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.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.
View Solution
2.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals
- \(2.5\) cm
- \(10.5\) cm
- \(3.5\) cm
- \(\frac{14}{3}\) cm
View Solution
3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
View Solution
4.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.
View Solution
5.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
View Solution
6.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).
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.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

6.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

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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.D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

3.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

6.If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

Comments

SUBSCRIBE TO OUR NEWS LETTER

Download: NCERT Solutions for Class 7 Mathematics Chapter 15 pdf

1.
D is the mid-point of side BC of \(\triangle ABC\). CE and BF intersect at O, a point on AD. AD is produced to G such that \(OD = DG\). Prove that OBGC is a parallelogram.

2.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

3.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

4.
A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is \(3/5\), then find the number of yellow balls.

5.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

6.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).