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**Surface Area And Volume Class 10 Formulas Download pdf** Studying in class 10th and searching for all the formulas to solve mensuration questions. So you are at the right page. In this page, we are providing you all with all the surface area and volume formulas for class 10th, that will help you solving your questions easily.

The best way to remember these surface area and volume formulas is that keep practicing a lot of questions from them, so that these formulas will be in your mind and you can start using these surface area and volume formulas as per the demand of the questions. Along with this post, we have also provided you some sample questions where you will get to know how to use these surface area and volume formulas.

As we know that surface area and volume formulas are used to calculate the surface area and volume of 3-D figures. Now let us help you understand what do we mean by 3-D figures, so these are the figures that lies on the x-axis, y-axis, and z-axis. These figures have length, width, and thickness. So, here we will provide you surface area and volume formulas of cube, cuboid, hemisphere, sphere, cone, etc.

### Surface Area And Volume Class 10 Formulas Download Pdf

The surface area and volume formulas for 3-D figures in mensuration that we will provide here will not only help you in getting good grades in your school level examinations, but these formulas will also be helpful when you will prepare for your competitive examinations. We hope that this post on surface area and volume will be really beneficial for you and will help you in solving a lot of questions.

### Surface Area And Volume Formulas Class 10

In this particular section, we are providing you a table below that will provide you all the important surface area and volume formulas that will help you in solving these 3-D figures. Go through these formulas and practice as many questions as you can based on these formulas so that you will get perfect in solving the question:

Figure |
Curved Surface Area |
Total Surface Area |
Volume |

Cube |
4a^{2} |
6a^{2} |
a^{3} |

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

Cylinder |
2πrh | 2πr(h + r) | πr^{2}h |

Cone |
πrl | πr (r + l) | 1/3hπr² |

Sphere |
4πr² | 4πr² | 4/3πr^{3 } |

Hemisphere |
2πr^{2} |
3πr² | 2/3πr^{3} |

### Some Other Important Formulas

Volume Of The Spherical Shell Whose Outer and Inner Radii Are ‘R’ and ‘r’ |
4/3π(R^{3}-r^{3}) |

### Formula For Frustum

Total Surface Area Of Frustum Of A Cone |
πL (R + r) + π (R^{2} + r^{2}) |

Curved Surface Area Of Frustum Of A Cone |
πL (R + r) |

Volume Of Frustum Of A Cone |
πH/3 (R^{2} + Rr + r^{2}) |

Slant Height Of Frustum Of A Cone |
√h^{2}+(R−r)^{2} |

### Formula For Square Regular Pyramid

Total Surface Area Of Square Pyramid |
a^{2} + 2al |

Curved Surface Area Of Square Pyramid |
2al |

Volume Of Square Pyramid |
a^{2}h/3 |

Slant Height Of Square Pyramid |
√(h^{2} + (1/4)a^{2}) |

So, these were some of the formulas related to the square and volume of 3-D figures. In case of queries you may drop down your questions in the comment section below. We will provide you with some of the solutions to your questions.

Mensuration Formulas for 2D and 3D Shapes

**Q. A circle has a radius of 21 cm. Find its circumference and area. (Use π = 22/7)**

**Solution:** Circumference of circle = 2πr = 2 x (22/7) x 21 = 2 x 22 x 3 = 132 cm

Area of circle = πr^{2} = (22/7) x 21^{2} = 22/7 x 21 x 21 = 22 x 3 x 21

Area of circle with radius, 21cm = 1386 cm^{2}

**Q. A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.**

**Solution:** Length of the paper will be the perimeter of the base of the cylinder and width will be its height.

Circumference of base of cylinder = 2πr = 11 cm

2 x 22/7 x r = 11 cm

r = 7/4 cm

Volume of cylinder = πr^{2}h = (22/7) x (7/4)^{2} x 4

= 38.5 cm^{3}

**Q. A copper sphere of diameter 12 cm is drawn into a wire of diameter 8 mm. Find the length of the wire.**

**Solution : **Here 4/3 πR^{3} = πr^{2}h where R is the radius of sphere, r is the radius of the wire and h is the length of the wire.

Hence 4/3 (12/2)^{3} = (.8/2)^{2} h

So h =1800cm

**Q. A solid hemisphere of radius ‘a’ that is made of a certain material weighs 60 kg. The figure is given below. What is the weight, in kg, of the entire hemispherical shell if the outer radius is ‘7a’ and the inner radius is 3a?**

**Solution :** V = 2/3πa^{3} and this volume weighs 60 kgs.

If it is a shell, you can find V of solid part by subtracting inner empty hemisphere from total.

V of entire sphere 2/3π(7a)^{3} and V of inner empty space = 2/3π(3a)^{3}

Therefore, V of solid = 2/3π(7a)^{3} – 2/3π(3a)^{3} = 2/3πa^{3} (343 – 27) = 2/3πa^{3} x 316

Now 2/3πa^{3} weighs 60 kg,

so, 2/3πa^{3} x 316 will weigh 60 x 316=18960kg.