Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Short Notes: Surface Areas and Volumes

Surface Area and Volumes Class 10 Notes Maths Chapter 12

Formulas for Surface Area and Volume of various Solids

Surface Area and Volume of Combinations

1. Cone on a Cylinder


Surface Area and Volumes Class 10 Notes Maths Chapter 12
r = radius of cone & cylinder;
h1 = height of cone
h2 = height of cylinder
Total Surface area = Curved surface area of cone + Curved surface area of cylinder + area of circular base
= πrl + 2πrh2 +πr2; 
Slant height, l = Surface Area and Volumes Class 10 Notes Maths Chapter 12
Total Volume = Volume of cone + Volume of cylinder 
Surface Area and Volumes Class 10 Notes Maths Chapter 12

2. Cone on a Hemisphere

Surface Area and Volumes Class 10 Notes Maths Chapter 12
h = height of cone; 
I = slant height of cone = Surface Area and Volumes Class 10 Notes Maths Chapter 12
r = radius of cone and hemisphere 
Total Surface area = Curved surface area of cone + Curved surface area of hemisphere = πrl + 2πr2 
Volume = Volume of cone + Volume of hemisphere = Surface Area and Volumes Class 10 Notes Maths Chapter 12

Question for Short Notes: Surface Areas and Volumes
Try yourself:In the context of a cone placed on top of a hemisphere, which term describes the distance from the apex of the cone to a point on the outer edge of the hemisphere's base?
View Solution

3. Conical Cavity in a Cylinder

Surface Area and Volumes Class 10 Notes Maths Chapter 12r = radius of cone and cylinder;
h = height of cylinder and conical cavity;
l = Slant height
Total Surface area = Curved surface area of cylinder + Area of the bottom face of cylinder + Curved surface area of cone = 2πrh + πr2 + πrl
Volume = Volume of cylinder – Volume of cone = Surface Area and Volumes Class 10 Notes Maths Chapter 12

4. Cones on Either Side of the Cylinder

Surface Area and Volumes Class 10 Notes Maths Chapter 12r = radius of cylinder and cone;
h1 = height of the cylinder
h2 = height of cones
Slant height of cone, l = Surface Area and Volumes Class 10 Notes Maths Chapter 12
Surface area = Curved surface area of 2 cones + Curved surface area of cylinder = 2πrl + 2πrh1 
Volume = 2(Volume of cone) + Volume of cylinder = Surface Area and Volumes Class 10 Notes Maths Chapter 12

5. Cylinder with Hemispherical Ends

Surface Area and Volumes Class 10 Notes Maths Chapter 12r = radius of cylinder and hemispherical ends;
h = height of cylinder
Total surface area= Curved surface area of cylinder + Curved surface area of 2 hemispheres = 2πrh + 4πr2
Volume = Volume of cylinder + Volume of 2 hemispheres =  πr2h+4/3πr3

6. Hemisphere on Cube or Hemispherical Cavity on Cube

a = side of cube;

r = radius of hemisphere.

Surface area = Surface area of cube – Area of hemisphere face + Curved surface area of hemisphere

= 6a2 – πr+ 2πr= 6a2 + πr2

Volume = Volume of cube + Volume of hemisphere = Surface Area and Volumes Class 10 Notes Maths Chapter 12

7. Hemispherical Cavity in a Cylinder

Surface Area and Volumes Class 10 Notes Maths Chapter 12r = radius of hemisphere;
h = height of cylinder
Total surface area = Curved surface area of cylinder + Surface area of base + Curved surface area of hemisphere
= 2πrh + πr2 + 2πr2 = 2πrh + 3πr2
Volume = Volume of cylinder – Volume of hemisphere =Surface Area and Volumes Class 10 Notes Maths Chapter 12

Question for Short Notes: Surface Areas and Volumes
Try yourself:What is the formula for the total surface area of a cylinder with hemispherical ends?
View Solution

Some Solved Examples

Example 1: A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Sol:

The figure drawn below of the vessel and lead shot is to visualize it.

Surface Area and Volumes Class 10 Notes Maths Chapter 12

As the water is filled up to the brim in the vessel

Volume of water in the vessel = Volume of the conical vessel

On dropping a certain number of lead shots (sphere) into the vessel one-fourth of the water flows out.

Volume of all lead shots dropped into the vessel = 1/4 × Volume of the water in the vessel

Hence,

Number of lead shots = 1/4 × volume of the water in the vessel ÷ volume of each lead shot

We will find the volume of the water in the vessel and lead shot by using formulae;

Volume of the sphere = 4/3 πr3

where r is the radius of the sphere

Volume of the cone = 1/3 πR2h

where R and h are the radius and height of the cone respectively

Height of the conical vessel, h = 8 cm

Radius of the conical vessel, R = 5 cm

Radius of the spherical lead shot, r = 0.5 cm

Number of lead shots = 1/4 × volume of the water in the vessel ÷ volume of each lead shot

= 1 /4 × (1/3) πR2h × 3/4 πr3

= πR2h/12 × 3/4πr3

= R2h / 16r3

= (5cm × 5 cm × 8 cm) / (16 × 0.5 cm × 0.5 cm × 0.5 cm)

= 100

Thus, the number of lead shots dropped in the vessel is 100.

Example 2: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Sol:

The visualization of the solid figure is drawn below.

Surface Area and Volumes Class 10 Notes Maths Chapter 12

The volume of a solid is the space occupied inside it or the capacity that an object holds.

As the solid is made up of a conical part and a hemispherical part,

Volume of the solid = volume of the conical part + volume of the hemispherical part

Let us find the volume of the solid by using formulae;

Volume of the hemisphere = 2/3 πr3 where r is the radius of the hemisphere

Volume of the cone = 1/3 πr2h where r and h are the radius and height of the cone respectively.

Radius of hemispherical part = Radius of conical part = r = 1 cm

Height of conical part = h = r = 1 cm

Volume of the solid = volume of the conical part + volume of the hemispherical part

= 1/3 πr2h + 2/3 πr3

= 1/3 πr3 + 2/3 πr3 [Since, h = r]

= πr3

= π (1cm)3

= π cm3

Thus, the volume of the solid is π cm3.

The document Surface Area and Volumes Class 10 Notes Maths Chapter 12 is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10
126 videos|457 docs|75 tests

Top Courses for Class 10

126 videos|457 docs|75 tests
Download as PDF
Explore Courses for Class 10 exam

Top Courses for Class 10

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Free

,

past year papers

,

pdf

,

Important questions

,

study material

,

ppt

,

Surface Area and Volumes Class 10 Notes Maths Chapter 12

,

Extra Questions

,

Semester Notes

,

Surface Area and Volumes Class 10 Notes Maths Chapter 12

,

practice quizzes

,

Sample Paper

,

Viva Questions

,

Surface Area and Volumes Class 10 Notes Maths Chapter 12

,

video lectures

,

mock tests for examination

,

Summary

,

Objective type Questions

,

Previous Year Questions with Solutions

,

shortcuts and tricks

,

Exam

,

MCQs

;