Class 10 Maths Surface Areas and Volumes – Get here the Notes for Class 10 Maths Surface Areas and Volumes. Candidates who are ambitious to qualify the Class 10 with good score can check this article for Notes. This is possible only when you have the best CBSE Class 10 Maths study material and a smart preparation plan. To assist you with that, we are here with notes. Hope these notes will help you understand the important topics and remember the key points for exam point of view. Below we provided the Notes of Class 10 Maths for topic Surface Areas and Volumes.

**Class:**10th**Subject:**Maths**Topic:**Surface Areas and Volumes**Resource:**Notes

## CBSE Notes Class 10 Maths Surface Areas and Volumes

Candidates who are pursuing in Class 10 are advised to revise the notes from this post. With the help of Notes, candidates can plan their Strategy for particular weaker section of the subject and study hard. So, go ahead and check the Important Notes for CBSE Class 10 Maths Surface Areas and Volumes from this article.

**(A) RIGHT CIRCULAR CYLINDER:**

A right circular cylinder is solid generated by the revolution of a rectangle about of its sides.

NOTE : If a paper, cylinder open at both the ends is cut along a vertical line on the curved surface and stretched on a plane surface, we obtain a rectangle of length i.e., 27πr and breadth= Height of cylinder h.

So, curved surface area (C.S.A) or lateral surface area = 2πr * height

**Important Formula For Cylinder**

1. C. S. A of cylinder = ( Perimeter of base) * Height = 2πrh

2. Area of each end of cylinder = 2πr^{2}

3. Total surface area (including both circular ends) = 2πrh + 2πr^{2} = 27πr(h + r)

4. Volume of cylinder — πr^{2}h = **[(Area of base) * height]**

**Hollow Cylinder’s formulae e.g., (Rubber tubes pipes, etc.)**

1. Volume of material = Exterior volume — Interior

volume = πR^{2}h — πr^{2}h = πh(R^{2} – r^{2})

2. C. S. A or L. S. A = external surface area + internal surface area

= 2πRh + 2πrh

3. T. S . A. of hollow cylinder = C. S. A+ 2 ( area of base ring )

= (2πRh + 2πrh) + 2(πR^{2} – πr^{2})

**NOTE:**

1. Two end faces of right circular cylinder are circles having each area = πr^{2}

2. Mass of cylinder = Volume * density

3. When rectangular sheet of paper is rolled along its length , we get a cylinder whose base circumference is length of sheet and height is same as breadth of sheet.

**(B) CONE**

From figure, AO = height of cone and is denoted by ‘h’

OB = radius of the base of cone, AB = slant height of a cone (l)

**Important Formula Of rt. Circular Cone :**

1. Volume of cone = 1 / 3 πr^{2}h

2 C. S. A or L. S. A=πrl where slant height

= l =√ r^{2} + hr^{2}

3. T. S. A of cone = πrl + πr^{2}

**(C) FRUSTUM OF A CONE**

**FRUSTUM :** A cone is cut by a plane parallel to the base of the cone,

then the portion between the plane and base is called frustum of the cone

**Important Formulae for Frustum :**

1. Volume of frustum of cone

= πh / 3[R^{2} + r^{2} + Rr] cubic unit

2. L. S. A or C. S. A = πl(R + r) Sq units where l^{2} = h^{2} + (R – r)^{2}

3. T. S. A = πR^{2} + πr^{2} + πl(R + r) Sq. units.

(Area of base + Area of top + Area of lateral )

4. Slant height (l) = √h^{2}2 + (R – r)^{2}

**(D) IMPORTANT FORMULA FOR SPHERE AND HEW-SPHERE**

(a) Surface area of sphere = 4πr^{2}

(b) Volume of sphere = 4 / 3 πr^{3}

(c) Volume of hemisphere = 2 / 3 πr^{3}

(d) C.S.A. of hemisphere = 2πr^{2}

(e) Total surface area of Hemi-sphere = 2πr^{2} + πr^{2} =3πr^{2}

**(E) IMPORTANT FORMULA FUR SPHERICAL SHELL/ HEMILSPHERICAL SHELL**

(a) Outer surface area of spherical shell =4πR^{2}

(b) Inner S.A. of spherical shell = 4πr^{2}

(c) Total surface area of spherical shell = 4π(R^{2}+ r^{2})

(d) Volume of spherical shell of external radius R and internal

radius ‘r’ = 4 / 3π(R^{3} – r^{3})

(e) Outer curved surface area hemispherical shell = 2πR^{2}

(f) Inner curved surface area of hemispherical shell = 2πr^{2}

(g) Thick hemispherical bowl of external and internal radii R and r,

Total S.A. = π(3R^{2}+ r^{2})

(h) Volume of hemispherical shell of external radius ‘R’ and internal radius ‘r’

= 2 / 3π(R^{3} — r^{2}).

### Class 10 Key Points, Important Questions & Practice Papers

Hope these notes helped you in your schools exam preparation. Candidates can also check out the Key Points, Important Questions & Practice Papers for various Subjects for Class 10 in both Hindi and English language form the link below.

Class 10 Maths | कक्षा 10 गणित |

Class 10 Science | कक्षा 10 विज्ञान |

Class 10 Social Science | कक्षा 10 सामाजिक विज्ञान |

Class 10 English |

### Class 10 NCERT Solutions

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### Class 10 Mock Test / Practice

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### Class 10 Exemplar Questions

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