Motion and Time Class 7 Notes Science Chapter 13

On this page, you will find Motion and Time Class 7 Notes Science Chapter 13 Pdf free download. CBSE NCERT Class 7 Science Notes Chapter 13 Motion and Time will seemingly help them to revise the important concepts in less time.

CBSE Class 7 Science Chapter 13 Notes Motion and Time

Motion and Time Class 7 Notes Understanding the Lesson

1. The act or process of moving or of changing position or change of posture or a gesture is called motion.

2. Motion of objects are of two types-motion of some objects are slow while that of some others are fast.

3. Objects which take a longer time to cover a certain distance is called slow whereas another object which takes a shorter time to cover the same distance is said to be

4. The speed of an object is defined as the distance travelled by the object in unit time.

Speed = \(\frac{\text { Distance travelled }}{\text { Time taken }}\)

5. The SI unit of speed is metre per second (m/s).

6. If the speed of an object moving along a straight line keeps changing, its motion is said to be non-uniform motion.

7. If an object moving along a straight line with a constant speed, its motion is said to be in uniform motion.

8. The average speed of a body can be defined as the total distance travelled divided by the total time taken.                                                                                     ,

Thus, Average speed = \(\frac{\text { Total Distance travelled }}{\text { Total Time taken }}\)

9. A simple pendulum consists of a small metallic ball, called the bob, suspended from a rigid stand by a thread.

10. The time taken by the pendulum to complete one oscillation is called its time period.

11. Speedometer is an instrument on a vehicle’s dashboard which indicates the speed of the vehicle when it is running. It records the speed directly in km/h.

12. Odometer is an instrument for measuring the distance travelled by the vehicle. It records the distance travelled by the vehicle in kilometres.

13. The basic unit of time is second.

14. A graph is used to study the relation between two inter-dependent physical quantities.

15. The quantity in the graph that is made to alter at will is called independent variable and the other quantity which varies as a result of this change is known as dependent variable. The graph may be a straight or curved line.

16. The distance-time graph for the motion of an object moving with a constant speed is a straight line.

Class 7 Science Chapter 13 Notes Important Terms

Bar graph: A graph in which statistical data are represented in form of bars of different heights.

Graphs: A graph is a mathematical relation between two inter-dependent physical quantities.

Non-uniform motion: When a body covers unequal distances in equal intervals of time or equal distances in unequal intervals of time, then it is said to be in non-uniform motion.

Oscillatory motion: The to and fro motion of a simple pendulum about its mean position is called oscillatory motion.

Speed: The distance travelled per unit time is known as speed.

Speed = \(\frac{\text { Distance Covered }}{\text { Time taken }}\)

Time period: The time taken by a pendulum for one oscillation is called its time period.

Uniform motion: Uniform motion can be defined as the motion in which body travels equal distances in equal intervals of time.

Unit of time: The SI unit of time is second.

Nutrition in Plants Class 7 Notes Science Chapter 1

On this page, you will find Nutrition in Plants Class 7 Notes Science Chapter 1 Pdf free download. CBSE NCERT Class 7 Science Notes Chapter 1 Nutrition in Plants will seemingly help them to revise the important concepts in less time.

CBSE Class 7 Science Chapter 1 Notes Nutrition in Plants

Nutrition in Plants Class 7 Notes Understanding the Lesson

1. Vitamins, carbohydrates, proteins, fats and minerals are the components of food. These components of food are necessary for our body and are called nutrients.

2. The process of procuring and utilization of food by the body is called nutrition.

3. The mode of nutrition in which organisms make food themselves from simple substances is called autotrophic nutrition.

4. The organisms which cannot make their food and depend on others for their food are called heterotrophs and this type of nutrition is called heterotrophic nutrition.

5. Carbon dioxide from air is taken in through the tiny pores present on the surface of the leaves. These pores are surrounded by ‘guard cells’ such pores are called stomata.

6. The leaves have a green pigment and they are known as chlorophyll.

7. Plants use sunlight, carbon dioxide and water in presence of chlorophyll to synthesise their food. This process is known as photosynthesis.
Nutrition in Plants Class 7 Notes Science Chapter 1
8. The sun is the ultimate source of energy for all living organisms.

9. The bodies of living organisms are made of tiny units which are called cells.

10. The organisms either plants or animals that derive nutrients from another organisms are known as parasites. For example, Cuscuta, lice, etc.

11. The organisms which provide nutrients to another organisms without being benefitted are known as hosts.

12. The type of plants which trap insects and digest them by producing digestive juices are called insectivorous plants. For example, pitcher plant, sundew, venus fly trap, etc.

13. The mode of nutrition in which organisms take in nutrients in form of solution from dead and decaying matter is called saprophytic nutrition.

14. Plants which use saprophytic mode of nutrition are called saprotrophs.

15. The organisms which live together and share shelter and nutrients are said to be in symbiotic relationship. For example, certain fungi live in the roots of trees.

Class 7 Science Chapter 1 Notes Important Terms

Autotrophic: It is the mode of nutrition in which organisms prepare its own food by using sunlight, air and other essential substances.

Chlorophyll: Chlorophyll is a green pigment present in leaves which absorb solar energy from sunlight, and enables the plant to prepare their food through the process of photosynthesis.

Heterotrophs: The organisms which cannot prepare their food and depends on other organisms for their food are known as heterotrophs.

Host: A plant or an animal which support other organisms by giving food, shelter, etc., is known as host.

Insectivorous: The plant or animal which eat insect for their survival is known as insectivorous.

Nutrients: Vitamins, carbohydrates, proteins, fats and minerals are the components of food. These components of food are essential for our body and are called nutrients.

Parasite: The organism which depends on other organisms for deriving nutrients and their survival is known as parasite.

Photosynthesis: The process by which plants prepare their own food in the presence of sunlight, water, carbon dioxide and chlorophyll is called photosynthesis.

Saprotrophs: Organisms which take their nutrients from dead and decaying matter are called saprotrophs.

Saprotrophic: The mode of nutrition in which an organism take its food from dead and decaying matter is known as saprotrophic mode of nutrition.

Stomata: The tiny pores which are present on the surface of the leaves are called stomata.

CBSE Class 7 Science Notes | NCERT Class 7 Science Revision Notes

We are providing students with chapter-wise CBSE class 7 Science notes. These Class 7 Science Notes are put together by subject experts and based on the latest CBSE syllabus. The free NCERT Class 7 Science Notes PDF available here come with detailed explanations of important topics to further make learning easy for students.

CBSE Class 7 Science Notes

Below shows the CBSE Class 7 Science Notes for all the chapters. Students can check out these notes by clicking on the link below.

  1. Nutrition in Plants Class 7 Notes
  2. Nutrition in Animals Class 7 Notes
  3. Fibre to Fabric Class 7 Notes
  4. Heat Class 7 Notes
  5. Acids, Bases and Salts Class 7 Notes
  6. Physical and Chemical Changes Class 7 Notes
  7. Weather, Climate and Adaptations of Animals to Climate Class 7 Notes
  8. Winds, Storms and Cyclones Class 7 Notes
  9. Soil Class 7 Notes
  10. Respiration in Organisms Class 7 Notes
  11. Transportation in Animals and Plants Class 7 Notes
  12. Reproduction in Plants Class 7 Notes
  13. Motion and Time Class 7 Notes
  14. Electric Current and Its Effects Class 7 Notes
  15. Light Class 7 Notes
  16. Water: A Precious Resource Class 7 Notes
  17. Forests: Our Lifeline Class 7 Notes
  18. Wastewater Story Class 7 Notes

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 6
Chapter Name The Triangle and its Properties
Exercise Ex 6.5
Number of Questions Solved 8
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

Question 1.
PQR is a triangle right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 1
Solution:
QR2 = 102 + 242 By Pythagoras Property
⇒ = 100 + 576 = 676
⇒ QR = 26 cm.

Question 2.
ABC is a triangle right-angled at C. If AB – 25 cm and AC = 7 cm, find BC.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 2
Solution:
AC2 + BC2 = AB2 By Pythagoras Property
⇒ 72 + BC2 = 252
⇒ 49 + BC2 = 625
⇒ BC2 = 625 – 49
⇒ BC2 = 576
⇒ BC = 24 cm.

Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 3
Solution:
Let the distance of the foot of the ladder from the wall be a m. Then,
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 4
Hence, the distance of the foot of the ladder from the wall is 9 m.

Question 4.
Which of the following can be the sides of a right triangle ?

  1. 2.5 cm, 6.5 cm, 6 cm.
  2. 2 cm, 2 cm, 5 cm.
  3. 1.5 cm, 2 cm, 2.5 cm.

In the case of right-angled triangles, identify the right angles.
Solution:
1. 2.5 cm, 6.5 cm, 6 cm We see that
(2.5)2 + 62 = 6.25 + 36 = 42.25 = (6.5)2
Therefore, the given lengths can be the sides of a right triangle. Also, the angle between the lengths, 2.5 cm and 6 cm is a right angle.

2. 2 cm, 2 cm, 5 cm
∵ 2 + 2 = 4 \(\ngtr\) 5
∴ The given lengths cannot be the sides of a triangle
The sum of the lengths of any two sides of a triangle is greater than the third side

3. 1.5 cm, 2 cm, 2.5 cm We find that
1.52 + 22 = 2.25 + 4 = 6.25 = 2.52
Therefore, the given lengths can be the sides of a right triangle.
Also, the angle between the lengths 1.5 cm and 2 cm is a right angle.

Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Solution:
AC = CD Given
In right angled triangle DBC, DC2 = BC2 + BD2
by Pythagoras Property = 52 + 122 = 25 + 144 = 169
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 5
⇒ DC = 13 ⇒ AC = 13
⇒ AB = AC + BC = 13 + 5 = 18
Therefore, the original height of the tree = 18 m.

Question 6.
Angles Q and R of a ∆ PQR are 25° and 65°. Write which of the following is true:
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 6
(i) PQ2 + QR2 = RP2
(ii) PQ2 + RP2 = QR2
(iii) RP2 + QR2 = PQ2
Solution:
(ii) PQ2 + RP2 = QR2 is true.

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 7
Solution:
In right-angled triangle DAB, AB2 + AD2 = BD2
⇒ 402 + AD2 = 412 ⇒ AD2 = 412 – 402
⇒ AD2 = 1681 – 1600
⇒ AD2 = 81 ⇒ AD = 9
∴ Perimeter of the rectangle = 2(AB + AD) = 2(40 + 9) = 2(49) = 98 cm
Hence, the perimeter of the rectangle is 98 cm.

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 8
Solution:
Let ABCD be a rhombus whose diagonals BD and AC are of lengths 16 cm and 30 cm respectively.
Let the diagonals BD and AC intersect each other at O.
Since the diagonals of a rhombus bisect each other at right angles. Therefore
BO = OD = 8 cm,
AO = OC = 15 cm,
∠AOB = ∠BOC
= ∠COD = ∠DOA = 90°
In right-angled triangle AOB.
AB2 = OA2 + OB2
By Pythagoras Property
⇒ AB2 = 152 + 82
⇒ AB2 = 225 + 64
⇒ AB2 = 289
⇒ AB = 17cm
Therefore, perimeter of the rhombus ABCD = 4 side = 4 AB = 4 × 17 cm = 68 cm
Hence, the perimeter of the rhombus is 68 cm.

We hope the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 15
Chapter Name Visualising Solid Shapes
Exercise Ex 15.2
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2

Question 1.
Use isometric dot paper and make an isometric sketch for each one of the given shapes :
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 8
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 9
Solution:
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 10
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 11

Question 2.
The dimensions of a cuboid are 5 cm, 3cm and 2cm. Draw three different isometric sketches of this cuboid
Solution:
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 12
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 13

Question 3.
Three cubes each with 2 cm edge are placed side by side to for,n a cuboid. Sketch an oblique or isometric sketch of this cuboid.
Solution:
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 14

Question 4.
Make an oblique sketch for each one of the given isometric shapes :
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 15
Solution:
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 16

Question 5.
Give (i) an oblique sketch and (ii) an isometric sketch for each of the following:
(a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique?)
(b) A cube with an edge 4 cm long.
Solution:
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 17
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 18
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 19

We hope the NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 15
Chapter Name Visualising Solid Shapes
Exercise Ex 15.3
Number of Questions Solved 1
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3

Question 1.
What cross-sections do you get when you give a
(i) vertical cut
(ii) horizontal cut to the following solids?
(a) A brick
(b) A round apple
(c) A dice
(d) A circular pipe
(e) An ice cream cone.
Solution:

S.No. Vertical Cut Horizontal Cut
(a) (i) rectangle (ii) rectangle
(b) (i) circle (ii) circle
(c) (i) square (ii) square
(d) (i) circle (ii) rectangle
(e) (i) triangle (ii) circle

We hope the NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 15
Chapter Name Visualising Solid Shapes
Exercise Ex 15.4
Number of Questions Solved 3
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4

Question 1.
A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions):
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 21
Solution:

  1. a circle.
  2. a rectangle.
  3. a rectangle.

Question 2.
Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these!)
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 22
NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes 23

Solution:

  1. → Sphere, cylinder
  2. → Cube, cuboid (of the square end)
  3. → Cone, the prism of the triangular base
  4. → Cylinder, the prism of the square base

Question 3.
Examine if the following are true statements:

  1. The cube can cast a shadow in the shape of a rectangle.
  2. The cube can cast a shadow in the shape of a hexagon.

Solution:

  1. True, the cube can cast a shadow in the shape of a rectangle.
  2. False, the cube can not cast a shadow in the shape of a hexagon.

We hope the NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 15 Visualising Solid Shapes Ex 15.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 14
Chapter Name Symmetry
Exercise Ex 14.2
Number of Questions Solved 2
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2

Question 1.
Which of the following figures have rotational symmetry of order more than 1?
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 18

Solution:
Figures (a), (b), (d), (e) and (f) have rotational symmetry of order more than 1.

Question 2.
Give the order of rotational symmetry for each figure :
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 19
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 20
Solution:

(a) → 2
(b) → 2
(c) → 3
(d) → 4
(e) → 4
(f) → 5
(g) → 6
(h) → 3

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 21

Mark a point P as shown in figure (i). We see that in a full turn, there are two positions (on rotation through the angles 180° and 360°) when the figure looks exactly the same. Because of this, it has rotational symmetry of order 2.
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 21

Mark a point P as shown in figure (i). We see that in a full turn, there are two positions (on rotation through the angles 180° and 360°) when the figure looks exactly the same.

Because of this, it has rotational symmetry of order 2.
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 23

Mark a point P as shown in figure (i). We see that in a full turn, there are three positions (on rotation through the angles 120°, 240°, and 360°) when the figure looks exactly the same. Because of this, it has rotational symmetry of order 3.
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 24

Mark a point P as shown in figure (i). We see that in a full turn, there are four positions (on rotation through the angles 90°, 180°, 270°, and 360°) when the figure looks exactly the same. Because of this, we say that it has rotational symmetry of order 4.

Similarly,

(e) In a full turn, there are four positions (on rotation through the angles 90°, 180°, 270°, and 360°) when the figure looks exactly the same.

∴ It has rotational symmetry of order 4.

(f) The figure is a regular pentagon. In a full turn, there are five positions (on rotation through the angles 72°, 144°, 216°, 288°, and 360°) when the figure looks exactly the same.

∴ It has rotational symmetry of order 5.

(g) In a full turn, there are six positions (on rotation through the angles 60°, 120°, 180°, 240°, 300°, and 360°) when the figure looks exactly the same.

It has rotational symmetry of order 6.

(h) In a full turn, there are three positions (on rotation through the angles 120°, 240°, and 360°) when the figure looks exactly the same.

∴ It has rotational symmetry of order 3.

We hope the NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 14
Chapter Name Symmetry
Exercise Ex 14.3
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3

Question 1.
Name any two figures that have both line symmetry and rotational symmetry.
Solution:
Two figures that have both line symmetry and rotational symmetry are an equilateral triangle and a circle.

Question 2.
Draw, wherever possible, a rough sketch of
(i) a triangle with both line and rotational symmetry of order more than 1.
(ii) a triangle with only line symmetry and no rotational symmetry of order more than 1.
(iii) a quadrilateral with rotational symmetry of order more than 1 but not a line symmetry.
(iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Solution:
(i) An equilateral triangle has 3 lines of symmetry and rotational symmetry of order 3.
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 25
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 26
(ii) An isosceles triangle has only one line symmetry but no rotational symmetry of order more than 1.
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 27
(iii) A parallelogram has no line of symmetry but has a rotational symmetry of order 2.
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 28
(iv) An isosceles trapezium has one line of symmetry but no rotational symmetry of order more than 1.
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 29

Question 3.
If a figure has two or more lines of symmetry, should it have rotational symmetry of order more than 1?
Solution:
When a figure has two or more lines of symmetry, then the figure should have rotational symmetry of order more than 1.

Question 4.
Fill in the blanks:
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 30
Solution:
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 31

Question 5.
Name the quadrilaterals which have both line and rotational symmetry of order more than 1.
Solution:
The name of quadrilaterals having both line and rotational symmetry is square.

Question 6.
After rotating by 60° about a centre, a figure looks exactly the same as its original position. At what other angles will this happen for the figure?
Solution:
The other angles are 120°, 180°, 240°, 300°, and 360°.
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 44

Question 7.
Can we have a rotational symmetry of order more than 1 whose angle of rotation is
(i) 45°?
(ii) 17°?
Solution:
(i) Yes
NCERT Solutions for Class 7 Maths Chapter 14 Symmetry 32
(ii) No

We hope the NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 13
Chapter Name Exponents and Powers
Exercise Ex 13.3
Number of Questions Solved 4
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3

Question 1.
Write the following numbers in the expand forms :
(i) 279404
(ii) 3006194
(iii) 2806196
(iv) 120719
(v) 20068.
Solution:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 34
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 35

Question 2.
Find the number from each of the following expanded forms :
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 36
Solution:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 37
Question 3.
Express the following numbers in standard form:

  1. 5,00,00,000
  2. 70,00,000
  3. 3,18,65,00,000
  4. 3,90,878
  5. 39087.8
  6. 3908.78

Solution:

  1. 5,00,00,000 = 5 × 107
  2. 70,00,000 = 7 × 106
  3. 3,18,65,00,000 = 3.1865 × 109
  4. 3,90,878 = 3.90878 × 105
  5. 39087.8 = 3.90878 × 104
  6. 3908.78 = 3.90878 × 103

Question 4.
Express the number appearing in the following statements in standard form :

  1. The distance between Earth and Moon is 384,000,000 m.
  2. The speed of light in a vacuum is 300,000,000 miles.
  3. The diameter of the Earth is 1,27,56,000 m.
  4. Diameter of the Sun is 1,400,000,000 m.
  5. In a galaxy, there are on average 100,000,000,000 stars.
  6. The universe is estimated to be about 12,000,000,000 years old.
  7. The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.
  8. 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
  9. The earth has 1,353,000,000 cubic km of seawater.
  10. The population of India was about 1,027,000,000 in March 2001.

Solution:

  1. The mean distance between Earth and Moon is 3.84 × 108 m.
  2. The speed of light in a vacuum is 3 × 108 m/s.
  3. The diameter of the Earth is 1.2756 × 107 m.
  4. The diameter of the Sun is 1.4 × 109 m.
  5. In a galaxy, there are on average 1 × 1011 stars.
  6. The universe is estimated to be about 1.2 × 1010 years old.
  7. The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 3 × 1020 m.
  8. 6.023 × 1022 molecules are contained in a drop of water weighing 1.8 gm.
  9. The earth has 1.353 × 109 cubic km of seawater.
  10. The population of India was about 1.027 × 109 in March 2001.

We hope the NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2.

Board CBSE
Textbook NCERT
Class Class 7
Subject Maths
Chapter Chapter 13
Chapter Name Exponents and Powers
Exercise Ex 13.2
Number of Questions Solved 5
Category NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2

Question 1.
Using laws of exponents, simplify and write the answer in exponential form :
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 18
Solution:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 19
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 20
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 21

Question 2.
Simplify and express each of the following in exponential form:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 22
Solution:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 23
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 24
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 25

Question 3.
Say true or false and justify your answer:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 26
Solution:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 27

Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768.
Solution:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 28
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 29
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 30

Question 5.
Simplify
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 31
Solution:
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 32
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers 33

We hope the NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2, drop a comment below and we will get back to you at the earliest.