is a rectangular solid which have six rectangular faces. For example- matchbox, a book, a brick etc.

-It contains 12 edges, 18 vertices, and 6 rectangular faces.

-Volume of cuboid refers to the space enclosed by the cuboid

Diagonal refers to the line joining the opposite corners. It is the length of the longest rod. It has four diagonals.

Diagonals -D1 AG, D2 HB, D3 EC, and D4 DF

-Total surface area refers to the sum total of all the areas of four faces.

-Lateral faces refer to the four faces which meet the base of a cuboid.

-Lateral surface area refers to the sum of areas of four walls of a cuboid.

**Formulae used to solve cuboid problems**

Length=l

Breath = b

Height =h

Sum of length of all edges =4(l+h+b) units

Diagonal of cuboid= √l^{2} +h^{2} +b ^{2}

Total Surface Area=2 (lb + bh+ lh) square units

Lateral surface Area= [2 (l+b) ×h] square units

Area of four walls= [2 (l+b) ×h] square unitss

Volume of a cuboid= l×b×h cubic units

**Example- Cuboid**

A rectangular solid length, breadth and height is in the ratio = 6:5:4 and total surface area =5328cm ^{2}

Find length, breadth, and height?

Let length, breadth and height = 6a, 5a, 4a

Total surface area = 2 (lb + bh+ lh) square units= 5328

= [ 2(6a × 5a+5a× 4a+ 4a× 6a)]= [ 2(30a ^{2}+20a ^{2 } + 24a^{2 } )]

=148a^{2}= 5328/148 = 36

= a^{2 } =√36

a=6

Length= 6a = 6 x6 =36 cm, breadth=5a =5×5=30 cm and height = 4a= 4×6=24 cm

__CUBE__

Cube refers to the cuboid whose length; breadth and height are all equal. For example – ice cube, sugar cube etc.

**Formulae used to solve cube problems**

Sun of length of all edges= =12a units

Diagonal of a cube= a√3units

Lateral surface area= 6a^{2}units

Total Surface area=4a^{2} units

Volume= a^{3} cubic units

**Cross section**

It is cut which is made through a solid perpendicular on the length.

**Uniform cross section **

When the cut is of the same size at every point of its length.

The lateral surface area of a uniform cross-section solid = Perimeter of cross section x length

Volume of a uniform cross-section solid= Area of cross section x length

*Find out examples and questions for practice, NCERT Class 10 Maths.*

__RIGHT CIRCULAR CYLINDER__

It is solid which is created by the revolution of a rectangle.

Solids like circular pillars, measuring glass, pencil etc. are the form of right circular cylinders.

__REMARK- UNLESS STATED CYLINDER IS NOT RIGHT CIRCULAR, A WPRD CYLINDER WILL BE CONSIDERED RIGHT CIRCULAR CYLINDER__

Case-1 – a cylinder with not circular base.

Case-2- a cylinder is circular but not a right angle.

Case 3- a cylinder is not a right circular cylinder if the lines joining a centre of a circle of a cylinder is **not perpendicular.**

**Formulae used to solve Cylinder problems**

Diameter = 2r

Perimeter= 2πr

Area of each circular end =πr^{2}sq units

Total surface area = Curved surface area + Area of two circular ends

2πr + 2πr^{2}sq units = {2πr (h + r) }sq. units

Volume = 2πr^{2}h cubic units

__HOLLOW RIGHT CIRCULAR CYLINDER__

The hollow right circular cylinder is a solid which is bounded by two coaxial cylinders of the same height. For example-a hollow pipe.

**Formulae used to solve hollow right circular cylinder problems**

**The thickness of a cylinder= R-r units**

Area of cross-sections **= π(R ^{2} – r^{2} )**square units

Lateral/curved surface area= External curved surface area + Internal curved surface area.

=2πRh+ 2πrh = 2πh(R+r) square units

Total surface area-= Curved surface area +2 (area of the base ring)

=2πRh+ 2πrh + 2(πR^{2} – πr^{2})

=2π( Rh+ rh+R^{2} – r ^{2})square units

Volume = π*h* (*R*^{2}– *r*^{2}) cubic units

A volume of hollow region= πr^{2}h cubic units

*To know about the Sphere volume and area, click Class 10 Maths and study with animated learning videos.*

*Practice more and more with the practical numerical zone, click CBSE Class 10th Maths for more details.*

Browse the class 10 related video lecture on Trigonometric Ratio

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