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RS Agarwal Solution | Class 9th | Chapter-6 | Introduction to Euclid’s Geometry | Edugrown

Exercise MCQ

Question 1

In ancient India, the shapes of altars used for household rituals were

(a) squares and rectangles

(b) squares and circles

(d) trapeziums and pyramidsSolution 1

Correct option: (b)

Squares and circular altars were used for household rituals.

Whereas altars having shapes as combinations of rectangles, triangles and trapeziums were used for public worship.Question 2

In ancient India, altars with combination of shapes like rectangles, triangles and trapeziums were used for

(a) household rituals

(b) public rituals

(d) none of (a), (b) and (c)Solution 2

Correct option: (b)

In ancient India, altars with combination of shapes like rectangles, triangles and trapeziums were used for public rituals. Question 3

The number of interwoven isosceles triangles in Sriyantra is

(a) five

(b) seven

(d) elevenSolution 3

Correct option: (c)

The Sriyantra consists of nine interwoven isosceles triangles.Question 4

In Indus Valley Civilization (about 300 BC) the bricks used for construction work were having dimensions in the ratio

(a) 5:3:2

(b) 4:2:1

(d) 6:4:2Solution 4

Correct option: (b)

In Indus Valley Civilization (about 300 BC) the bricks used for construction work were having dimensions in the ratio is 4:2:1.Question 5

Into how many chapters was the famous treatise, ‘The Elements’ divided by Euclid?

(a) 13

(b) 12

(d) 9Solution 5

Correct option: (a)

The famous treatise ‘The Elements’ was divided into 13 chapters by Euclid.Question 6

Euclid belongs to the country

(a) India

(b) Greece

(d) EgyptSolution 6

Correct option: (b)

Euclid belongs to the country, Greece.Question 7

Thales belongs to the country

(a) India

(b) Egypt

(d) BabyloniaSolution 7

Correct option: (c)

Thales belongs to the country, Greece.Question 8

Pythagoras was a student of

(a) Euclid

(b) Thales

(d) BhaskaraSolution 8

Correct option: (b)

Pythagoras was a student of Thales.Question 9

Which of the following needs a proof?

(a) axiom

(b) postulate

(d) theoremSolution 9

Correct option: (d)

A statement that requires a proof is called a theorem.Question 10

‘Lines are parallel if they do not intersect’ is started in the form of

(a) a definition

(b) an axiom

(d) a theoremSolution 10

Correct option: (a)

‘Lines are parallel if they do not intersect’ is started in the form of a definition.Question 11

Euclid stated that ‘All right angles are equal to each other’ in the form of

(a) a definition

(b) an axiom

(d) a proofSolution 11

Correct option: (c)

Euclid stated that ‘All right angles are equal to each other’ in the form of a postulate.

This is Euclid’s Postulate 4.

Note: The answer in the book is option (a). But if you have a look at the Euclid’s postulate, the answer is a postulate.Question 12

A pyramid is a solid figure, whose base is

(a) only a triangle

(b) only a square

(d) any polygonSolution 12

Correct option: (d)

A pyramid is a solid figure, whose base is any polygon.Question 13

The side faces of a pyramid are

(a) triangles

(b) squares

(d) polygonsSolution 13

Correct option: (a)

The side faces of a pyramid are triangles.Question 14

The number of dimensions of a solid are

(a) 1

(b) 2

(d) 5Solution 14

Correct option: (c)

A solid has 3 dimensions.Question 15

The number of dimensions of a surface is

(a) 1

(b) 2

(d) 0Solution 15

Correct option: (b)

A surface has 2 dimensions.Question 16

How many dimensions dose a point have

(a) 0 dimension

(b) 1 dimension

(d) 3 dimensionSolution 16

Correct option: (a)

A point is an exact location. A fine dot represents a point. So, a point has 0 dimensions.Question 17

Boundaries of solids are

(a) lines

(b) curves

(d) none of theseSolution 17

Correct option: (c)

Boundaries of solids are surfaces.Question 18

Boundaries of surfaces are

(a) lines

(b) curves

(d) none of theseSolution 18

Correct option: (b)

Boundaries of surfaces are curves.Question 19

The number of planes passing through three non-collinear points is

(a) 4

(b) 3

(d) 1Solution 19

Correct option: (d)

The number of planes passing through three non-collinear points is 1.Question 20

Axioms are assumed

(a) definitions

(b) theorems

(d) universal truths in all branches of mathematicsSolution 20

Correct option: (d)

Axioms are assumed as universal truths in all branches of mathematics because they are taken for granted, without proof.Question 21

Which of the following is a true statement?

(a) The floor and a wall of a room are parallel planes

(b) The ceiling and a wall of a room are parallel planes.

(d) Two adjacent walls of a room are the parallel planes.Solution 21

Correct option: (c)

Two lines are said to be parallel, if they have no point in common.

Options (a), (b) and (d) have a common point, hence they are not parallel.

In option (c), the floor and the ceiling of a room are parallel to each other is a true statement.Question 22

Which of the following is true statement?

(a) Only a unique line can be drawn to pass through a given point

(b) Infinitely many lines can be drawn to pass through two given points

(d)A line has a definite length.Solution 22

Correct option: (c)

In option (a), infinite number of line can be drawn to pass through a given point. So, it is not a true statement.

In option (b), only one line can be drawn to pass through two given points. So, it is not a true statement.

In option (c),

‘If two circles are equal, then their radii are equal’ is the true statement.

In option (d), A line has no end points. A line has an indefinite length. So, it is not a true statement.Question 23

Which of the following is a false statement?

(a) An infinite number of lines can be drawn to pass through a given point.

(b) A unique line can be drawn to pass through two given points.

(c)

(d)A ray has one end point.Solution 23

Correct option: (c)

Option (a) is true, since we can pass an infinite number of lines through a given point.

Option (b) is true, since a unique line can be drawn to pass through two given points.

Consider option (c).

As shown in the above diagram, a ray has only one end-point. So, option (d) is true.

Hence, the only false statement is option (c).Question 24

A point C is called the midpoint of a line segment , if

(a) C is an interior point of AB

(b) AC = CB

(d) AC + CB = ABSolution 24

Correct option: (c)

A point C is called the midpoint of a line segment , if C is an interior point of AB such that =.

Question 25

A point C is said to lie between the points A and B if

(a) AC = CB

(b) AC + CB = AB

(d) options (b) and (c)

* Options modifiedSolution 25

Correct option: (d)

Observe the above figure. Clearly, C lies between A and B if AC + CB = AB.

That means, points A, B, C are collinear.Question 26

Euclid’s which axiom illustrates the statement that when x + y = 15, then x + y + z = 15 + z?

(a) first

(b) second

(d) fourthSolution 26

Correct option: (b)

Euclid’s second axiom states that ‘If equals are added to equals, the wholes are equal’.

Hence, when x + y = 15, then x + y + z = 15 + z. Question 27

A is of the same age as B and C is of the same age as B. Euclid’s which axiom illustrates the relative ages of A and C?

(a) First axiom

(b) Second axiom

(d) Fourth axiomSolution 27

Correct option: (a)

Euclid’s first axiom states that ‘Things which are equal to the same thing are equal to one another’.

That is,

A’s age = B’s age and C’s age = B’ age

⇒ A’s age = C’s age

Exercise Ex. 6

Question 1

What is the difference between a theorem and an axiom?Solution 1

A theorem is a statement that requires a proof. Whereas, a basic fact which is taken for granted, without proof, is called an axiom.

Example of Theorem: Pythagoras Theorem

Example of axiom: A unique line can be drawn through any two points.
Question 2

Define the following terms:

(i) Line segment (ii) Ray (iii) Intersecting lines (iv) Parallel lines (v) Half-line (vi) Concurrent lines (vii) Collinear points (viii) PlaneSolution 2

(i) Line segment: The straight path between two points is called a line segment.

(ii) Ray: A line segment when extended indefinitely in one direction is called a ray.

(iii) Intersecting Lines: Two lines meeting at a common point are called intersecting lines, i.e., they have a common point.

(iv) Parallel Lines: Two lines in a plane are said to be parallel, if they have no common point, i.e., they do not meet at all.

(v) Half-line: A ray without its initial point is called a half-line.

(vi) Concurrent lines: Three or more lines are said to be concurrent, if they intersect at the same point.

(vii) Collinear points: Three or more than three points are said to be collinear, if they lie on the same line.

(viii) Plane: A plane is a surface such that every point of the line joining any two points on it, lies on it.Question 3

In the adjoining figure, name:

(i) Six points

(ii) Five line segments

(iii) Four rays

(iv) Four lines

(v) Four collinear points

Solution 3

(i) Six points: A,B,C,D,E,F

(ii) Five line segments:

(iii) Four rays:

(iv) Four lines:

(vi) Four collinear points: M,E,G,BQuestion 4

In the adjoining figure, name:

(i) Two pairs of intersecting lines and their corresponding points of intersection

(ii) Three concurrent lines and their points of intersection

(iii) Three rays

(iv) Two line segments

Solution 4

(i) and their corresponding point of intersection is R.

and their corresponding point of intersection is P.

(ii) and their point of intersection is R.

(iii) Three rays are:

(iv) Two line segments are:

.Question 5

From the given figure, name the following:

(i) Three lines

(ii) One rectilinear figure

(iii) Four concurrent pointsSolution 5

(i) Three lines: Line AB, Line PQ and Line RS

(ii) One rectilinear figure: EFGC

(iii) Four concurrent points: Points A, E, F and BQuestion 6

(i) How many lines can be drawn to pass through a given point?

(ii) How many lines can be drawn to pass through two given points?

(iii) In how many points can the two lines at the most intersect?

(iv) If A, B, C are three collinear points, name all the line segments determined by them.Solution 6

(i) An infinite number of lines can be drawn to pass through a given point.

(ii) One and only one line can pass through two given points.

(iii) Two given lines can at the most intersect at one and only one point.

(iv) Question 7

Which of the following statements are true?

(i) A line segments has no definite length.

(ii) A ray has no end point.

(iii) A line has a definite length.

(iv) A line is the same as line .

(v) A ray is the same as ray .

(vi) Two distinct points always determine a unique line.

(vii) Three lines are concurrent if they have a common point.

(viii) Two distinct lines cannot have more than one point in common.

(ix) Two intersecting lines cannot be both parallel to the same line.

(x) Open half-line OA is the same thing as ray

(xi) Two lines may intersect in two points.

(xii) Two lines l and m are parallel only when they have no point in common.Solution 7

(i) False

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

(vii) True

(viii) True

(ix) True

(x) True

(xi) False

(xii) TrueQuestion 8

In the given figure, L and M are mid-points of AB and BC respectively.

(i) If AB = BC, prove that AL = MC.

(ii) If BL = BM, prove that AB = BC.Solution 8

(ii) BL = BM

⇒ 2BL = 2BM

⇒ AB = BC

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RS Agarwal Solution | Class 9th | Chapter-7 | Lines and Angles | Edugrown

Exercise MCQ

Question 1

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is

(a) an isosceles triangle

(b) an obtuse triangle

(d) a right triangleSolution 1

Correct option: (d)

In a right triangle, one angle is 90° and the sum of acute angles of a right triangle is 90°.Question 2

An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Each of these equal angles is

(a) 70°

(b) 55°

(d) Solution 2

Correct option: (b)

Let each interior opposite angle be x.

Then, x + x = 110° (Exterior angle property of a triangle)

⇒ 2x = 110°

⇒ x = 55° Question 3

The angles of a triangle are in the ratio 3:5:7 The triangle is

Acute angled
Obtuse angled
Right angled
an isosceles triangle

Solution 3

Question 4

If one of the angles of triangle is 130° then the angle between the bisector of the other two angles can be

(a) 50°

(b) 65°

(d) 155Solution 4

Correct option: (d)

Let ∠A = 130°

In ΔABC, by angle sum property,

∠B + ∠C + ∠A = 180°

⇒ ∠B + ∠C + 130° = 180°

⇒ ∠B + ∠C = 50°

Question 5

In the given figure, AOB is a straight line. The value of x is

(a) 12

(b) 15

(d) 25Solution 5

Correct option: (b)

AOB is a straight line.

⇒ ∠AOB = 180°

⇒ 60° + 5x° + 3x° = 180°

⇒ 60° + 8x° = 180°

⇒ 8x° = 120°

⇒ x = 15° Question 6

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is

(a) 120°

(b) 100°

(d) 60° Solution 6

Correct option: (c)

By angle sum property,

2x + 3x + 4x = 180°

⇒ 9x = 180°

⇒ x = 20°

Hence, largest angle = 4x = 4(20°) = 80° Question 7

In the given figure, ∠OAB = 110° and ∠BCD = 130° then ∠ABC is equal to

(a) 40°

(b) 50°

(d) 70° Solution 7

Correct option: (c)

Through B draw YBZ ∥ OA ∥ CD.

Now, OA ∥ YB and AB is the transversal.

⇒ ∠OAB + ∠YBA = 180° (interior angles are supplementary)

⇒ 110° + ∠YBA = 180°

⇒ ∠YBA = 70°

Also, CD ∥ BZ and BC is the transversal.

⇒ ∠DCB + ∠CBZ = 180° (interior angles are supplementary)

⇒ 130° + ∠CBZ = 180°

⇒ ∠CBZ = 50°

Now, ∠YBZ = 180° (straight angle)

⇒ ∠YBA + ∠ABC + ∠CBZ = 180°

⇒ 70° + x + 50° = 180°

⇒ x = 60°

⇒ ∠ABC = 60° Question 8

If two angles are complements of each other, then each angle is

An acute angle
An obtuse angle
A right angle
A reflex angle

Solution 8

Correct option: (a)

Two angles are said to be complementary, if the sum of their measures is 90°.

Clearly, the measures of each of the angles have to be less than 90°.

Hence, each angle is an acute angle.Question 9

An angle which measures more than 180° but less than 360°, is called

An acute angle
An obtuse angle
A straight angle
A reflex angle

Solution 9

Correct option: (d)

An angle which measures more than 180^o but less than 360^ois called a reflex angle.Question 10

The measure of an angle is five times its complement. The angle measures

25°
35°
65°
75°

Solution 10

Question 11

Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The larger of the two measures

72°^o
54°
63°
36°

Solution 11

Question 12

In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC =?

Solution 12

Question 13

In the given figure, AOB is a straight line. If ∠AOC = (3x + 10) ° and ∠BOC = (4x – 26) °, then ∠BOC =?

96°
86°
76°
106°

Solution 13

Question 14

In the given figure, AOB is a straight line. If ∠AOC = (3x – 10) °, ∠COD = 50° and ∠BOD = (x +20) °, then ∠AOC =?

40°
60°
80°
50°

Solution 14

Question 15

Which of the following statements is false?

Through a given point, only one straight line can be drawn
Through two given points, it is possible to draw one and only one straight line.
Two straight lines can intersect only at one point
A line segment can be produced to any desired length.

Solution 15

Correct option: (a)

Option (a) is false, since through a given point we can draw an infinite number of straight lines.Question 16

An angle is one-fifth of its supplement. The measure of the angle is

15°
30°
75°
150°

Solution 16

Question 17

In the adjoining figure, AOB is straight line. If x:y:z = 4:5:6, then y = ?

60°
80°
48°
72°

Solution 17

Question 18

In the given figure, straight lines AB and CD intersect at O. If ∠AOC =ϕ, ∠BOC = θ and θ = 3 ϕ, then ϕ =?

30°
40°
45°
60°

Solution 18

Question 19

In the given figure, straight lines AB and CD intersect at O. If ∠AOC + ∠BOD = 130°, then ∠AOD =?

65°
115°
110°
125°

Solution 19

Question 20

In the given figure AB is a mirror, PQ is the incident ray and and QR is the reflected ray. If ∠PQR = 108°, then ∠ AQP =?

72°
18°
36°
54°

Solution 20

Question 21

In the given figure, AB ∥ CD, If ∠BAO = 60° and ∠OCD = 110° then ∠AOC = ?

(a) 70°

(b) 60°

(d) 40° Solution 21

Correct option: (c)

Let ∠AOC = x°

Draw YOZ ∥ CD ∥ AB.

Now, YO ∥ AB and OA is the transversal.

⇒ ∠YOA = ∠OAB = 60° (alternate angles)

Again, OZ ∥ CD and OC is the transversal.

⇒ ∠COZ + ∠OCD = 180° (interior angles)

⇒ ∠COZ + 110° = 180°

⇒ ∠COZ = 70°

Now, ∠YOZ = 180° (straight angle)

⇒ ∠YOA + ∠AOC + ∠COZ = 180°

⇒ 60° + x + 70° = 180°

⇒ x = 50°

⇒ ∠AOC = 50° Question 22

In the given figure, AB ‖ CD. If ∠AOC = 30° and ∠OAB = 100°, then ∠OCD =?

130°
150°
80°
100°

Solution 22

Question 23

In the given figure, AB ‖ CD. If ‖CAB = 80^o and ∠EFC= 25°, then ∠CEF =?

65°
55°
45°
75°

Solution 23

Question 24

In the given figure, AB ‖ CD, CD ‖ EF and y:z = 3:7, then x = ?

108°
126°
162°
63°

Solution 24

Question 25

In the given figure, AB ‖ CD. If ∠APQ = 70° and ∠RPD = 120°, then ∠QPR =?

50°
60°
40°
35°

Solution 25

Question 26

In the given figure AB ‖ CD. If ∠EAB = 50° and ∠ECD=60°, then ∠AEB =?

50°
60°
70°
50°

Solution 26

Question 27

In the given figure, ∠OAB = 75°, ∠OBA=55° and ∠OCD = 100°. Then ∠ODC=?

20°
25°
30°
35°

Solution 27

Question 28

In the adjoining figure y =?

36°
54°
63°
72°

Solution 28

Exercise Ex. 7A

Question 1

Define the following terms:

(i) Angle (ii) Interior of an angle (iii) Obtuse angle (iv) Reflex angle (v) Complementary angles (vi) Supplementary anglesSolution 1

(i) Angle: Two rays having a common end point form an angle.

(ii) Interior of an angle: The interior of AOB is the set of all points in its plane, which lie on the same side of OA as B and also on same side of OB as A.

(iii) Obtuse angle: An angle whose measure is more than 90^o but less than 180^o, is called an obtuse angle.

(iv) Reflex angle: An angle whose measure is more than 180^o but less than 360^o is called a reflex angle.

(v) Complementary angles: Two angles are said to be complementary, if the sum of their measures is 90^o.

(vi) Supplementary angles: Two angles are said to be supplementary, if the sum of their measures is 180^o.
Question 2(ii)

Find the complement of each of the following angles.

16^oSolution 2(ii)

Complement of 16^o = 90 – 16^o = 74^oQuestion 2(iv)

Find the complement of each of the following angles.

46^o 30Solution 2(iv)

Complement of 46^o 30′ = 90^o – 46^o 30′ = 43^o 30’Question 2(i)

Find the complement of each of the following angle:

55° Solution 2(i)

Complement of 55° = 90° – 55° = 35° Question 2(iii)

Find the complement of each of the following angle:

90° Solution 2(iii)

Complement of 90° = 90° – 90° = 0° Question 3(iv)

Find the supplement of each of the following angles.

75^o 36’Solution 3(iv)

Supplement of 75^o 36′ = 180^o – 75^o 36′ = 104^o 24’Question 3(i)

Find the supplement of each of the following angle:

42° Solution 3(i)

Supplement of 42° = 180° – 42° = 138° Question 3(ii)

Find the supplement of each of the following angle:

90° Solution 3(ii)

Supplement of 90° = 180° – 90° = 90° Question 3(iii)

Find the supplement of each of the following angle:

124° Solution 3(iii)

Supplement of 124° = 180° – 124° = 56° Question 4

Find the measure of an angle which is

(i) equal to its complement, (ii) equal to its supplement.Solution 4

(i) Let the required angle be x^o

Then, its complement = 90^o – x^o

The measure of an angle which is equal to its complement is 45^o.

(ii) Let the required angle be x^o

Then, its supplement = 180^o – x^o

The measure of an angle which is equal to its supplement is 90^o.Question 5

Find the measure of an angle which is 36^o more than its complement.Solution 5

Let the required angle be x^o

Then its complement is 90^o – x^o

The measure of an angle which is 36^o more than its complement is 63^o.Question 6

Find the measure of an angle which is 30° less than its supplement.Solution 6

Let the measure of the required angle = x°

Then, measure of its supplement = (180 – x)°

It is given that

x° = (180 – x)° – 30°

⇒ x° = 180° – x° – 30°

⇒ 2x° = 150°

⇒ x° = 75°

Hence, the measure of the required angle is 75°. Question 7

Find the angle which is four times its complement.Solution 7

Let the required angle be x^o

Then, its complement = 90^o – x^o

The required angle is 72^o.Question 8

Find the angle which is five times its supplement.Solution 8

Let the required angle be x^o

Then, its supplement is 180^o – x^o

The required angle is 150^o.Question 9

Find the angle whose supplement is four times its complement.Solution 9

Let the required angle be x^o

Then, its complement is 90^o – x^o and its supplement is 180^o – x^o_{That is we have,}

The required angle is 60^o.
Question 10

Find the angle whose complement is one-third of its supplement.Solution 10

Let the required angle be x^o

Then, its complement is 90^o – x^o and its supplement is 180^o – x^o

The required angle is 45^o.Question 11

Two complementary angles are in the ratio 4: 5. Find the angles.Solution 11

Let the two required angles be x^o and 90^o – x^o.

Then

5x = 4(90 – x)

5x = 360 – 4x

5x + 4x = 360

9x = 360

Thus, the required angles are 40^o and 90^o – x^o = 90^o – 40^o = 50^o.
Question 12

Find the value of x for which the angles (2x – 5)° and (x – 10)° are the complementary angles.Solution 12

(2x – 5)° and (x – 10)° are complementary angles.

∴ (2x – 5)° + (x – 10)° = 90°

⇒ 2x – 5° + x – 10° = 90°

⇒ 3x – 15° = 90°

⇒ 3x = 105°

⇒ x = 35°

Exercise Ex. 7B

Question 1

In the given figure, AOB is a straight line. Find the value of x.

Solution 1

Since BOC and COA form a linear pair of angles, we have

BOC + COA = 180^o

x^o+ 62^o = 180^o

x = 180 – 62

x = 118^oQuestion 2

In the adjoining figure, AOB is a straight line. Find the value of x. Hence, find ∠AOC and ∠BOD.

Solution 2

∠AOB is a straight angle.

⇒ ∠AOB = 180°

⇒ ∠AOC + ∠COD + ∠BOD = 180°

⇒ (3x – 7)° + 55° + (x + 20)° = 180°

⇒ 4x + 68° = 180°

⇒ 4x = 112°

⇒ x = 28°

Thus, ∠AOC = (3x – 7)° = 3(28°) – 7° = 84° – 7° = 77°

And, ∠BOD = (x + 20)° = 28° + 20° = 48° Question 3

In the given figure, AOB is a straight line. Find the value of x. Hence, find AOC, COD and BOD.

Solution 3

Since BOD and DOA from a linear pair of angles.

BOD + DOA = 180^o

BOD + DOC + COA = 180^o

x^o + (2x – 19)^o + (3x + 7)^o = 180^o

6x – 12 = 180

6x = 180 + 12 = 192

x = 32

AOC = (3x + 7)^o = (3 32 + 7)^o = 103^o

COD = (2x – 19)^o = (2 32 – 19)^o = 45^o

and BOD = x^o = 32^o
Question 4

In the given figure, x: y: z = 5: 4: 6. If XOY is a straight line, find the values of x, y and z.

Solution 4

x: y: z = 5: 4: 6

The sum of their ratios = 5 + 4 + 6 = 15

But x + y + z = 180^o

[Since, XOY is a straight line]

So, if the total sum of the measures is 15, then the measure of x is 5.

If the sum of angles is 180^o, then, measure of

And, if the total sum of the measures is 15, then the measure of y is 4.

If the sum of the angles is 180^o, then, measure of

And z = 180^o – x – y

= 180^o – 60^o – 48^o

= 180^o – 108^o = 72^o

x = 60, y = 48 and z = 72.
Question 5

In the given figure, what value of x will make AOB, a straight line?

Solution 5

AOB will be a straight line, if two adjacent angles form a linear pair.

BOC + AOC = 180^o

(4x – 36)^o + (3x + 20)^o = 180^o

4x – 36 + 3x + 20 = 180

7x – 16 = 180^o

7x = 180 + 16 = 196

The value of x = 28.Question 6

Two lines AB and CD intersect at O. If AOC = 50^o, find AOD, BOD and BOC.

Solution 6

Since AOC and AOD form a linear pair.

AOC + AOD = 180^o

50^o + AOD = 180^o

AOD = 180^o – 50^o = 130^o

AOD and BOC are vertically opposite angles.

AOD = BOC

BOC = 130^o

BOD and AOC are vertically opposite angles.

BOD = AOC

BOD = 50^oQuestion 7

In the given figure, three coplanar lines AB,CD and EF intersect at a point O, forming angles as shown. Find the values of x,y,z and t.

Solution 7

Since COE and DOF are vertically opposite angles, we have,

COE = DOF

z = 50^o

Also BOD and COA are vertically opposite angles.

So, BOD = COA

t = 90^o

As COA and AOD form a linear pair,

COA + AOD = 180^o

COA + AOF + FOD = 180^o [t = 90^o]

t + x + 50^o = 180^o

90^o + x^o + 50^o = 180^o

x + 140 = 180

x = 180 – 140 = 40

Since EOB and AOF are vertically opposite angles

So, EOB = AOF

y = x = 40

Thus, x = 40 = y = 40, z = 50 and t = 90Question 8

In the given figure, three coplanar lines AB,CD and EF intersect at a point O. Find the value of x. Hence, find AOD, COE and AOE.

Solution 8

Since COE and EOD form a linear pair of angles.

COE + EOD = 180^o

COE + EOA + AOD = 180^o

5x + EOA + 2x = 180

5x + BOF + 2x = 180

[EOA and BOF are vertically opposite angles so, EOA = BOF]

5x + 3x + 2x = 180

10x = 180

x = 18

Now AOD = 2x^o = 2 18^o = 36^o

COE = 5x^o = 5 18^o = 90^o

and, EOA = BOF = 3x^o = 3 18^o = 54^oQuestion 9

Two adjacent angles on a straight line are in the ratio 5 : 4. Find the measure of each one of these angles.Solution 9

Let the two adjacent angles be 5x and 4x.

Now, since these angles form a linear pair.

So, 5x + 4x = 180^o

9x = 180^o

The required angles are 5x = 5x = 5 20^o = 100^o

and 4x = 4 20^o = 80^oQuestion 10

If two straight lines intersect each other in such a way that one of the angles formed measures 90^o, show that each of the remaining angles measures 90^o.Solution 10

Let two straight lines AB and CD intersect at O and let AOC = 90^o.

Now, AOC = BOD [Vertically opposite angles]

BOD = 90^o

Also, as AOC and AOD form a linear pair.

90^o + AOD = 180^o

AOD = 180^o – 90^o = 90^o

Since, BOC = AOD [Verticallty opposite angles]

BOC = 90^o

Thus, each of the remaining angles is 90^o.Question 11

Two lines AB and CD intersect at a point O such that BOC +AOD = 280^o, as shown in the figure. Find all the four angles.

Solution 11

Since, AOD and BOC are vertically opposite angles.

AOD = BOC

Now, AOD + BOC = 280^o [Given]

AOD + AOD = 280^o

2AOD = 280^o

AOD =

BOC = AOD = 140^o

As, AOC and AOD form a linear pair.

So, AOC + AOD = 180^o

AOC + 140^o = 180^o

AOC = 180^o – 140^o = 40^o

Since, AOC and BOD are vertically opposite angles.

AOC = BOD

BOD = 40^o

BOC = 140^o, AOC = 40^o , AOD = 140^o and BOD = 40^o.Question 12

Two lines AB and CD intersect each other at a point O such that ∠AOC : ∠AOD = 5 : 7. Find all the angles.

Solution 12

Let ∠AOC = 5x and ∠AOD = 7x

Now, ∠AOC + ∠AOD = 180° (linear pair of angles)

⇒ 5x + 7x = 180°

⇒ 12x = 180°

⇒ x = 15°

⇒ ∠AOC = 5x = 5(15°) = 75° and ∠AOD = 7x = 7(15°) = 105°

Now, ∠AOC = ∠BOD (vertically opposite angles)

⇒ ∠BOD = 75°

Also, ∠AOD = ∠BOC (vertically opposite angles)

⇒ ∠BOC = 105° Question 13

In the given figure, three lines AB, CD and EF intersect at a point O such that ∠AOE = 35° and ∠BOD = 40°. Find the measure of ∠AOC, ∠BOF, ∠COF and ∠DOE.

Solution 13

∠BOD = 40°

⇒ AOC = ∠BOD = 40° (vertically opposite angles)

∠AOE = 35°

⇒ ∠BOF = ∠AOE = 35° (vertically opposite angles)

∠AOB is a straight angle.

⇒ ∠AOB = 180°

⇒ ∠AOE + ∠EOD + ∠BOD = 180°

⇒ 35° + ∠EOD + 40° = 180°

⇒ ∠EOD + 75° = 180°

⇒ ∠EOD = 105°

Now, ∠COF = ∠EOD = 105° (vertically opposite angles)Question 14

In the given figure, the two lines AB and CD intersect at a point O such that ∠BOC = 125°. Find the values of x, y and z.

Solution 14

∠AOC + ∠BOC = 180° (linear pair of angles)

⇒ x + 125 = 180°

⇒ x = 55°

Now, ∠AOD = ∠BOC (vertically opposite angles)

⇒ y = 125°

Also, ∠BOD = ∠AOC (vertically opposite angles)

⇒ z = 55° Question 15

If two straight lines intersect each other then prove that the ray opposite to the bisector of one of the angles so formed bisects the vertically opposite angle.Solution 15

Given : AB and CD are two lines which are intersecting at O. OE is a ray bisecting the BOD. OF is a ray opposite to ray OE.

To Prove: AOF = COF

Proof : Since are two opposite rays, is a straight line passing through O.

AOF = BOE

and COF = DOE

[Vertically opposite angles]

But BOE = DOE (Given)

AOF = COF

Hence, proved.Question 16

Prove that the bisectors of two adjacent supplementary angles include a right angle.Solution 16

Given: is the bisector of BCD and is the bisector of ACD.

To Prove: ECF = 90^o

Proof: Since ACD and BCD forms a linear pair.

ACD + BCD = 180^o

ACE + ECD + DCF + FCB = 180^o

ECD + ECD + DCF + DCF = 180^o

because ACE = ECD

and DCF = FCB

2(ECD) + 2 (CDF) = 180^o

2(ECD + DCF) = 180^o

ECD + DCF =

ECF = 90^o (Proved)

Exercise Ex. 7C

Question 1

In the given figure, l ∥ m and a transversal t cuts them. If ∠1 = 120°, find the measure of each of the remaining marked angles.

Solution 1

Given, ∠1 = 120°

Now, ∠1 + ∠2 = 180° (linear pair)

⇒ 120° + ∠2 = 180°

⇒ ∠2 = 60°

∠1 = ∠3 (vertically opposite angles)

⇒ ∠3 = 120°

Also, ∠2 = ∠4 (vertically opposite angles)

⇒ ∠4 = 60°

Line l ∥ line m and line t is a transversal.

⇒ ∠5 = ∠1 = 120° (corresponding angles)

∠6 = ∠2 = 60° (corresponding angles)

∠7 = ∠3 = 120° (corresponding angles)

∠8 = ∠4 = 60° (corresponding angles)Question 2

In the given figure, l ∥ m and a transversal t cuts them. If ∠7 = 80°, find the measure of each of the remaining marked angles.

Solution 2

Given, ∠7 = 80°

Now, ∠7 + ∠8 = 180° (linear pair)

⇒ 80° + ∠8 = 180°

⇒ ∠8 = 100°

∠7 = ∠5 (vertically opposite angles)

⇒ ∠5 = 80°

Also, ∠6 = ∠8 (vertically opposite angles)

⇒ ∠6 = 100°

Line l ∥ line m and line t is a transversal.

⇒ ∠1 = ∠5 = 80° (corresponding angles)

∠2 = ∠6 = 100° (corresponding angles)

∠3 = ∠7 = 80° (corresponding angles)

∠4 = ∠8 = 100° (corresponding angles) Question 3

In the given figure, l ∥ m and a transversal t cuts them. If ∠1 : ∠2 = 2 : 3, find the measure of each of the marked angles.

Solution 3

Given, ∠1 : ∠2 = 2 : 3

Now, ∠1 + ∠2 = 180° (linear pair)

⇒ 2x + 3x = 180°

⇒ 5x = 180°

⇒ x = 36°

⇒ ∠1 = 2x = 72° and ∠2 = 3x = 108°

∠1 = ∠3 (vertically opposite angles)

⇒ ∠3 = 72°

Also, ∠2 = ∠4 (vertically opposite angles)

⇒ ∠4 = 108°

Line l ∥ line m and line t is a transversal.

⇒ ∠5 = ∠1 = 72° (corresponding angles)

∠6 = ∠2 = 108° (corresponding angles)

∠7 = ∠3 = 72° (corresponding angles)

∠8 = ∠4 = 108° (corresponding angles) Question 4

For what value of x will the lines l and m be parallel to each other?

Solution 4

Lines l and m will be parallel if 3x – 20 = 2x + 10

[Since, if corresponding angles are equal, lines are parallel]

3x – 2x = 10 + 20

x = 30Question 5

For what value of x will the lines l and m be parallel to each other?

*Question modified, back answer incorrect.Solution 5

For lines l and m to be parallel to each other, the corresponding angles (3x + 5)° and (4x)° should be equal.

⇒ (3x + 5)° = 4x°

⇒ x = 5° Question 6

In the given figure, AB || CD and BC || ED. Find the value of x.

Solution 6

Since AB || CD and BC is a transversal.

So, BCD = ABC = x^o [Alternate angles]

As BC || ED and CD is a transversal.

BCD + EDC = 180^o

BCD + 75^o =180^o

BCD = 180^o – 75^o = 105^o

ABC = 105^o [since BCD = ABC]

x^o = ABC = 105^o

Hence, x = 105.
Question 7

In the given figure, AB || CD || EF. Find the value of x.

Solution 7

Since AB || CD and BC is a transversal.

So, ABC = BCD [atternate interior angles]

70^o = x^o + ECD(i)

Now, CD || EF and CE is transversal.

So,ECD + CEF = 180^o [sum of consecutive interior angles is 180^o]

ECD + 130^o = 180^o

ECD = 180^o – 130^o = 50^o

Putting ECD = 50^o in (i) we get,

70^o = x^o + 50^o

x = 70 – 50 = 20Question 8

In the give figure, AB ∥ CD. Find the values of x, y and z.

Solution 8

AB ∥ CD and EF is transversal.

⇒ ∠AEF = ∠EFG (alternate angles)

Given, ∠AEF = 75°

⇒ ∠EFG = y = 75°

Now, ∠EFC + ∠EFG = 180° (linear pair)

⇒ x + y = 180°

⇒ x + 75° = 180°

⇒ x = 105°

∠EGD = ∠EFG + ∠FEG (Exterior angle property)

⇒ 125° = y + z

⇒ 125° = 75° + z

⇒ z = 50°

Thus, x = 105°, y = 75° and z = 50° Question 9(i)

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 9(i)

Through E draw EG || CD. Now since EG||CD and ED is a transversal.

So,GED = EDC = 65^o[Alternate interior angles]

Since EG || CD and AB || CD,

EG||AB and EB is transversal.

So,BEG = ABE = 35^o[Alternate interior angles]

So,DEB = x^o

BEG + GED = 35^o + 65^o = 100^o.

Hence, x = 100.Question 9(ii)

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 9(ii)

Through O draw OF||CD.

Now since OF || CD and OD is transversal.

CDO + FOD = 180^o

[sum of consecutive interior angles is 180^o]

25^o + FOD = 180^o

FOD = 180^o – 25^o = 155^o

As OF || CD and AB || CD [Given]

Thus, OF || AB and OB is a transversal.

So,ABO + FOB = 180^o[sum of consecutive interior angles is 180^o]

55^o + FOB = 180^o

FOB = 180^o – 55^o = 125^o

Now, x^o = FOB + FOD = 125^o + 155^o = 280^o.

Hence, x = 280.Question 9(iii)

Ineach of the figures given below, AB || CD. Find the value of x in each case.

Solution 9(iii)

Through E, draw EF || CD.

Now since EF || CD and EC is transversal.

FEC + ECD = 180^o

[sum of consecutive interior angles is 180^o]

FEC + 124^o = 180^o

FEC = 180^o – 124^o = 56^o

Since EF || CD and AB ||CD

So, EF || AB and AE is a trasveral.

So,BAE + FEA = 180^o

[sum of consecutive interior angles is 180^o]

116^o + FEA = 180^o

FEA = 180^o – 116^o = 64^o

Thus,x^o = FEA + FEC

= 64^o + 56^o = 120^o.

Hence, x = 120.Question 10

In the given figure, AB || CD. Find the value of x.

Solution 10

Through C draw FG || AE

Now, since CG || BE and CE is a transversal.

So, GCE = CEA = 20^o [Alternate angles]

DCG = 130^o – GCE

= 130^o – 20^o = 110^o

Also, we have AB || CD and FG is a transversal.

So, BFC = DCG = 110^o [Corresponding angles]

As, FG || AE, AF is a transversal.

BFG = FAE [Corresponding angles]

x^o = FAE = 110^o.

Hence, x = 110Question 11

In the given figure, AB || PQ. Find the values of x and y.

Solution 11

Since AB || PQ and EF is a transversal.

So, CEB = EFQ [Corresponding angles]

EFQ = 75^o

EFG + GFQ = 75^o

25^o + y^o = 75^o

y = 75– 25 = 50

Also, BEF + EFQ = 180^o [sum of consecutive interior angles is 180^o] BEF = 180^o – EFQ

= 180^o – 75^o

BEF = 105^o

FEG + GEB = BEF = 105^o

FEG = 105^o – GEB = 105^o – 20^o = 85^o

In EFG we have,

x^o + 25^o + FEG = 180^o

Hence, x = 70.
Question 12

In the given figure, AB || CD. Find the value of x.

Solution 12

Since AB || CD and AC is a transversal.

So, BAC + ACD = 180^o [sum of consecutive interior angles is 180^o]

ACD = 180^o – BAC

= 180^o – 75^o = 105^o

ECF = ACD [Vertically opposite angles]

ECF = 105^o

Now in CEF,

ECF + CEF + EFC =180^o 105^o + x^o + 30^o = 180^o

x = 180 – 30 – 105 = 45

_{Hence, x = 45.}
Question 13

In the given figure, AB || CD. Find the value of x.

Solution 13

Since AB || CD and PQ a transversal.

So, PEF = EGH [Corresponding angles]

EGH = 85^o

EGH and QGH form a linear pair.

So, EGH + QGH = 180^o

QGH = 180^o – 85^o = 95^o

Similarly, GHQ + 115^o = 180^o

GHQ = 180^o – 115^o = 65^o

In GHQ, we have,

x^o + 65^o + 95^o = 180^o

x = 180 – 65 – 95 = 180 – 160

x = 20
Question 14

In the given figure, AB || CD. Find the values of x, y and z.

Solution 14

Since AB || CD and BC is a transversal.

So, ABC = BCD

x = 35

Also, AB || CD and AD is a transversal.

So, BAD = ADC

z = 75

In ABO, we have,

x^o + 75^o + y^o = 180^o

35 + 75 + y = 180

y = 180 – 110 = 70

x = 35, y = 70 and z = 75.
Question 16

In the given figure, AB || CD. Prove that p + q – r = 180.

Solution 16

Through F, draw KH || AB || CD

Now, KF || CD and FG is a transversal.

KFG = FGD = r^o (i)

[alternate angles]

Again AE || KF, and EF is a transversal.

So,AEF + KFE = 180^o

KFE = 180^o – p^o (ii)

Adding (i) and (ii) we get,

KFG + KFE = 180 – p + r

EFG = 180 – p + r

q = 180 – p + r

i.e.,p + q – r = 180Question 17

In the given figure, AB || CD and EF || GH. Find the values of x, y, z and t.

Solution 17

PRQ = x^o = 60^o [vertically opposite angles]

Since EF || GH, and RQ is a transversal.

So, x = y [Alternate angles]

y = 60

AB || CD and PR is a transversal.

So, [Alternate angles]

[since ]

x + QRD = 110^o

QRD = 110^o – 60^o = 50^o

In QRS, we have,

QRD + t^o + y^o = 180^o

50 + t + 60 = 180

t = 180 – 110 = 70

Since, AB || CD and GH is a transversal

So, z^o = t^o = 70^o [Alternate angles]

x = 60 , y = 60, z = 70 and t = 70
Question 18

In the given figure, AB ∥ CD and a transversal t cuts them at E and F respectively. If EG and FG are the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF = 90°.

Solution 18

AB ∥ CD and a transversal t cuts them at E and F respectively.

⇒ ∠BEF + ∠DFE = 180° (interior angles)

⇒ ∠GEF + ∠GFE = 90° ….(i)

Now, in ΔGEF, by angle sum property

∠GEF + ∠GFE + ∠EGF = 180°

⇒ 90° + ∠EGF = 180° ….[From (i)]

⇒ ∠EGF = 90° Question 19

In the given figure, AB ∥ CD and a transversal t cuts them at E and F respectively. If EP and FQ are the bisectors of ∠AEF and ∠EFD respectively, prove that EP ∥ FQ.

Solution 19

Since AB ∥ CD and t is a transversal, we have

∠AEF = ∠EFD (alternate angles)

⇒ ∠PEF = ∠EFQ

But, these are alternate interior angles formed when the transversal EF cuts EP and FQ.

∴ EP ∥ FQQuestion 20

In the given figure, BA ∥ ED and BC ∥ EF. Show that ∠ABC = ∠DEF.

Solution 20

Construction: Produce DE to meet BC at Z.

Now, AB ∥ DZ and BC is the transversal.

⇒ ∠ABC = ∠DZC (corresponding angles) ….(i)

Also, EF ∥ BC and DZ is the transversal.

⇒ ∠DZC = ∠DEF (corresponding angles) ….(ii)

From (i) and (ii), we have

∠ABC = ∠DEF Question 21

In the given figure, BA ∥ ED and BC ∥ EF. Show that ∠ABC + ∠DEF = 180°.

Solution 21

Construction: Produce ED to meet BC at Z.

Now, AB ∥ EZ and BC is the transversal.

⇒ ∠ABZ + ∠EZB = 180° (interior angles)

⇒ ∠ABC + ∠EZB = 180° ….(i)

Also, EF ∥ BC and EZ is the transversal.

⇒ ∠BZE = ∠ZEF (alternate angles)

⇒ ∠BZE = ∠DEF ….(ii)

From (i) and (ii), we have

∠ABC + ∠DEF = 180° Question 22

In the given figure, m and n are two plane mirrors perpendicular to each other. Show that the incident ray CA is parallel to the reflected ray BD.

Solution 22

Let the normal to mirrors m and n intersect at P.

Now, OB ⊥ m, OC ⊥ n and m ⊥ n.

⇒ OB ⊥ OC

⇒ ∠APB = 90°

⇒ ∠2 + ∠3 = 90° (sum of acute angles of a right triangle is 90°)

By the laws of reflection, we have

∠1 = ∠2 and ∠4 = ∠3 (angle of incidence = angle of reflection)

⇒ ∠1 + ∠4 = ∠2 + ∠3 = 90°

⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°

⇒ ∠CAB + ∠ABD = 180°

But, ∠CAB and ∠ABD are consecutive interior angles formed, when the transversal AB cuts CA and BD.

∴ CA ∥ BD Question 23

In the figure given below, state which lines are parallel and why?

Solution 23

In the given figure,

∠BAC = ∠ACD = 110°

But, these are alternate angles when transversal AC cuts AB and CD.

Hence, AB ∥ CD. Question 24

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.Solution 24

Let the two parallel lines be m and n.

Let p ⊥ m.

⇒ ∠1 = 90°

Let q ⊥ n.

⇒ ∠2 = 90°

Now, m ∥ n and p is a transversal.

⇒ ∠1 = ∠3 (corresponding angles)

⇒ ∠3 = 90°

⇒ ∠3 = ∠2 (each 90°)

But, these are corresponding angles, when transversal n cuts lines p and q.

∴ p ∥ q.

Hence, two lines which are perpendicular to two parallel lines, are parallel to each other.

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Chapter – 3 Fibre to Fabric | Class 7th | NCERT Science Solutions | Edugrown

NCERT solutions for Class 7 Science have been provided below to aid the students with answering the questions correctly, using a logical approach and methodology. CBSE Class 7th Science solutions provide ample material to enable students to form a good base with the fundamentals of the NCERT Class 7 Science textbook.

Chapter 3 Fibre to Fabric

Q.1. You must be familiar with the following nursery rhymes:
(i) ‘Baa baa black sheep, have you any Wool.’
(ii) ‘Mary had a little lamb, whose fleece was white as snow.’
Answer the following:
(a) Which parts of the black sheep have wool?
(b) What is meant by the white fleece of the lamb?
Ans. (a) The hairy skin called fleece has wool in black sheep.
(b) White fleece of the lamb means the white coloured hairy skin.

Q.2. The silkworm is (a) a caterpillar (b) a larva. Choose the correct option.
(i) (a) (ii) (b) (iii) both (a) and (b) (iv) neither (a) nor (b)
Ans. (iii) both (a) and (b).

Q.3. Which of the following does not yield wool?
(i) Yak (ii) Camel (iii) Goat (iv) Woolly dog
Ans. (iv) Woolly dog

Q.4. What is meant by the following terms?
(i) Rearing (ii) Shearing (iii) Sericulture
Ans. (i) Rearing: The process of keeping, feeding, breeding and medical care of useful animals is called rearing of animals. These animals produce one or more useful products for htiman beings.
(ii) Shearing: The process of removing the fleece of the sheep alongwith thin layer of skin is called shearing.
(iii) Sericulture: The rearing of silkworms for obtaining silk is called sericulture.

Q.5. Given below is a sequence of steps in the processing of wool. Which are the missing steps? Add them.
Shearing, ________ , sorting,________, __________
Ans. Shearing, scouring, sorting, picking out of burrs, colouring, rolling.
Q.6 Make sketches of the two stages in the life history of the silk moth which are directly related to the production of silk
Ans.

Q.7. Out of the following, which are the two terms related to silk production? Sericulture, floriculture, moriculture, apiculture and silviculture.
Hints: (i) Silk production involves cultivation of mulberry leaves and rearing silkworms. (ii) Scientific name of mulberry is Morus alba.
Ans. (i) Sericulture (ii) Moriculture

Q.8. Match the words of Column I with those given in Column II:

Ans

Q.9. Given below is a crossword puzzle based on this lesson. Use hints to fill in the blank spaces with letters that complete the words.

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Chapter 2 Nutrition in Animals | Class 7th | NCERT Science Solutions | Edugrown

Chapter 2 Nutrition in Animals

Q.1.Fill in the blanks:
(a) The main steps of nutrition in humans are ___________, ___________, ___________ , ___________ and ___________
(b) The largest gland in the human body is ___________.

(c) The stomach releases hydrochloric acid and ___________ juices which act on food.

(d) The inner wall of the small intestine has many finger-like outgrowths called ___________
(e) Amoeba digests its food in the ___________ .
Ans.(a) ingestion, digestion, absorption, assimilation, egestion
(b) liver (c) digestive (d) villi (e) food vacuole.

Q.2. Mark ‘T’ if the statement is true and ‘F’ if it is false:
(a) Digestion of starch starts in the stomach. (T/F)
(b) The tongue helps in mixing food with saliva. (T/F)
(c) The gall bladder temporarily stores bile. (T/F)
(d) The ruminants bring back swallowed grass into their mouth and chew it for sometime. (T/F)
Ans. a) F (b) T (c) T (d) T

Q.3.Tick (S) mark the correct answer in each of the following:
(a) Fat is completely digested in the
(i) stomach (ii) mouth (in) small intestine (iv) large intestine

(b) Water from the undigested food is absorbed mainly in the:
(i) Stomach (ii) Food pipe (iii) Small intestine (iv) Large intestine

Ans.(a) (iii) Small intestine (b) (iv) Large intestine

Q.4.Match the items of column I with those given in column II:

Q.5. What are villi? What is their location and function?
Ans. The finger like projections in the inner walls of the small intestine is called villi. These are found in small intestine.
Function: The villi increase the surface area for absorption of the digested food

Q.6. Where is the bile produced? Which component of the food does it help to digest?
Ans. Bile is produced in liver. The bile juice stored in sac called the gall bladder. It helps in the digestion of fats.

Q.7. Name the type of carbohydrate that can be digested by ruminants but not by humans. Give the reason also.
Ans. Cellulose is the carbohydrate that can be digested by ruminants. Ruminants have large sac like structure between the small intestine and large intestine. The cellulose of the food is digested by the action of certain bacteria which are not present in humans.

Q.8. Why do we get instant energy from glucose?
Ans. Because it easily breaks down in the cell with the help of oxygen which provides instant energy to the organism. Glucose does not need digestion, it is directly absorbed into the blood.

Q.9. Which part of the digestive canal is involved in:
(i) Absorption of food ________ .
(ii) Chewing of food ________ .
(iii) Killing of bacteria ________ .
(iv) Complete digestion of food ________ .
(v) Formation of faeces ________ .
Ans. (i) Small intestine (ii) Mouth (iii) Stomach (iv) Small intestine (v) Large intestine

Q. 10. Write one similarity and one difference between the nutrition in amoeba and human beings.
Ans. Similarity: The digestive juices in amoeba are secreted into food vacuole and in human beings the digestive juices are secreted in stomach and small intestine. Then the juices convert complex food into simpler soluble and absorbable substances. ‘
Difference: Amoeba captures the food with help of pseudopodia and engulf it. In human beings food is taken by the mouth.

Q.11. Match the items of Column I with suitable items in Column II.

Ans.

Q.12.Label Fig. 2.11 of the digestive system (as given in the NCERT Textbook Exercise)

Ans.

Q.13. Can we survive only on raw, leafy vegetables/grass? Discuss.
Ans. We know that the animals, fungi, bacteria, non-green plants and human being do not have the ability to make their own food. They depend upon autotrophs for their food directly or indirectly. The green plant (leafy vegetables/grass) trap solar energy and make their own food in the form of glucose. So, we can say that leafy vegetables and grass can provide sufficient energy to help us survive.

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Chapter -1 Nutrition in Plants | Class 7th | NCERT Science Solutions | Edugrown

Chapter 1 Nutrition in Plants | Science

Q.1. Why do organisms need to take food?
Ans. Food is needed by all organisms for many purposes:
(a) The main function of food is to help in growth.

(b) Food provides energy for movements such as running, walking or raising our arm.

(c) Food is also needed for replacement and repairing damaged parts of body.
(d) Food gives us resistance to fight against diseases and protects us from infections

Q.3. How would you test the presence of starch in leaves?
Ans. The presence of starch in leaves can be tested by Iodine test. When we remove chlorophyll from leaf by boiling it in alcohol and then put 2 drops of iodine solution, its colour change to blue indicates the presence of starch.

Q.4. Give a brief description of the process of synthesis of food in green plants.
Ans. The green plants have chlorophyll in the leaves. The leaves use C02 and water to make food in presence of sunlight.

Q.5. Show with the help of a sketch that the plants are the ultimate source of food.

Q.6. Fill in the blanks:
(a) Green plants are called ________________ since they synthesise their own food.
(b) The food synthesised by the plants is stored as ________________ .
(c) In photosynthesis solar energy is captured by the pigment called ________________
(d) During photosynthesis plants take in ________________ and release ________________ .
Ans. (a) autotrophs (b) starch (c) chlorophyll (d) carbon dioxide, oxygen

Q.7. Name the following:
(i) A parasitic plant with yellow, slender and tubular stem.
(ii) A plant that has both autotrophic and heterotrophic mode of nutrition.
(iii) The pores through which leaves exchange gases.
Ans. (i) cuscuta (ii) Insectivorous plant (iii) Stomata

Q.8. Tick the correct answer:
(a) Amarbel is an example of:
(i) Autotroph (ii) Parasite (iii) Saprotroph (iv) Host

(b) The plant which traps and feeds on insects is:
(a) Cuscuta (ii) China rose {iii) Pitcher plant (iu) Rose
Ans. (a) (ii) Parasite (b) (iii) Pitcher plant

Q.9. Match the items given in Column I with those in Column II:

Q.10. Mark T’ if the statement is true and ‘F’ if it is false:

(i) Carbon dioxide is released during photosynthesis. (T/F)
(ii) Plants which synthesise their food themselves are called saprotrophs. (T/F)
(iii) The product of photosynthesis is not a protein. (T/F)
(iv) Solar energy is converted into chemical energy during photosynthesis. (T/F)
Ans. (a) F (ii) F (iii) T (iu) T

Q.11.Choose the correct option from the following.
Which part of the plant takes in carbon dioxide from the air for photosynthesis?
(i) Root hair (ii) Stomata (iii) Leaf veins (iv) Sepals
Ans. (ii) Stomata

Q.12. Choose the correct option from the following:
Plants take carbon dioxide from the atmosphere mainly through their:
(i) Roots (ii) Stem (iii) Flowers (iv) Leaves
Ans. (iv) Leaves

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Chapter -15 Visualising Solid Shapes | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Maths are solved by experts in order to help students to obtain excellent marks in their annual examination. All the questions and answers that are present in the CBSE NCERT Books has been included in this page. We have provided all the Class 7 Maths NCERT Solutions with a detailed explanation i.e., we have solved all the question with step by step solutions in understandable language. So students having great knowledge over NCERT Solutions Class 7 Maths can easily make a grade in their board exams.

Chapter - 15 Visualising Solid Shapes

Chapter 15 Visualising Solid Shapes Ex 15.1

Question 1.
Identify the nets which can be used to make cubes (cut out copies of the nets and try them):
(i)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 164
(ii)

(iii)

(iv)

(v)

(vi)

Solution:
Cube’s nets are (ii), (iii), (iv) and (vi).

Question 2.
Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them.
NCERT Solutions for Class 7 maths Algebraic Expreesions img 171
Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.
(i)

(ii)

Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total 7.
Solution:
According to the given condition, that opposite faces of a die always have a total of 7 dots on them, we insert the suitable numbers in the blanks as follows:
(i)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 174
(ii)

Question 3.
Can this be a net for a die? Explain your answer.
NCERT Solutions for Class 7 maths Algebraic Expreesions img 176
Solution:
No; Because one pair of opposite faces will have 1 and 4 on them whose total is not 7 and another pair of opposite faces will have 3 and 6 on them whose total is also not 7.

Question 4.
Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there on the net here? (Give two-separate diagrams. If you like, you may use a squared sheet for easy manipulation.)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 177
Solution:
A cube has six faces, so the complete net for making a cube in at least two different ways are as follows :
(i)

(ii)

Question 5.
Match the nets with appropriate solids:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 180
Solution:
After matching the nets with appropriate solids, we get the following pairs:
(a) – (ii)
(b) – (iii)
(c) – (iv)
(d) – (i)

Chapter 15 Visualising Solid Shapes Ex 15.2

Question 1.
Use isometric dot paper and make an isometric sketch to teach one of the given shapes :
(i)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 181
Solution:

(ii)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 182
Solution:

(iii)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 183
Solution:

(iv)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 184
Solution:

Question 2.
The dimensions of a cuboid are 5 cm, 3 cm, and 2 cm. Draw three different isometric sketches of this cuboid.
Solution:
Given, the dimensions of a cuboid are 5 cm, 3 cm, and 2 cm. Three different isometric sketches of a cuboid are as follows :

(i) When the length and height of the front face are 5 cm and 3 cm, respectively.
NCERT Solutions for Class 7 maths Algebraic Expreesions img 189
(ii) When the length and height of the front face are 2 cm and 5 cm, respectively.

(iii) Make an oblique sketch for each one of the given isometric shapes:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 191

Question 3.
There cubes each with a 2 cm edge are placed side by side to form a cuboid. Sketch an oblique or isometric sketch of this cuboid.
Solution:
Oblique sketch:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 192
Isometric sketch:

Question 4.
Make an oblique sketch for each one of the given isometric shapes.
(i)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 194
Solution:
We know that an oblique sketch does not have proportional lengths but it conveys all important aspects of the appearance of the solid. So, oblique sketch for each one of the appearances of the solid. So, the oblique sketch for each one of the given isometric sketch is as follows:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 196
(ii)

Solution:
We know that an oblique sketch does not have proportional lengths but it conveys all important aspects of the appearance of the solid. So, oblique sketch for each one of the appearances of the solid. So, the oblique sketch for each one of the given isometric sketch is as follows :
NCERT Solutions for Class 7 maths Algebraic Expreesions img 197

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?
Solution:
(a)
(i) Oblique sketch of a cuboid of dimensions 5 cm, 3 cm, and 2 cm is as follows:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 198
(ii) An isometric sketch of this cuboid is as follows :

No, these sketches are not unique.

(b) A cube with an edge 4 cm long.
Solution:
(i) Oblique sketch of the cube of edge 4 cm is as follows :
NCERT Solutions for Class 7 maths Algebraic Expreesions img 200
(ii) An isometric sketch of this cube is as follows:

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 die
(d) A circular pipe
(e) An ice cream cone
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 202

Chapter 15 Visualising Solid Shapes Ex 15.4

Question 1.
A bulb is kept burnings 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).
(i)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 203
(ii)

(iii)

Solution:

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)
(i)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 207
(ii)

(iii)

(iv)

Solution:

Question 3.
Examine if the following are true statements :

(i) The cube can cast a shadow in the shape of a rectangle.
Solution:
True, the cube can cast a shadow in the shape of a rectangle.

(ii) The cube can cast a shadow in the shape of a hexagon.
Solution:
False, the cube can not cast a shadow in the shape of a hexagon.

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Chapter -14 Symmetry | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 14 Symmetry

Chapter 14 Symmetry Ex 14.1

Question 1.
Copy the figures with punched holes and find the axes symmetry for the following:

(a)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 61
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(b)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 63
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(c)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 65
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(d)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 67
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(e)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 69
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(f)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 71
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(g)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 72
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(h)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 73
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(i)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 74
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(j)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 75
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(k)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 76
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

(l)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 77
Solution:
The axis of symmetry corresponding to the given punched holes are shown by dotted lines in the figures as under :

Question 2.
Given the lines of symmetry, find the order holes:

(a)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 85
Solution:
With the respect to the given lines of symmetry, the order holes are marked in the given figures as order:

(b)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 86
Solution:
With the respect to the given lines of symmetry, the order holes are marked in the given figures as order:

(c)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 87
Sol:
With the respect to the given lines of symmetry, the order holes are marked in the given figures as order:

(d)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 89
Solution:
With the respect to the given lines of symmetry, the order holes are marked in the given figures as order:

(e)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 90
Solution:
With the respect to the given lines of symmetry, the order holes are marked in the given figures as order:

Question 3.
In the following figures, the mirror line (i.e., the line of symmetry) is given as a dotted line. Complete each figure performing reflection in the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image) Are you able to recall the name of the figure you complete?
Solution:
(a)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 96
Solution:
Corresponding to the given line of symmetry, the completed figure is as under. Their respective names are also given the figures:

(b)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 98
Solution:
Corresponding to the given line of symmetry, the completed figure is as under. Their respective names are also given the figures:

(c)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 97
Solution:
Corresponding to the given line of symmetry, the completed figure is as under. Their respective names are also given the figures :

(d)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 99
Solution:
Corresponding to the given line of symmetry, the completed figure is as under. Their respective names are also given the figures :

(e)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 101
Solution:
Corresponding to the given line of symmetry, the completed figure is as under. Their respective names are also given the figures :

(f)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 100
Solution:
Corresponding to the given line of symmetry, the completed figure is as under. Their respective names are also given the figures :

Question 4.
The following figures have more than one line of symmetry. Such figures are said to have multiple lines of symmetry.
NCERT Solutions for Class 7 maths Algebraic Expreesions img 108

Identify, multiple lines of symmetry, if any, in each of the following figures :
Solution:

(a)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 111
solution:
The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

(b)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 112
solution:
The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

(c)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 113
solution:
The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

(d)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 114
solution:
The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

(e)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 115
solution:
The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

(f)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 116
solution:
The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

(g)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 117
solution:
The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

(h)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 118
solution:
The multiple lines of symmetry in respect of the given figures are shown by dotted lines as under :

Question 5.
Copy the figure given here. Take any one diagonal as a line of symmetry and shade a few more squares to make the figure symmetric about a diagonal. Is there more than one way to do that? Will the figure be symmetric about both the diagonals?
NCERT Solutions for Class 7 maths Algebraic Expreesions img 127
Solution:

Let us mark the vertices of the square as A, B, C and D. Take the diagonal BD as a line of symmetry and shade a few more squares as shown to make the figure symmetric about this diagonal. There is only one way to do it. Clearly, the figure is symmetric about the second diagonal AC. Hence, the figure is symmetric about both the diagonals.

Question 6.
Copy the diagram and complete each shape to be symmetric about the mirror line(s):

(a)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 129
Solution:
Completed shape symmetric about the mirror lines are as under :

(b)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 130
Solution:
Completed shape symmetric about the mirror lines are as under :

(c)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 131
Solution:
Completed shape symmetric about the mirror lines are as under :

(d)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 132
Solution:
Completed shape symmetric about the mirror lines are as under :

Question 7.
State the number of lines of symmetry for the following figures:
(a) An equilateral triangle
(b) An isosceles triangle
(c) A scalene triangle
(d) A square
(e) A rectangle
(f) A rhombus
(g) A parallelogram
(h) A quadrilateral
(i) A regular hexagon
(j) A circle
Solution:

	Figure	Number of lines of symmetry
(a)	An equilateral triangle	3
(b)	An isosceles triangle	1
(c)	A scalene triangle	0
(d)	A square	4
(e)	A rectangle	2
(f)	A rhombus	2
(g)	A parallelogram	0
(h)	A quadrilateral	0
(i)	A regular hexagon	6
(j)	A circle	Infinite

Question 8.
What letters of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection).

(a) a vertical mirror
Solution:
The English alphabet letters having reflectional symmetry about a vertical mirror are A, H, I, M, 0, T, U, V, W, X, Y

(b) a horizontal mirror
Solution:
The English alphabet having reflectional symmetry about a horizontal mirror are B, C, D, E, H, I, O, and X

(c) both horizontal and vertical mirrors.
Solution:
The English alphabet having reflectional symmetry about both horizontal and vertical mirrors are H, I, O, and X

Question 9.
Give three examples of shapes with no line of symmetry.
Solution:
Three examples of shapes with no line of symmetry are

A scalene triangle
A parallelogram
An irregular quadrilateral

Question 10.
What another name can you give to the line of symmetry of

(a) an isosceles triangle?
Solution:
Another name for the line of symmetry of an isosceles triangle is median.

(b) a circle?
Solution:
Another name for the line of symmetry of a circle is the diameter.

Chapter 14 Symmetry Ex 14.2

Question 1.
Which of the following figures have rotational symmetry of order more than 1:
(a)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 137
(b)

(c)

(d)

(e)

(f)

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 :
(a)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 143
Solution:
Let us mark a point A on each figure and also indicate the angle through which the figure is to be rotated as under :

(b)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 144
Solution:
Let us mark a point A on each figure and also indicate the angle through which the figure is to be rotated as under :

(c)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 145
Solution:
Let us mark a point A on each figure and also indicate the angle through which the figure is to be rotated as under :

(d)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 146
Solution:
Let us mark a point A on each figure and also indicate the angle through which the figure is to be rotated as under :

(e)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 148
Solution:
Let us mark a point A on each figure and also indicate the angle through which the figure is to be rotated as under :

(f)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 149
Solution:
Let us mark a point A on each figure and also indicate the angle through which the figure is to be rotated as under :

(g)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 147
Solution:
Let us mark a point A on each figure and also indicate the angle through which the figure is to be rotated as under :

(h)
NCERT Solutions for Class 7 maths Algebraic Expreesions img 150
Solution:
Let us mark a point A on each figure and also indicate the angle through which the figure is to be rotated as under :

Now to find the rotational symmetry, we proceed as under :

In figure (a): it requires two rotations, each through an angle of 180°, about the marked point (x) to come back to its original position. So, its rotational symmetry is of order 2.

In figure (b): It requires two rotations, each through an angle of 180°, about the marked point (x) to come back to its original position. So, its rotational symmetry is of order 2.

In figure (c): The triangle requires three rotations, each through an angle of 120° about the marked point to come back to its original position. So, it has rotational symmetry of order 3.

In figure (d): The figure requires four rotations, each through an angle of 90°, about the marked point (x) to come back to its original position. So, its rotational symmetry is of order 4.

n figure (e): The figure requires four rotations, each through an angle of 90°, about the marked point (x) to come back to its original position. So, its rotational symmetry is of order 4.

In figure (f): The regular pentagon requires five rotations, each through an angle of 72°, about the marked point to come back to its original position. So, it has rotational symmetry of order 5.

In figure (g): The figure requires six rotations, each through an angle of 60°, about the marked point to come back to its original position. So, it has rotational symmetry of order 6.

In figure (h): The figure requires three rotations each through an angle of 120°, about the marked point to come back to its original position. So, it has rotational symmetry of order 3.

Chapter 14 Symmetry Ex 14.3

Question 1.
Name any two figures that have both line symmetry and rotational symmetry.
Solution:

Circle
Equilateral triangle

Question 2.
Draw, wherever possible, a rough sketch of
(i) a triangle with both line and rotational symmetries of order more than 1.
Solution:
Three lines of symmetry
NCERT Solutions for Class 7 maths Algebraic Expreesions img 159
Also, an equilateral triangle has rotational symmetry of order 3 as shown below:

(ii) a triangle with only line symmetry and no rotational symmetry of order more than 1.
Solution:
One line of symmetry but no rotational symmetry of order more than 1.
NCERT Solutions for Class 7 maths Algebraic Expreesions img 161

(iii) a quadrilateral with rotational symmetry of order more than 1 but not a line symmetry.
Solution:
No line of symmetry but have rotational symmetry of order more than 1.
NCERT Solutions for Class 7 maths Algebraic Expreesions img 162

(iv) a quadrilateral with line symmetry but not a rotational symmetry or order more than 1.
Solution:
One line of symmetry but no rotational symmetry of order more than 1.
NCERT Solutions for Class 7 maths Algebraic Expreesions img 163

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:

Shape	Centre of Rotation	Order of Rotation	Angle of Rotation
Square
Rectangle
Rhombus
Equilateral Triangle
Regular Hexagon
Circle
Semicircle

Solution:

Shape	Centre of Rotation	Order of Rotation	Angle of Rotation
Square	Yes	4	90°
Rectangle	Yes	4	90°
Rhombus	Yes	4	90°
Equilateral Triangle	Yes	3	120°
Regular Hexagon	Yes	6	60°
Circle	Yes	Infinite	Any angle
Semicircle	Yes	4	90°

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

Square
Rectangle

Question 6.
After rotating by 60° about a center, 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°, 360°.

Question 7.
Can we have rotational symmetry of order more than 1 whose angle of rotation is

(i) 45°?
Solution:
Yes

(ii) 17°?
Solution:
No

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Chapter – Chapter 13 Exponents and Powers | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 13 Exponents and Powers

Chapter 13 Exponents and Powers Ex 13.1

Question 1.
Find the value of:

(i) 2⁶
Solution:
2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

(ii) 9³
Solution:
9² = 9 × 9 × 9 = 729

(iii) 11²
Solution:
11² = 11 × 11 = 121

(iv) 5⁴
Solution:
5⁴ = 5 × 5 × 5 × 5 = 625

Question 2.
Express the following in exponential form :

(i) 6 × 6 × 6 × 6
Solution:
6 × 6 × 6 × 6 = 6⁴

(ii) t x t
Solution:
t × t = t²

(iii) b × b × b × b
Solution:
b × b × b × b = b⁴

(iv) 5 × 5 × 7 × 7 × 7
Solution:
5 × 5 × 7 × 7 × 7 = 5² × 7³

(v) 2 × 2 × a × a
Solution:
2 × 2 × a × a = 2² × a²

(vi) a × a × a × c × c × c × c × d
Solution:
a × a × a × c × c × c × c d = a³ × c⁴ × d

Question 3.
Express each of the following numbers using the exponential notation:

(i) 512
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 29

(ii) 343
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 30

(iii) 729
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 31

(vi) 3125
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 32

Question 4.
Identify the greater number, wherever possible, in each of the following?

(i) 4³ or 3⁴
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 38

(ii) 5³ or 3⁵
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 34

(iii) 2⁸ or 8²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 35

(iv) 100² or 2¹⁰⁰
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 36

(v) 2¹⁰ or 10²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 37

Question 5.
Express each of the following as a product of powers of their prime factors:

(i) 648
Solution:
To determine the prime factorization of 648, we use the division method, as shown below:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 25

(ii) 405
Solution:
We use the division method as shown under:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 26

(iii) 540
Solution:
We use the division method as shown under:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 27

(iv) 3600
Solution:
We use the division method as shown under:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 28

Question 6.
Simplify:

(i) 2 × 10³
Solution:
2 × 10³ = 2 × 1000 = 2000

(ii) 7² x 2²
Solution:
7² × 2² = (7 × 2)² = (14)³ = 14 × 14 = 196

(iii) 2³ × 5
Solution:
2³ × 5 = 2 × 2 × 2 × 5
= 8 × 5 = 40

(iv) 3 × 4⁴
Solution:
3 × 4⁴ = 3 × 4 × 4 × 4 × 4
= 3 × 256 = 768

(v) 0 × 10²
Solution:
0 × 10² = 0

(vi) 5² × 3³
Solution:
5² × 3³ = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675

(vii) 24 × 32
Solution:
2⁴ × 3² = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144

(viii) 3² × 10⁴
Solution:
3² × 10⁴ = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000

Question 7.
Simplify:

(i) (-4)³
Solution:
(- 4)³ = (- 4) × (- 4) × (- 4) = – 64

(ii) (-3) × (-2)³
Solution:
(-3) × (-2)³ = (-3) × (- 2) × (- 2) × (- 2) = (- 3) × (- 8) = 24

(iii) (- 3)² × (- 5)2
Solution:
(- 3)² × (- 5)² = (- 3) × (- 3) × (- 5) × (- 5) = 9 × 25 = 225

(iv) (-2)³ × (-10)²
Solution:
(-2)³ × (-10)³ = (-2) × (-2) × (-2) x (-10) × (-10) × (-10) = (-8) × (-1000) = 8000

Question 8.
Compare the following numbers:

(i) 2.7 × 10¹²; 1.5 × 10⁸
Solution:
We have,
2.7 × 10¹² = 2.7 × 10 × 10¹¹
= 27 × 10¹¹, contains 13 digits and 1.5 × 10⁸ = 1.5 × 10 × 10⁷
= 15 × 10⁷ contains 9 digits
Clearly, 2.7 × 10¹² > 1.5 × 10⁸

(ii) 4 × 10¹⁴; 3 × 10¹⁷
Solution:
We have, 4 × 10¹⁴, contains 15 digits and, 3 × 10¹⁷, contains 18 digits 4 × 10¹⁴ < 3 × 10¹⁷

Chapter 13 Exponents and Powers Ex 13.2

Question 1.
Using the law of exponents, simplify and write the answer in exponential form:

(i) 3² × 3⁴ × 3⁸
Solution:
3² × 3⁴ × 3⁸ = 3² + 4 + 8

(ii) 6¹⁵ ÷ 6¹⁰
Solution:
6¹⁵ ÷ 6¹⁰ = 6¹⁵ ^–¹⁰ = 6⁵

(iii) a³ × a²
Solution:
a³ × a² = a³ ⁺ ² = a⁵

(iv) 7^x x 7²
Solution:
7^x × 7² = 7^x ⁺ ²

(v) (5²)³ ÷ 5³
Solution:
(5²)³ ÷ 5³ = 5² ^× ³ ÷ 5³
= 5⁶ ÷ 5³ = 5⁶ ^– ³ = 5³

(vi) 2⁵ × 5⁵
Solution:
2⁵ × 5⁵ = (2 × 5)⁵ = (10)⁵

(vii) a⁴ × b⁴
Solution:
a⁴ × b⁴ = (ab)⁴

(viii) (3⁴)³
Solution:
(3⁴)³ = 3⁴ ^× ³ = 3¹²

(ix) (2²⁰ ÷ 2¹⁵) × 2³
Solution:
(2²⁰ ÷ 2¹⁵) × 2³ = (2¹⁰ ^– ¹⁵) × 2³
= 2⁵ × 2³ = 2⁵ ⁺ ³ = 2⁸

(x) 8^t ÷ 8²
Solution:
8^t ÷ 8² = 8^t ^– ²

Question 2.
Simplify and express each of the following in exponential form:
(i) 23×34×43×32
Solution:
23×34×43×32 = 23×34×223×25
[∵ 4 = 2 × 2 = 2² , 32 = 2 × 2 × 2 × 2 × 2 = 2⁵]
= 23+2×3431×25 = 25×3431×25
= 2⁵ ^– ⁵ x 3⁴ ^– ¹ = 2⁰ × 3³ = 1 × 3³ = 3³

(ii) [(5²)³ × 5⁴] ÷ 5⁷
Solution:
[(5²)³ × 5⁴] ÷ 5⁷ = (5 ^{2 ×} ³ × 5⁴) ÷ 5⁷
=(5⁶ × 5⁴) ÷ 5⁷ = 5⁶ ⁺ ⁴ ÷ 5⁷
= 5¹⁰ ÷ 5⁷ = 5¹⁰ ^– ⁷ = 5³

(iii) 25⁴ ÷ 5³
Solution:
25⁴ ÷ 5³ = (5²)⁴ ÷ 5³
= 5² ^× ⁴ ÷ 5³
= 5⁸ ÷ 5³ = 5⁸ ^– ³
= 5⁵

(iv) 3×72×11821×113
Solution:
3×72×11821×113 = 3×72×1183×7×113 = 3¹ ^– ¹ × 7² ^– ¹ × 11⁸ ^– ³
= 3⁰ × 7¹ × 11⁵
= 1 × 7 × 11⁵
= 7 × 11⁵
= 7 × 11⁵

(v) 3734×33
Solution:
3734×33 = 3734+3 = 3737
3⁷ ^– ⁷ = 3⁰ = 1

(vi) 2⁰ + 3⁰ + 4⁰
Solution:
2⁰ + 3⁰ + 4⁰ = 1 + 1 + 1 = 3

(vii) 2⁰ x 3⁰ x 4⁰
Solution:
2⁰ × 3⁰ × 4⁰ = 1 × 1 × 1 = 3

(viii) (3⁰ + 2⁰) × 5⁰
Solution:
(3⁰ + 2⁰) × 5⁰ = (1 + 1) × 1 = 2 × 1 = 2

(ix) 28×a543×a3
Solution:
28×a543×a3 = 28×a5(22)3×a3 = 28×a526×a3
= 2⁸ ^– ⁶ × a⁵ ^– ³ = 2² × a² = (2a)²

(x) (a5a3) x a⁸
Solution:
(a5a3) × a⁸ = (a⁵ ^– ³) × a⁸
= a² × a⁸
= a² ⁺ ⁸
= a¹⁰

(xi) 45×a8b345×a5b2
Solution:
45×a8b345×a5b2 = 4⁵ ^– ⁵ × a⁸ ^– ⁵ × b³ ^– ²
= 4⁰ × a³ × b¹
= 1 × a × b
= a³b

(xii) (2³ × 2)²
Solution:
(2³ × 2)² = (2³ + 1)² = (2⁴)² = 2⁸

Question 3.
Say true or false and justify your answer:

(i) 10 × 10¹¹ = 100¹¹
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 58

(ii) 2³ > 5²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 55

(iii) 2³ × 3² = 6⁵
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 56

(iv) 3⁰ = (1000)⁰
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 57

Question 4.
Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 50
∴ 108 × 192 = (2 × 2 × 3 × 3 × 3) x (2 × 2 × 2 × 2 × 2 × 2 × 3)
= 2² × 3³ × 2⁶ × 3¹
= 2² ⁺ ⁶ × 3³ ⁺ ¹
= 2⁸ × 3⁴

(ii) 270
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 51
∴ 270 = 2 × 3 × 3 × 3 × 5 = 2 × 3³ × 5

(iii) 729 x 64
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 52
∴ 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3²
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶
∴ 729 × 64 = 3⁶ × 2⁶

(iv) 768
Solution:
We have
NCERT Solutions for Class 7 maths Algebraic Expreesions img 53
∴ 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3

Question 5.
Simplify:

(i) (25)2×7383×7
Solution:
(25)2×7383×7 = 210×73(23)3×7 = 210×7329×7
= 2¹⁰ ^– ⁹ × 7³ ^– ¹
= 2¹ × 7²
= 2 × 49
= 98

(ii) 25×52×t8103×t4
Solution:
25×52×t8103×t4 = 52×52×t8(2×5)3×t4 = 54×t823×53×t4 = 54−3×t8−423 = 5×t423 = 5t48

(iii) 35×105×2557×65
Solution:
35×105×2557×65 = 35×(2×5)5×5257×(2×3)5 = 35×25×55×5257×25×35
= 3⁵ ^– ⁵ × 2⁵ ^– ⁵ × 5⁵ ⁺ ² ^– ⁷= 30 × 20 × 50
= 1 × 1 × 1
= 1

Chapter 13 Exponents and Powers Ex 13.3

Question 1.
Write the following numbers in the expanded forms :
Solution:

(i) 279404 = 2 × 100000 + 7 × 10000 x 9 × 1000 + 4 × 100 + 0 × 10 + 4 × 1
= 2 × 10⁵ + 7 × 10⁴ +9 × 10³ + 4 × 10² + 0 × 10¹ + 4 × 10⁰

(ii) 3006194 = 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4 × 1
= 3 × 10⁶ + 0 × 10⁵ + 0 × 10⁴ + 6 × 10³
+ 1 × 10² + 9 × 10¹ + 4 × 10°

(iii) 2806196 = 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 +1 × 100 + 9 × 10¹ + 6 × 10⁰
= 2 × 10⁶ + 8 × 10⁵ + 0 × 10⁴ + 6 × 10³

(iv) 120719 = 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9 × 1
= 1 × 10⁵ + 2 × 10⁴ + 0 × 10³ + 7 × 10² +1 × 10¹ + 9 × 10⁰

(v) 20068 = 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1
= 2 × 10⁴ + 0 × 10³ + 0 × 10² + 6 × 10¹ + 8 × 10⁰

Question 2.
Find the number from each of the following expanded forms:

(a) 8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
Solution:
8 × 10⁴ + 6 × 10³ + 0 × 10² + 4 × 10¹ + 5 × 10⁰
= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 86045

(b) 4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰
Solution:
4 × 10⁵ + 5 × 10³ + 3 × 10² + 2 × 10⁰
= 4 × 100000 + 5 × 1000 + 3 × 100 + 2
= 405302

(d) 9 × 10⁵ + 2 × 10² + 3 × 10¹
Solution:
9 × 10⁵ + 2 × 10² + 3 × 10¹
= 9 × 100000 + 2 × 100 + 30
= 900230

Question 3.
Express the following numbers in standard form:

(i) 5,00,00,000
Solution:
5,00,00,000 = 5 × 10000000 = 5 × 10⁷

(ii) 70,00,000
Solution:
70,00,000 = 7 × 1000000 = 7 × 10⁶

(iii) 3,18,65,00,000
Solution:
3,18,65,00,000 = 3.1865 × 1000000000 = 3.1865 × 10⁹

(iv) 3,90,878
Solution:
3,90,878 = 3.90878 × 100000 = 3.90878 × 10⁵

(v) 39087.8
Solution:
39087.8 = 3.90878 × 10000 = 3.90878 × 10⁴

(vi) 3908.78
Solution:
3908.78 = 3.90878 × 1000 = 3.90878 × 10³

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

(a) The distance between Earth and Moon is 384,000,000 m.
Solution:
The mean distance between Earth and Moon = 384,000,000 m = 3.84 × 100000000 m = 3.84 × 10⁸ m

(b) Speed of light in vaccum is 300,000,000 m/s.
Solution:
Speed of light in vaccum
= 300,000,000 m/s
= 3.0 × 100000000 m/s
= 3.0 × 10⁸ m/s

(c) Diameter of the Earth is 1,27,56,000 m.
Solution:
Diameter of the Earth 1,27,56,000 m
= 1.2756 × 10000000 m = 1.2756 × 10⁷

(d) Diameter of the Sun is 1,400,000,000 m.
Solution:
Diameter of the Sun
= 1,400,000,000 m = 1.4 × 1000000000 m = 1.4 × 10⁹ m

(e) In a galaxy there are on4m average 100,000,000,000 stars.
Solution:
In a galaxy there are on an average
= 100,000,000,000 stars = 1 × 100000000000 stars = 1 × 10¹¹stars

(f) The universe is estimated to be 12,000,000,000 years old.
Solution:
The universe is estimated to be 12,000,000,000 years old
= 12 × 1,000,000,000 = 1.2 × 10¹⁰ year

(g) The distance of the Sun from the center of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.
Solution:
The distance of Sun from the centre of the Milky Way Galaxy is estimated to be
= 300,000,000,000,000,000,000
= 3 × 100000000000000000000
= 3 × 10²⁰ m

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
Solution:
Number of molecules contained in a drop of water weighing 1.8 gm = 60,230,000,000,000,000,000,000
= 6.023 × 10000000000000000000000 = 6.023 × 10²⁰

(i) The Earth has 1,353,000,000 cubic km of seawater.
Solution:
The Earth has 1,353,000,000 cubic km of seawater i.e.,
1.353 × 1,000,000,000 = 1.353 × 10⁹ cubic km.

(j) The population of India was about 1,027,000,000 in March, 2001.
Solution:
The population of India was about in March 2001 = 1,027,000,000 = 1.027 × 1000000000 = 1.027 × 10⁹

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Chapter – 12 Algebraic Expressions | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 12 Algebraic Expressions

Chapter 12 Algebraic Expressions Ex 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.

Subtraction of z from y.
One-half of the sum of numbers x and y.
The number z multiplied by itself
One-fourth of the product of numbers p and q.
Numbers x and y both squared and added.
Number 5 added to three times the product of numbers m and n.
Product of numbers y and z subtracted from 10.
Sum of numbers a and b subtracted from their product.

Solution:

y – z
12 ( x + y )
z²
14 pq
x² + y²
3mn + 5
10 – yz
ab – ( a + b )

Question 2.
(i) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams.

(a) x – 3
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 2

(b) 1 + x + x²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 3

(d) 5xy²+ 7x²y
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 5

(e) – ab + 2b² – 3a²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 6

(ii) Identify terms and factors in the expressions given below :
(a) -4x + 5
(b) -4x + 5y
(c) 5y + 3y²
(d) xy + 2x² y²
(e) pq + q
(f) 1.2ab – 2.4b + 3.6a
(g) 34 x + 14
(h) 0.1p² + 0.2q²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 12

Question 3.
Identify the numerical coefficients terms (other than constants) in the following expressions :
(i) 5 – 3t²
(ii) 1 + t + t² + f³
(iii) x + 2xy + 3y
(iv) 100m + 1000n
(v) – p²q² + 7 pq
(vi) 1.2a + 0.8b
(vii) 3.14r²
(viii) 2 (l + b)
(ix) 0.1y + 0.01y²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 11

Question 4.
(a) Identify terms that contain x and give the coefficient of x.
(i) y²x + y
(ii) 13y² – 8yx
(iii) x + y + 2
(iv) 5 + z + zx
(v) 1 + x + xy
(vi) 12xy² + 25
(vii) 7x + xy²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 10

(b) Identify terms that contain y² and give the coefficient of y².
(i) 8 – xy²
(ii) 5y²+ 7x
(iii) 2x²y – 15xy² + 7y²
Solution:
NCERT Solutions for Class 7 maths Algebraic Expreesions img 9

Question 5.
Classify into monomials, binomials, and 4y trinomials
(i) 4y – 7z
(ii) y²
(iii) x + y – xy
(iv) 100
(v) ab – a – b
(vi) 5 – 3t
(vii) 4p²q – 4pq²
(viii) 7mn
(ix) z² – 3z + 8
(x) a² + b²
(xi) z² + z
(xii) 1 + x + x²
Solution:
We know that an algebraic expression containing only one term is called a monomial. So, the monomials are : (ii), (iv), and (viii).

We know that an algebraic expression containing two terms is called a binomial. So, the binomials are : (i), (vi), (vii), (x) and (xi).

We know that an algebraic expression containing three terms is called a trinomial. So, the trinomial are : (iii), (v), (ix) and (xii).

Question 6.
State whether a given pair of terms j is of like or unlike terms:

1,100
-7x, 52 x
-29 x, -29 y
14 xy, 42 yx
4m²p, 4mp²
12xz, 12x² z².

Solution:

Like
Like
Unlike
Like
Unlike
Unlike

Question 7.
Identify like terms in the following:
(a) – xy², -4yx², 8x², 2xy², 7y, – 11x², – 100x, – 11yx, 20x²y, – 6x², y, 2xy, 3x
Solution:
In the given terms, like terms are grouped as under :
NCERT Solutions for Class 7 maths Algebraic Expreesions img 13

(b) 10pq, 7p, 8q, – p²q², – 7qp, – 100q,- 23, 12q²p², – 5p², 41, 2405p, 78qp, 13p²q, qp², 701p²
Solution:
In the given terms, like terms are grouped as under :
NCERT Solutions for Class 7 maths Algebraic Expreesions img 14

Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms:
(i) 21b -32 + 7b- 206
(ii) -z² + 13z2 -5z + 7z³ – 15²
(iii) p – (p – q) – q – (q – p)
(iv) 3a – 2b — ab – (a – b + ab) + 3ab + 6 – a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² -3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
Solution:
(i) 21b – 32 + 7b – 206
Re-arranging the like terms, we get
216 + 7b – 206 – 32
= (21 + 7 – 20)b – 32
= 8b – 32 which is required.

(ii) -z² + 13z² – 5z – 15z
Re-arranging the like terms, we get
7z³ – z² + 13z² – 5z + 5z – 15z
= 7z³ + (-1 + 13)z² + (-5 – 15)z
= 7z³ + 12z² – 20z which is required.

(iii) p – (p – q) – q – (q – p)
=p – p + q – q – q + p
Re-arranging the like terms, we get

= p – q which is required.

(iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
Re-arranging the like terms, we get
= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a + ab which is required.

(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
Re-arranging the like terms, we get
5x²y + 3x²y + 8xy² – 5x² + x² – 3y² – y² – 3y²
= 8x²y + 8xy² – 4x² – 7y² which is required.

(vi) (3y² + 5y – 4) – (8y – y² – 4)
= 3y² + 5y – 4 – 8y + y² + 4 (Solving the brackets)
Re-arranging the like terms, we get
= 3y² + y² + 5y – 8y – 4 + 4
= 4y² – 3y which is required.

Ex 12.2 Class 7 Maths Question 2.
Add:
(i) 3mn, -5mn, 8mn, -4mn
(ii) t – 8tz, 3tz, -z, z – t
(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
(iv) a + b – 3, b – a + 3, a – b + 3
(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(vii) 4x²y, -3xy², -5xy², 5x²y
(viii) 3p²q² – 4pq + 5, -10p²q², 15 + 9pq + 7p²q²
(ix) ab – 4a, 4b – ab, 4a – 4b
(x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
Solution:
(i) 3mn, -5mn, 8mn, -4mn
= (3 mn) + (-5 mn) + (8 mn) + (- 4 mn)
= (3 – 5 + 8 – 4)mn
= 2mn which is required.

(ii) t – 8tz, 3tz – z, z – t
t – 8tz + 3tz – z + z – t
Re-arranging the like terms, we get
t – t – 8tz + 3tz – z + z
⇒ 0 – 5 tz + 0
⇒ -5tz which is required.

(iii) -7mn + 5, 12mn + 2, 9mn – 8, -2mn – 3
= -7mn + 5 + 12mn + 2 + 9mn – 8 + (-2mn) – 3
Re-arranging the like terms, we get
-7mn + 12 mn + 9mn – 2 mn + 5 + 2 – 8 – 3

= 12 mn – 4 which is required.

(iv) a + b – 3, b – a + 3, a – b + 3
⇒ a + b – 3 + b – a + 3 + a – b + 3
Re-arranging the like terms, we get
a – a + a + b + b – b – 3 + 3 + 3
⇒ a + b + 3 which is required.

(v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
∴ 14x + 14y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy
Re-arranging the like terms, we get
-12xy + 8xy + 4xy + 14x – 7x + 10y – 10y – 13 + 18

= 0 + 7x + 0 + 5
= 7x + 5 which is required

(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5 5m -In + 3n — 4m + 2 + 2m – 3mn – 5
Re-arranging the like terms, we get
5m – 4m + 2m – 7n + 3n- 3mn + 2 – 5
= 3m – 4n – 3mn – 3 which is required.

(vii) 4x²y, -3xy², -5xy², 5x²y
Re-arranging the like terms and adding, we get
4x²y – 5xy² – 3xy² + 5x²y
=9x²y — 8xy² which is required.

(viii) 3p²q² – 4pq + 5, -10p²q², 15 + 9pq + 7p²q²
= (3p²q² – 4pq + 5) + (-10p²q²) + (15 + 9pq + 7p²q²)
= 3p²q² – 4pq + 5 – 10p²q² + (15 + 9pq + 7p²q² )
= 3p²q² + 7p²q² – 10p²q² – 4pq + 9pq + 5 + 15
= 10p²q² – 10p²q² + 5pq + 20
= 0 + 5pq + 20
= 5pq + 20 which is required.

(ix) ab – 4a, 4b – ab, 4a – 4b
= ab – 4a + 4b – ab + 4a – 4b

= 0 + 0 + 0 = 0 which is required.

(x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
= x² – y² – 1 + y² – 1 – x² + 1 – x² – y²

= -x² – y² – 1
= -(x² + y² + 1) which is required.

Ex 12.2 Class 7 Maths Question 3.
Subtract:
(i) -5y² from y²
(ii) 6xy from -12xy
(iii) (a – b) from (a + b)
(iv) a(b – 5) from b(5 – a)
(v) -m² + 5mn from 4m² – 3mn + 8
(vi) -x² + 10x – 5 from 5x – 10
(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Solution:
(i) -5y² from y² = y² – (-5y²)
= y² + 5y² = 6y²

(ii) 6ry from -12ry = -12xy – 6xy = -18xy which is required.
(iii) (a – b) from (a + b)
= (a + b) – (a – b)
= a + b – a + b = 2b which is required

(iv) a(b – 5) from b(5 – a)
= b(5 – a) – a(b – 5)
= 5b – ab – ab + 5a
= 5a – 2ab + 5b
= 5a + 5b – 2ab which is required.

(v) -m² + 5mn from 4m² – 3mn + 8
= (4m² – 3mn + 8) – (-m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= 4m² + m² – 3mn – 5mn + 8
= 5m² – 8mn + 8 which is required.

(vi) -x² + 10x – 5 from 5x – 10
= (5x – 10) – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= x² + 5x – 10x – 10 + 5
= x² – 5x – 5 which is required.

(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
= (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= 3ab + 7ab – 2a² – 5a² – 2b² – 5b²
= 10ab – 7a² – 7b²
which is required.

(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
= (5p² + 3q² – pq) – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq – 4pq + 5q² + 3p²
= 5p² + 3p² + 3q² + 5q² – pq – 4pq
= 8p² + 8q² – 5pq
which is required.

Ex 12.2 Class 7 Maths Question 4.
(a) What should be added to x² + xy + y²to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16?
Solution:
(a) (2x² + 3xy) – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y²
= 2x² – x² + 3xy – xy – y²
= x² + 2xy – y² is required expression.

(b) (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
= 5a + b – 6 is required expression.

Ex 12.2 Class 7 Maths Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:
Let A be taken away.
∴ (3x² – 4y² + 5 xy + 20)-A
= -x² – y² + 6xy + 20
⇒ A = (3x² – 4y² + 5xy + 20) – (-x²2 – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= 3x² + x² – 4y² + y² + 5xy – 6xy + 20 – 20
= 4x² – 3y² – xy is required expression.

Ex 12.2 Class 7 Maths Question 6.
(a) From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
(b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:
(a) Sum of 3x – y + 11 and -y – 11
= (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
∴ 3x – 2y – (3x – 2y) – (3x – y – 11)
= 3x – 2y – 3x + y + 11
= 3x – 3x – 2y + y + 11
= -y + 11 is required solution.

(b) Sum of (4 + 3x) and (5 – 4x + 2x²)
= 4 + 3x + 5 – 4x + 2x²
= 2x² – 4x + 3x + 9 = 2x² – x + 9
Sum of (3x² – 5x) and (-x² + 2x + 5)
= (3x² – 5x) + (-x² + 2x + 5)
= 3x² – 5x – x² + 2x + 5 = 2x² – 3x + 5
Now (2x² – x + 9) – (2x² – 3x + 5)
= 2x² – x + 9 – 2x² + 3x – 5
= 2x² – 2x² + 3x – x + 4
= 2x + 4 is required expression.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Q1

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Q2

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Q3

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Q4

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Q5

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 Q6

NCERT Solutions

Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m² – 2m – 7
(v) 5m2−4
Solution:
(i) m – 2
Putting m = 2, we get
2 – 2 = 0

(ii) 3m – 5
Putting m = 2, we get
3 × 2 – 5 = 6 – 5 = 1

(iii) 9 – 5m
Putting m = 2, we get
9 – 5 × 2 = 9 – 10 = -1

(iv) 3m² – 2m – 7 Putting m = 2, we get
3(2)² – 2(2) – 7 = 3 × 4 – 4 – 7
=12 – 4 – 7 = 12 – 11 = 1

(v) 5m2−4
Putting m = 2, we get
5×22−4=5−4=1

Ex 12.3 Class 7 Maths Question 2.
If p = -2, find the value of:
(i) 4p + 7
(ii) -3p² + 4p + 7
(iii) -2p³ – 3p² + 4p + 7
Solution:
(i) 4p + 7
Putting p = -2, we get 4(-2) + 7 = -8 + 7 = -1

(ii) -3p² + 4p + l
Putting p = -2, we get
-3(-2)² + 4(-2) + 7
= -3 × 4 – 8 + 7 = -12 – 8+ 7 = -13

(iii) -2p³ – 3p² + 4p + 7
Putting p = -2, we get
– 2(-2)³ – 3(-2)² + 4(-2) + 7
= -2 × (-8) – 3 × 4 – 8 + 7
= 16 – 12 – 8 + 7 = 3

Ex 12.3 Class 7 Maths Question 3.
If a = 2, b = -2, find the value of:
(i) a² + b²
(ii) a² + ab + b²
(iii) a² – b²
Solution:
(i) a² + b²
Putting a = 2 and b = -2, we get
(2)² + (-2)² = 4 + 4 = 8

(ii) a² + ab + b²
Putting a = 2 and b = -2, we get
(2)² + 2(-2) + (-2)² = 4 – 4 + 4 = 4

(iii) a² – b²
Putting a = 2 and b = -2, we get
(2)² – (-2)² = 4 – 4 = 0

Ex 12.3 Class 7 Maths Question 4.
When a = 0, b = -1, find the value of the given expressions:
(i) 2a + 2b
(ii) 2a² + b² + 1
(iii) 2a²b + 2ab² + ab
(iv) a² + ab + 2
Solution:
(i) 2a + 2b = 2(0) + 2(-1)
= 0 – 2 = -2 which is required.

(ii) 2a² + b² + 1
= 2(0)² + (-1)² + 1 =0 + 1 + 1 = 2 which is required.

(iii) 2a²b + 2ab² + ab
= 2(0)² (-1) + 2(0)(-1)² + (0)(-1)
=0 + 0 + 0 = 0 which is required.

(iv) a² + ab + 2
= (0)² + (0)(-1) + 2
= 0 + 0 + 2 = 0 which is required.

Ex 12.3 Class 7 Maths Question 5.
Simplify the expressions and find the value if x is equal to 2.
(i) x + 7 +4(x – 5)
(ii) 3(x + 2) + 5x – 7
(iii) 6x + 5(x – 2)
(iv) 4(2x – 1) + 3x + 11
Solution:
(i) x + 7 + 4(x – 5) = x + 7 + 4x – 20 = 5x – 13
Putting x = 2, we get
= 5 × 2 – 13 = 10 – 13 = -3
which is required.

(ii) 3(x + 2) + 5x – 7 = 3x + 6 + 5x -7 = 8x – 1
Putting x = 2, we get
= 8 × 2 – 1 = 16 – 1 = 15
which is required.

(iii) 6x + 5(x – 2) = 6x + 5x – 10
= 11 × – 10
Putting x = 2, we get
= 11 × 2 – 10 = 22 – 10 = 12
which is required.

(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11
= 11x + 7
Putting x = 2, we get
= 11 × 2 + 7 = 22+ 7 = 29

Ex 12.3 Class 7 Maths Question 6.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
(i) 3x – 5 – x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 55
(v) 2a – 2b – 4 – 5 + a
Solution:
(i) 3x – 5 – x + 9 = 2x + 4
Putting x = 3, we get
2 × 3 + 4 = 6 + 4 = 10
which is required.

(ii) 2 – 8x + 4x + 4 = -8x + 4x + 2 + 4 = -4x + 6
Putting x = 2, we have
= -4 × 2 + 6 = -8 + 6 =-2
which is required.

(iii) 3a + 5 – 8a +1 = 3a – 8a + 5 + 1 = -5a + 6
Putting a = -1, we get
= -5(-1) + 6 = 5 + 6 = 11
which is required.

(iv) 10 – 3b – 4 – 5b = -3b – 5b + 10 – 4
= -8b + 6
Putting b = -2, we get
= -8(-2) + 6 = 16 + 6 = 22
which is required.

(v) 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5
= 3a – 26 – 9
Putting a = -1 and b = -2, we get
= 3(-1) – 2(-2) – 9
= -3 + 4 – 9 = 1 – 9 = -8
which is required.

Ex 12.3 Class 7 Maths Question 7.
(i) If z = 10, find the value of z² – 3(z – 10).
(ii) If p = -10, find the value of p² -2p – 100.
Solution:
(i) z² – 3(z – 10)
= z² – 3z + 30
Putting z = 10, we get
= (10)² – 3(10) + 30
= 1000 – 30 + 30 = 1000 which is required.

(ii) p² – 2p – 100
Putting p = -10, we get
(-10)² – 2(-10) – 100
= 100 + 20 – 100 = 20 which is required.

Ex 12.3 Class 7 Maths Question 8.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0?
Solution:
2x² + x – a = 5
Putting x = 0, we get
2(0)² + (0) – a = 5
0 + 0 – a = 5
-a = 5
⇒ a = -5 which is required value.

Ex 12.3 Class 7 Maths Question 9.
Simplify the expression and find its value when a = 5 and b = -3.
2(a² + ab) + 3 – ab
Solution:
2(a² + ab) + 3 – ab = 2a2 + 2ab + 3 – ab
= 2a² + 2ab – ab + 3
= 2ab + ab + 3
Putting, a = 5 and b = -3, we get
= 2(5)² + (5)(-3) + 3
= 2 × 25 – 15 + 3
= 50 – 15 + 3
= 53 – 15 = 38
Hence, the required value = 38.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Q1

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Q2

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Q3

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Q4

Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 1
If the number of digits formed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern. How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 3.
Solution:
(i) The number of line segments required to form
n digits is given by the expressions.

For 5 figures, the number of line segments = 5 × 5 + 1 = 25 + 1 = 26
For 10 figures, the number of line segments = 5 × 10 + 1
= 50 + 1 = 51
For 100 figures, the number of line segments = 5 × 100 + 1
= 500 + 1 = 501

(ii)
For 5 figures, the number of line segments =3 ×5 + 1
= 15 + 1 = 16
For 10 figures, the number of line segments = 3 × 10 + 1
= 30 + 1 = 31
For 100 figures, the number of line segments = 3 × 100 + 1
= 300 + 1 = 301

(iii)
For 5 figures, the number of line numbers = 5 × 5 + 2
= 25 + 2 = 27
For 10 figures, the number of line segments = 5 × 10 + 2
= 50 + 2 = 52
For 100 figures, the number of line segments = 5 × 100 + 2
= 500 + 2 = 502

Ex 12.4 Class 7 Maths Question 2.
Use the given algebraic expression to complete the table of number patterns:

S.No.	Expression	Terms
S.No.	Expression	1st	2^nd	3^rd	4th	5^th	…	10^th	…	100^th	…
(i)	2n – 1	1	3	5	7	9	–	19	–	–	–
(ii)	3n + 2	5	8	11	14	–	–	–	–	–	–
(iii)	4n + 1	5	9	13	17	–	–	–	–	–	–
(iv)	7n + 20	27	34	41	48	–	–	–	–	–	–
(v)	n² + 1	2	5	10	17	–	–	–	–	10,001	–

Solution:
(i) Given expression is 2n – 1
For n = 100, 2 × 100 – 1
= 200 – 1 = 199

(ii) Given expression is 3n + 2
For n = 5, 3 × 5 + 2 = 15 + 2 = 17
For n = 10, 3 × 10 + 2 = 30 + 2 = 32
For n = 100, 3 × 100 + 2 = 300 + 2 = 302

(iii) Given expression is 4n + 1
For n = 5, 4 × 5 + 1 = 20 + 1 = 21
For n = 10, 4 × 10 + 1 = 40 + 1 = 41
For n = 100, 4 × 100 + 1 = 400 + 1 = 401

(iv) Given expression is 7n + 20
For n = 5, 7 × 5 + 20 = 35 + 20 = 55
For n = 10, 7 × 10 + 20 = 70 + 20 = 90
For n = 100, 7 × 100 + 20 = 700 + 20 = 720

(v) Given expression is n² + 1
For n = 5, 5² + 1 = 25 + 1 = 26
For n = 10, 10² + 1 = 100 + 1 = 101

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 Q1

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 Q2

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 Q3

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Chapter – 11 Perimeter and Area | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 11 Perimeter and Area

Chapter 11 Perimeter and Area Exercise 11.1

Question 1.
The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find
(i) its area
(ii) the cost of the land if 1 m² of the land costs ₹ 10,000.
Solution:
(i) For a rectangular piece of land,
Length = 500 m
Breadth = 300 m
∴ Area of the rectangular piece of land = Length × Breadth , = 500 × 300 m² = 150,000 m²

(ii) 1 m² of the land costs = ₹ 10,000
∵ 150,000 m² of the land costs = ₹ 10,000 × 150,000 = ₹ 1,500,000,000.

Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:
We have, perimeter of a square park = 320 m
∴ Side of the square park = ([late×]\frac { 320 }{ 4 } [/late×]) m = 80 m
∴ Area of square park = (Side)²= (80)² m² = 6400 m²

Question 3.
Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.
Solution:
Let the breadth of the rectangular plot of land be b.
It is given that length of the plot = 22 m
and the area of the plot = 440 m²

Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also, find the area.
Solution:
Let the breadth of the rectangular sheet be b cm.
It is given that its length is 35 cm and perimeter is 100 cm.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 3

Question 5.
The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:

Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?
Solution:

Hence, the measure of the side of the square is 31 cm and the square shape encloses more area.

Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also find the area of the rectangle.
Solution:

Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing the wall is ₹ 20 per m².
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 8
Solution:

Chapter 11 Perimeter and Area Exercise 11.2

Question 1.
Find the area of each of the following parallelograms :
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 10
Solution:
(a) Area of the parallelogram = base × height = (7 × 4) cm² = 28 cm²

(b) Area of the parallelogram = base × height = (5 × 3) cm² = 15 cm²

(c) Area of the karallelogram = base × height = (2.5 × 3.5) cm² = 8.75 cm²

(d) Area of the parallelogram = base × height = (5 × 4.8) cm² = 24 cm²

(e) Area of the parallelogram = base × height = (2 × 4.4) cm2 = 8.8 cm²

Question 2.
Find the area of each of the following triangles :
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 11
Solution:

Question 3.
Find the missing values :
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 13
Solution:
We know that
Area of a parallelogram = Base × Height
Therefore, the missing values are calculated as shown :

Question 4.
Find the missing values :
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 15
Solution:

Question 5.
PQRS is a parallelogram (in the given figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find
(a) the area of the parallelogram PQRS.
(b) QN, if PS = 8 cm.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 17
Solution:

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (in the given figure). If the area of the parallelogram is 1470 cm², AB = 35 cm, AD = 49 cm, find the length of BM and DL.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 19
Solution:

Question 7.
∆ ABC is right angled at A (in the given figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ABC. Also find the length of AD.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 21
Solution:

Question 8.
∆ ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (in the given figure). The height AD from A to BC is 6 cm. Find the area of A ABC. What will be the height from C to AB i.e., CE ?
Solution:

Chapter 11 Perimeter and Area Exercise 11.3

Question 1.
Find the circumference of the circles with the following radius : ( Take π 227)
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution:

Question 2.
Find the area of the following circles, given that :
(a) radius = 14 mm (Take π = 227)
(b) diameter = 49 m
(c) radius = 5cm
Solution:

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet (Take π = 227)
Solution:

Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per metre (Take π = 227)
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 29
Solution:

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Solution:
Here, Outer radius, r = 4 cm
Inner radius, r = 3 cm
Area of the remaining sheet = Outer area – Inner area
= π (R² – r²) = 3.14 (4² – 3²) cm²
= 3.14 (16 – 9) cm²
= 3.14 × 7 cm² = 21.98 cm²

Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also its cost if one metre of the lace costs ₹ 15. (Take π = 3.14)
Solution:

Question 7.
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
Solution:

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m² (Take π = 3.14)
Solution:

Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle of the square ? (Take π = 227)
Solution:

Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoining figure). Find the area of the remaining sheet (Take π = 227)
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 36

Solution:

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
Solution:
Area of the square aluminium sheet = (6)² cm² = 36 cm²
Area of the circle cut out from the sheet = (3.14 × 2 × 2) cm² = 12.56 cm²
Area of the sheet left over = (36 – 12.56) cm² = 23.44 cm²

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
Solution:

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (π = 3.14)
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 40
Solution:

Question 14.
A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden? (Take π = 3.14)
Solution:
Circular area of the sprinkler = πr²
= 3.14 × 12 × 12
=3.14 × 144 = 452.16 m²
Area of the circular flower garden = 314 m²
Since, area of the circular flower garden is smaller than by sprinkler. Therefore, the sprinkler will water the entire garden.

Question 15.
Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take π = 3.14)
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 42
Solution:

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 227)
Solution:

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
Solution:
We know that the minute hand describes one complete revolution in one hour.
∴ Distance covered by its tip
= Circumference of the circle of radius 15 cm
= (2 × 3.14 × 15) cm
= 94.2 cm

Chapter 11 Perimeter and Area Exercise 11.4

Question 1.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.
Solution:

Question 2.
A 3 m wide path runs outside and around a rectangular-park of length 125 m and the breadth 65 m. Find the area of the path.
Solution:

Question 3.
A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:

Question 4.
A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find :
(i) the area of the verandah,
(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m².
Solution:

Question 5.
A path 1 m wide is built along the border inside a square garden of side 30 m. Find :
(i) the area of the path,
(ii) the cost of planting grass in the remaining portion of the garden at the rate of ₹ 40 per m².
Solution:

Question 6.
Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Solution:

Question 7.
Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find :
(i) the area covered by the roads,
(ii) the cost of constructing the roads at the rate of ₹ 110 per m².
Solution:

Question 8.
Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π = 3.14)
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 54
Solution:

Question 9.
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find :
(i) the area of the Whole land,
(ii) the area of the flower bed,
(iii) the area of the lawn excluding the area of the flower bed,
(iv) the circumference of the flower bed.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 56
Solution:

Question 10.
In the following figures, find the area of the shaded portions :
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 58
Solution:

Question 11.
Find the area of the quadrilateral ABCD.
Here, AC = 22 cm,
BM = 3 cm,
DN = 3 cm, and BM ⊥ AC, DN ⊥ AC.
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area 60
Solution:

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Chapter 2 Nutrition in Animals

Chapter -1 Nutrition in Plants | Class 7th | NCERT Science Solutions | Edugrown

Chapter 1 Nutrition in Plants | Science

Chapter -15 Visualising Solid Shapes | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 15 Visualising Solid Shapes

Chapter -14 Symmetry | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 14 Symmetry

Chapter – Chapter 13 Exponents and Powers | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 13 Exponents and Powers

Chapter – 12 Algebraic Expressions | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 12 Algebraic Expressions

Chapter – 11 Perimeter and Area | Class 7th |NCERT Maths Solutions | Edugrown

NCERT Solutions for Class 7 Maths

Chapter - 11 Perimeter and Area

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Exercise MCQ

Exercise Ex. 6

Exercise MCQ

Exercise Ex. 7A

Exercise Ex. 7B

Exercise Ex. 7C

Chapter 3 Fibre to Fabric

Chapter 2 Nutrition in Animals

Chapter 1 Nutrition in Plants | Science

NCERT Solutions for Class 7 Maths

Chapter - 15 Visualising Solid Shapes

NCERT Solutions for Class 7 Maths

Chapter - 14 Symmetry

NCERT Solutions for Class 7 Maths

Chapter - 13 Exponents and Powers

NCERT Solutions for Class 7 Maths

Chapter - 12 Algebraic Expressions

NCERT Solutions for Class 7 Maths

Chapter - 11 Perimeter and Area

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