NCERT Solutions For Class 7 Science Chapter 15 : Visualizing Solid Shapes

NCERT Solutions for class 7 Mathematics Chapter 15 Visualizing Solid Shapes are provided in the article below. Plane shapes have two measurements like length and breadth which are called dimensions of a plane and therefore they are called two-dimensional shapes whereas a solid object has three measurements like length, breadth, height or depth. Hence, they are called three-dimensional shapes or 3-D shapes. Some of the important topics in this chapter include:

Download: NCERT Solutions for Class 7 Mathematics Chapter 15 pdf

NCERT Solutions for Class 7 Mathematics Chapter 15

NCERT Solutions for Class 7 Mathematics Chapter 15 Visualizing Solid Shapes are given below.

NCERT Solution class 7 mathematics chapter 15

Class 7 Maths Chapter 15 Visualising Solid Shapes – Important Topics

Three dimensional figure: Three-dimensional figures are those which consist of length, breadth and height.

Some of the important three dimensional figures include:

Cube
Cuboid
Cone
Cylinder
Pyramid

Volume of Cube – a³

Volume of Cuboid – length x breadth x height

Volume of Cone – 1/3 $\pi$r²h

Volume of Cylinder – $\pi$r²h

NCERT Solutions for Class 7 Maths Chapter 15 Exercises

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Exercises are given below.

Read More:

Visualising Solid Shapes MCQ	Measurement of Objects	Isomeric Sketch
Rectangular Pyramid	3D Shapes	Hexagon Formula

Read More:

NCERT Solutions for Class 7 Maths	Number System	Geometry
Arithmetics	Basic Maths	Maths Formulas

CBSE X Related Questions

1.
The value of $\left(\frac{1}{2} \tan^2 45^\circ - \cos^2 60^\circ\right)$ is :
- $0$
- $-\frac{1}{2}$
- $\frac{1}{4}$
- $-\frac{1}{4}$
View Solution
2.
It is given that $\triangle ABC \sim \triangle EDF$. Which of the following is not true?
- $\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle EDF} = \frac{AB}{ED}$
- $\frac{AB}{ED} = \frac{AC}{EF}$
- $\angle A = \angle D, \angle C = \angle F$
- $\frac{AB + BC}{AC} = \frac{ED + DF}{EF}$
View Solution
3.
For an acute angle $\theta$, if $\sin \theta = \frac{1}{9}$, then value of $\frac{9 \csc \theta + 1}{9 \csc \theta - 1}$ is
- $0$
- $\frac{80}{81}$
- $1$
- $\frac{82}{80}$
View Solution
4.
If $ \alpha $ and $ \beta $ are two zeroes of a polynomial $ f(x) = px^2 - 2x + 3p $ and $ \alpha + \beta = \alpha\beta $, then value of p is :
- $ -\frac{2}{3} $
- $ \frac{2}{3} $
- $ \frac{1}{3} $
- $ -\frac{1}{3} $
View Solution
5.
Arc $PQ$ subtends an angle $\theta$ at the centre of the circle with radius $6.3 \text{ cm}$. If $\text{Arc } PQ = 11 \text{ cm}$, then the value of $\theta$ is
- $10^{\circ}$
- $60^{\circ}$
- $45^{\circ}$
- $100^{\circ}$
View Solution
6.
Which of the following graphs represents a polynomial with both zeroes being positive?
View Solution

View All

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NCERT Solutions For Class 7 Science Chapter 15 : Visualizing Solid Shapes

Download: NCERT Solutions for Class 7 Mathematics Chapter 15 pdf

NCERT Solutions for Class 7 Mathematics Chapter 15

Class 7 Maths Chapter 15 Visualising Solid Shapes – Important Topics

NCERT Solutions for Class 7 Maths Chapter 15 Exercises

CBSE X Related Questions

1.
The value of $\left(\frac{1}{2} \tan^2 45^\circ - \cos^2 60^\circ\right)$ is :

2.
It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

4.
If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

5.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

6.
Which of the following graphs represents a polynomial with both zeroes being positive?

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NCERT Solutions For Class 7 Science Chapter 15 : Visualizing Solid Shapes

​Download: NCERT Solutions for Class 7 Mathematics Chapter 15 pdf

NCERT Solutions for Class 7 Mathematics Chapter 15

Class 7 Maths Chapter 15 Visualising Solid Shapes – Important Topics

NCERT Solutions for Class 7 Maths Chapter 15 Exercises

CBSE X Related Questions

1.The value of $\left(\frac{1}{2} \tan^2 45^\circ - \cos^2 60^\circ\right)$ is :

2.It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

3.For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

4.If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

5.Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

6.Which of the following graphs represents a polynomial with both zeroes being positive?

Comments

SUBSCRIBE TO OUR NEWS LETTER

Download: NCERT Solutions for Class 7 Mathematics Chapter 15 pdf

1.
The value of $\left(\frac{1}{2} \tan^2 45^\circ - \cos^2 60^\circ\right)$ is :

2.
It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

4.
If \( \alpha \) and \( \beta \) are two zeroes of a polynomial \( f(x) = px^2 - 2x + 3p \) and \( \alpha + \beta = \alpha\beta \), then value of p is :

5.
Arc \(PQ\) subtends an angle \(\theta\) at the centre of the circle with radius \(6.3 \text{ cm}\). If \(\text{Arc } PQ = 11 \text{ cm}\), then the value of \(\theta\) is

6.
Which of the following graphs represents a polynomial with both zeroes being positive?