NCERT Solutions for Class 9 Maths Chapter 13 : Surface Areas And Volumes

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The NCERT Solutions for Class 9 Mathematics Chapter 13 Surface Areas And Volumes have been provided in this article. Surface Areas and Volumes deals with the calculation of surface area, total area and volume of various 3-D figures such as cube, cuboid, cone, cylinder, prism, etc. Surface area is the area that is occupied by the surface of the object while volume is the total space that is occupied by the whole object.

Class 9 Maths Chapter 13 Surface Areas and Volumes belong to Unit 5 Mensuration which has a weightage of 13 marks in the Class 9 Maths Examination. Class 9 Mathematics Chapter 13 has the following important concepts:

Download: NCERT Solutions for Class 9 Mathematics Chapter 13 pdf

NCERT Solutions for Class 9 Mathematics Chapter 13

NCERT Solution

Important Topics in Class 9 Maths Chapter 13 Surface Areas and Volumes

Important Topics in Class 9 Maths Chapter 13 Surface Areas and Volumes are given below:

Area of Prism

Prism refers to a three-dimensional solid figure which has two ends that are identical. A Prism is a combination of plane faces, identical bases and equivalent cross-sections.

Example: Express the Area of a Prism.

Solution: Considering the area of the lateral surface = S₁xh + S₂xh +S₃xh + .........+ S_nxh
= h (S₁+ S₂+ S₃…+ S_n)
Thus,
= height x perimeter of the base
Area of the total bases = 2 x Area of the base
Hence, the total area of the prism is = (height x perimeter of the base) + (2 x Area of the base)

Surface Area of a Cylinder

The total amount of area a cylinder is seen to cover is its surface are. It is further classified into two segments: The Curved Surface Area (CSA) and the Total Surface Area (TSA).

Important Formulae to Remember:

Surface Area of Cylinder = 2πr (h + r) sq. unit
Curved surface area (CSA) of cylinder = 2πrh sq. units.
Total surface area (TSA) = 2π × r × h + 2πr² = 2πr (h + r) sq. units
Area of the circular bases of cylinder = 2 (πr²)

Surface Area of a Cone Formula

A cone, in geometry, is a three-dimensional object which narrows gradually from a circular base to an upward point, otherwise known as the apex or vertex.

Total Surface Area of a Cone = πr² + πrl
Here,

I = Slant height
r = The base circle's radius
SA = Total surface area
h = Height of cone

Cuboid

A cuboid is a three-dimensional figure that comprises six rectangular faces placed at right angles. The volume of a cuboid is calculated considering its length, breadth and height.

Example: What is the total surface area of a plywood box which is 40 cm long, 35 cm wide, and 20 cm high?

Solution: We are already aware of the formula required to determine the total surface area of the plywood box,
Total surface area of cuboid = 2 (lb + bh + lh)
= 2 (40 x 35 + 35 x 20 + 40 x 20)
= 2 (1400 + 700 + 800)
= 2 (2900)
= 5800 sq. cm.

NCERT Solutions for Class 9 Maths Chapter 13 Exercises:

The detailed solutions for all the NCERT Solutions for Surface Areas and Volumes under different exercises are:

Also check:

Chapter Related Topics
Surface Area of Cuboid	Area of Hollow Cylinder	Sphere Formula
Volume of a Pyramid Formula	Surface Area of Hemisphere	Surface Areas and Volume MCQs

Also Read:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
Math Formula	NCERT Solutions for Class 9 Maths	Geometry

CBSE X Related Questions

1.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

View Solution

2.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . . .
(ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
(iv) – 10, – 6, – 2, 2, . . .
(v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
(vi) 0.2, 0.22, 0.222, 0.2222, . . . .
(vii) 0, – 4, – 8, –12, . . . .
(viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
(ix) 1, 3, 9, 27, . . . .
(x) a, 2a, 3a, 4a, . . . .
(xi) a, \(a^2, a^3, a^4,\) . . . .
(xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
(xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
(xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
(xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

View Solution

3.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
  \(\frac{s}{3} + \frac{t}{2}\) =6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)\(\sqrt2x\) + \(\sqrt3y\)=0
  \(\sqrt3x\) - \(\sqrt8y\) = 0
(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
  \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

View Solution

4.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

View Solution

5.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

View Solution

6.
Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

View Solution

View All

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NCERT Solutions for Class 9 Maths Chapter 13 : Surface Areas And Volumes

NCERT Solutions for Class 9 Mathematics Chapter 13

Important Topics in Class 9 Maths Chapter 13 Surface Areas and Volumes

Area of Prism

Surface Area of a Cylinder

Surface Area of a Cone Formula

Cuboid

NCERT Solutions for Class 9 Maths Chapter 13 Exercises:

CBSE X Related Questions

1.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

4.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

5.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

6.
Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 9 Maths Chapter 13 : Surface Areas And Volumes

NCERT Solutions for Class 9 Mathematics Chapter 13

Important Topics in Class 9 Maths Chapter 13 Surface Areas and Volumes

Area of Prism

Surface Area of a Cylinder

Surface Area of a Cone Formula

Cuboid

NCERT Solutions for Class 9 Maths Chapter 13 Exercises:

CBSE X Related Questions

1. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

4. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

5. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

6. Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

4.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

5.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

6.
Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).