Dattu

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

• Symmetry Class 7 Ex 14.1
• Symmetry Class 7 Ex 14.3
• Symmetry Class 7 MCQ

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

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.2

Question 1.
Which of the following figures have rotational symmetry of order more than 1?

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 :


Solution:

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

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

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

Because of this, it has rotational symmetry of order 2.

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

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

Similarly,

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

∴ It has rotational symmetry of order 4.

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

∴ It has rotational symmetry of order 5.

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

It has rotational symmetry of order 6.

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

∴ It has rotational symmetry of order 3.

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

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

• Symmetry Class 7 Ex 14.1
• Symmetry Class 7 Ex 14.2
• Symmetry Class 7 MCQ

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

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.3

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

Question 2.
Draw, wherever possible, a rough sketch of
(i) a triangle with both line and rotational symmetry of order more than 1.
(ii) a triangle with only line symmetry and no rotational symmetry of order more than 1.
(iii) a quadrilateral with rotational symmetry of order more than 1 but not a line symmetry.
(iv) a quadrilateral with line symmetry but not a rotational symmetry of order more than 1.
Solution:
(i) An equilateral triangle has 3 lines of symmetry and rotational symmetry of order 3.


(ii) An isosceles triangle has only one line symmetry but no rotational symmetry of order more than 1.

(iii) A parallelogram has no line of symmetry but has a rotational symmetry of order 2.

(iv) An isosceles trapezium has one line of symmetry but no rotational symmetry of order more than 1.

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:

Solution:

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

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

Question 7.
Can we have a rotational symmetry of order more than 1 whose angle of rotation is
(i) 45°?
(ii) 17°?
Solution:
(i) Yes

(ii) No

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

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

• Exponents and Powers Class 7 Ex 13.1
• Exponents and Powers Class 7 Ex 13.2
• Exponents and Powers Class 7 MCQ

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

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

Question 1.
Write the following numbers in the expand forms :
(i) 279404
(ii) 3006194
(iii) 2806196
(iv) 120719
(v) 20068.
Solution:

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

Solution:

Question 3.
Express the following numbers in standard form:

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

Solution:

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

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

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

Solution:

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

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

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

• Exponents and Powers Class 7 Ex 13.1
• Exponents and Powers Class 7 Ex 13.3
• Exponents and Powers Class 7 MCQ

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

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

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

Solution:

Question 2.
Simplify and express each of the following in exponential form:

Solution:

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

Solution:

Question 4.
Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768.
Solution:

Question 5.
Simplify

Solution:

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

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.4
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths 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.

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

Solution:

Let the number of digits formed be n, Then, the number of segments required to form n digits is given by the algebraic expression 5n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 1 = 25 + 1 = 26
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 1 = 50 + 1 = 51
3. the number of segments required to form 100 digits of this kind = 5 × 100 + 1 = 500 + 1 = 501.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 3n + 1.
So,

1. the number of segments required to form 5 digits of this kind = 3 × 5 + 1 = 15 + 1 = 16
2. the number of segments required to form 10 digits of this kind = 3 × 10 + 1 = 30 + 1 = 31
3. The number of segments required to form 100 digits of this kind = 3 × 100 + 1 = 300 + 301.

Let the number of digits formed be n. Then, the number of segments required to form n digits is given by the algebraic expression 5n + 2.
So,

1. the number of segments required to form 5 digits of this kind = 5 × 5 + 2 = 25 + 2 = 27
2. the number of segments required to form 10 digits of this kind = 5 × 10 + 2 = 50 + 2 = 52
3. the number of segments required to form loo digits of this kind = 5 × 100 + 2 = 500 + 2 = 502.

Question 2.
Use the given algebraic expression to complete the table of number patterns.

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.3
Number of Questions Solved	10
Category	NCERT Solutions

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

Question 1.
If m = 2, find the value of :

1. m – 2
2. 3m – 5
3. 9 – 5m
4. 3m² – 2m – 7
5. \(\frac { 5m }{ 2 } \) – 4

Solution:

Question 2.
If p = – 2, find the value of :

1. 4p + 7
2. – 3p² + 4p + 7
3. – 2p³ – 3p² + 4p + 7.

Solution:

Question 3.
Find the value of the following expressions, when x = – 1 :

1. 2x – 7
2. – x + 2
3. x² + 2x + 1
4. 2x² – x – 2.

Solution:

Question 4.
If a = 2, b = – 2, find the value of :

1. a² + b²
2. a² + ab + b²
3. a² – b².

Solution:

Question 5.
When a = 0, b = – 1, find the value of the given expressions :

1. 2a + 2b
2. 2a² + b² + 1
3. 2a²b + 2ab² + ab
4. a² + ab + 2.

Solution:

Question 6.
Simplify the expressions and find the value ifx is equal to 2.

1. x + 7 + 4 (x – 5)
2. 3 (x + 2) + 5x – 7
3. 6x + 5 (x – 2)
4. 4 (2x – 1) + 3x + 11.

Solution:

Question 7.
Simplify these expressions and find their values if x = 3, a = – 1, b = – 2.

1. 3x – 5 – x + 9
2. 2 – 8x + 4x + 4
3. 3a + 5 – 8a + 1
4. 10 – 3b – 4 – 5b
5. 2a – 2b – 4 – 5 + a.

Solution:

Question 8.
(i) If z = 10, find the value of z³ – 3(z – 10).
(ii) If p = -10, find the value of p² – 2p – 100.
Solution:

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Solution:

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3 2(a² + ab) + 3 – ab.
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2.

• Algebraic Expressions Class 7 Ex 12.1
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 12
Chapter Name	Algebraic Expressions
Exercise	Ex 12.2
Number of Questions Solved	6
Category	NCERT Solutions

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

Question 1.
Simplify combining like terms:

1. 21b – 32 + 7b – 20b
2. – z² + 13z² – 5z + 7z³ – 15z
3. p – (p – q) – q – (q – p)
4. 3a – 2b – ab – (a – b + ab) + 3ab + b – a
5. 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
6. (3y² + 5y – 4) – (8y – y² – 4).

Solution:

Question 2.
Add:

Solution:

Question 3.
Subtract:

1. -5y² from y²
2. 6xy from – 12xy
3. (a – b) from (a + b)
4. a (b – 5) from b (5 – a)
5. -m² + 5mn from 4m² – 3mn + 8
6. -x² + 10x – 5 from 5x – 10
7. 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
8. 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq.

Solution:

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy?
(b) What should be subtracted from 2a + 8b + 10 to get -3a + 76 + 16?
Solution:

Question 5.
What should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:

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:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2.

• Perimeter and Area Class 7 Ex 11.1
• Perimeter and Area Class 7 Ex 11.3
• Perimeter and Area Class 7 Ex 11.4
• Perimeter and Area Class 7 MCQ

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 11
Chapter Name	Perimeter and Area
Exercise	Ex 11.2
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

Question 1.
Find the area of the following parallelograms:


Solution:

Question 2.
Find the area of each of the following triangles:


Solution:

Question 3.
Find the missing values:

Solution:

Question 4.
Find the missing values:

Solution:

Question 5.
PQRS is a parallelogram (in 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.

Solution:
(a) Area of the parallelogram PQRS = base × height = SR × QM = 12 × 7.6 cm² = 91.2 cm²
(b) Area of the parallelogram PQRS = base × height = PS × QN
⇒ 91.2 = 8 × QN
⇒ QN = \(\frac { 912 }{ 8 } \) cm
⇒ QN = 11.4 cm.

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (in Figure). If the area of a parallelogram is 1470 cm2, AB = 35 cm, and AD = 49 cm, find the length of BM and DL.

Solution:

Question 7.
∆ ABC is right-angled at A (in 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.

Solution:

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

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.4
• Practical Geometry Class 7 Ex 10.5

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2

Question 1.
Construct A XYZ in which XY = 4.5 cm, YZ = 5 cm and, ZX = 6 cm.
Solution:
Steps of Construction

1. Draw a line segment YZ of length 5 cm.
   
2. With Y as centre, draw an arc of radius 4.5 cm.
3. With Z as centre, draw an arc of radius 6 cm,
4. Mark the point of intersection of arcs as X.
5. Join XY and XZ. ∆ XYZ is now ready.

Question 2.
Construct an equilateral triangle of side 5.5 cm.
Solution:
Steps of Construction:

1. Draw a line segment BC of length 5.5 cm.
2. With B as centre, draw an arc of radius 5.5 cm.
   
3. With C as centre, draw an arc of radius 5.5 cm.
4. Mark the point of intersection of arcs as A.
5. Join AB and AC. Equilateral ∆ ABC is now ready.

Question 3.
Draw ∆ PQR with PQ = 4 cm, QR =3.5 cm and PR = 4 cm. What type of triangle is this?
Solution:
Steps of Construction:

1. Draw a line segment QR of length 3.5 cm.
2. With Q as centre, draw an arc of radius 4 cm.
3. With R as centre, draw an arc of radius 4 cm.
   
4. Mark the point of intersection of arcs as P.
5. Join PQ and PR.

∆ PQR is now ready,
∵ PQ = PR
∴ ∆ PQR is isosceles.

Question 4.
Construct ∆ ABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.
Solution:
Steps of Construction

1. Draw a line segment BC of length 6 cm.
   
2. With B as centre, draw an arc of radius 2.5 cm.
3. With C as centre, draw an arc of a radius of 6.5 cm.
4. Mark the point of intersection of arcs as A.
5. Join AB and AC.
6. ∆ ABC is now ready. On measurement, ∠B = 90°.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.2 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.4
• Practical Geometry Class 7 Ex 10.5

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.3
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 1.
Construct ADEF such that DE = 5 cm, DF 3 cm, and m ∠EDF = 90°.
Solution:
Steps of Construction:

1. Draw a line segment DE = 5cm.
2. Draw ∠EDX = 90°.
3. With centre D and radius = 3 cm, draw an arc to intersect DX at F.
4. Join EF to obtain the required triangle DBF.

Question 2.
Construct an isosceles triangle in which the length of each of its equal sides is 6.5 cm and the angle between them is 110°.
Solution:
Steps of Construction

1. Draw a line segment QR of length 6.5 cm.
2. At Q, draw QX making 110° with QR, using a protractor.
   
3. With Q as centre, draw an arc of a radius of 6.5 cm. It cuts QX at P.
4. Join PR. ∆ PQR is now obtained.

Question 3.
Construct ∆ ABC with BC = 7.5 cm, AC = 5 cm and m ∠C = 60°.
Solution:
Steps of Construction:

1. Draw a line segment BC = 7.5 cm.
2. Draw ∠BCX = 60°.
3. With C as centre and radius = 5 cm, draw an arc intersecting CX at A.
4. Join AB to obtain the required ∆ABC.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4.

• Practical Geometry Class 7 Ex 10.1
• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.5

Board	CBSE
Textbook	NCERT
Class	Class 7
Subject	Maths
Chapter	Chapter 10
Chapter Name	Practical Geometry
Exercise	Ex 10.4
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

Question 1.
Construct ∆ ABC, given m ∠A = 60°, m ∠B = 30° and AB = 5.8 cm.
Solution:
Steps of Construction

1. Draw AB of length 5.8 cm.
2. At A, draw a ray AP making an angle of 60° with AB.
3. At B, draw a ray BQ making an angle of 30° with BA.
4. Mark the point of intersection of two rays as C.
5. ∆ ABC is now completed.

Question 2.
Construct ∆ PQR if PQ = 5 cm, m ∠PQR = 105° and m ∠QRP = 40°.
(Hint: Recall angle-sum property of a triangle).
Solution:
By angle-sum property of a triangle
m ∠RPQ + m ∠PQR + m ∠QRP = 180°
⇒ m ∠RPQ + 105° + 40° = 180°
⇒ m ∠RPQ + 145° = 180°
⇒ m ∠RPQ = 35°
Steps of Construction

1. Draw PQ of length 5 cm.
2. At Q, draw a ray QX making an angle of 105° with QP.
   
3. At P draw a ray PY making an angle of 35° with PQ.
4. Mark the point of intersection of two rays as R.

∆ PQR is now completed.

Question 3.
Examine whether you can construct ∆DEF such that EF = 7.2 cm, m ∠E = 110° and m ∠F = 80°. Justify your answer.
Solution:
m ∠E + m ∠F = 110° + 80° = 190° > 180°
This is not possible since the sum of the measures of the three angles of a triangle is 180°. As such, the sum of two angles of a triangle cannot exceed 180°.
Hence, ∆ DEF cannot be constructed.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.4, drop a comment below and we will get back to you at the earliest.