CBSE Class 7

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.1.

• Symmetry Class 7 Ex 14.2
• Symmetry Class 7 Ex 14.3
• Symmetry Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.1

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


Solution:

Question 2.
Given the line(s) of symmetry, find the other hole(s):

Solution:

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?
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Solution:

Question 4.
The following figures have more than one line of symmetry. Such figures are said to have multiple lines of symmetry:

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

Solution:

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?

Solution:


Yes! there is more than one way to make the figure symmetric.

• Let us take the diagonal BD and shade the squares as shown in the figure to make the figure symmetric about BD.
• Similarly, the figure is symmetric about the diagonal AC. Thus, the figure is symmetric about both the diagonals.
• The figure is symmetric about EF and GH also.

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

Solution:

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:

Question 8.
What letters of the English alphabet have reflectional symmetry (i.e., symmetry related to mirror reflection) about
(a) a vertical mirror
(b) a horizontal mirror
(c) both horizontal and vertical mirrors.
Solution:

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

(b) A scalene triangle,
(c) A parallelogram

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

1. an isosceles triangle?
2. a circle?

Solution:

1. Median
2. Diameter

We hope the NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.1 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 14 Symmetry Ex 14.1, 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 13.1 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 13.1.

• Exponents and Powers Class 7 Ex 13.2
• Exponents and Powers Class 7 Ex 13.3
• Exponents and Powers Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1

Question 1.
Find the value of:

1. 2⁶
2. 9³
3. 11²
4. 5⁴.

Solution:

1. 2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64
2. 9³ = 9 × 9 × 9 = 729
3. 11² = 11 × 11 = 121
4. 5⁴ = 5 × 5 × 5 × 5 = 625.

Question 2.
Express the following in exponential form:

1. 6 × 6 × 6 × 6
2. t × t
3. b × b × b × b
4. 5 × 5 × 7 × 7 × 7
5. 2 × 2 × a × a
6. a × a × a × c × c × c × c × d.

Solution:

1. 6 × 6 × 6 × 6 = 6⁴
2. t × t = t²
3. b × b × b × b = b⁴
4. 5 × 5 × 7 × 7 × 7 = 5² × 7³
5. 2 × 2 × a × a = 2² × a²
6. 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
(ii) 343
(iii) 729
(iv) 3125.
Solution:




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

Solution:

Question 5.
Express each of the following as a product of powers of their prime factors:
(i) 648
(ii) 405
(iii) 540
(iv) 3600
Solution:

Question 6.
Simplify:

Solution:

Question 7.
Simplify:

Solution:

Question 8.
Compare the following numbers:

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.1, 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 12.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1.

• Algebraic Expressions Class 7 Ex 12.2
• Algebraic Expressions Class 7 Ex 12.3
• Algebraic Expressions Class 7 Ex 12.4
• Algebraic Expressions Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants, and arithmetic operations.
(i) Subtraction of z from y.
Solution:
y – z

(ii) One-half of the sum of numbers x and y.
Solution:
\(\frac{1}{2}\)  (x -y)

(iii) The number z multiplied by itself.
Solution:
z × z i.e., z²

(iv) One-fourth of the product of numbers p and q.
Solution:
\(\frac{1}{4}\) pq

(v) Numbers x and y both squared and added.
Solution:
x² + y²

(vi) Number 5 added to three times the product of numbers m and n.
Solution:
3mn + 5

(vii) Product of numbers y and z subtracted from 10.
Solution:
10 – yz

(viii) Sum of numbers a and 6 subtracted from their product.
Solution:
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
(b) 1 + x + x²
(c) y – y³
(d) 5xy² + 7x²y
(e) -ab + 2b² -3a²

(ii) Identify terms and factors in the expressions given below :

(a) – 4x + 5
(b) – 4x + 5y
(c) 5y + 3y²
(d) xy + 2x²y²
(e) pq + q
(f) 1.2 ab -2.4 b + 3.6 a
(g) \(\frac { 3 }{ 4 } \) x + \(\frac { 1 }{ 4 } \)
(h) 0.1 p² + 0.2 q²

Solution:

Question 3.
Identify the numerical coefficients of terms (other than constants) in the following expressions :

1. 5 – 3t²
2. 1 + t + t² + t³
3. x + 2xy + 3y
4. 100m + 1000n
5. -p²q² + 7pq
6. 1.2 a + 0.8 b
7. 3.14 r²
8. 2 (l + b)
9. 0.1 y + 0.01 y².

Solution:

Question 4.
(a) Identify terms which contain x and give the coefficient of x.

1. y²x + y
2. 13y² – 8yx
3. x + y + 2
4. 5 + z + zx
5. 1 + x + xy
6. 12xy² + 25
7. 7x + xy².

(b) Identify terms which contain y² and give the coefficient of y².

1. 8 – xy²
2. 5y² + 7x
3. 2x²y – 15xy² + 7y²

Solution:

Question 5.
Classify into monomials, binomials and trinomials.

1. 4y – 7z
2. y²
3. x + y – xy
4. 100
5. ab – a – b
6. 5 – 3t
7. 4p²q – 4pq²
8. 7mn
9. z² – 3z + 8 a² + b²
10. z² + z
11. 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 is of like or unlike terms :
(i) 1, 100
Solution:
Like

(ii) -7x, \(\frac{5}{2}\)x
Solution:
Like

(iii) – 29x, – 29y
Solution:
Unlike

(iv)14xy, 42yx
Solution:
Like

(v) 4m²p, 4mp²
Answer:
Unlike

(vi) 12xz, 12x²z²
Solution:
Unlike

(i) 4y – 7z.
This expression is a binomial because it contains two terms: 4y and – Iz.
(ii) y².
This expression is a monomial because it contains only one term: y²
(iii) x + y – xy.
This expression is a trinomial because it contains three terms: x, y, and – xy.
(iv) 100.
This expression is a monomial because it contains only one term: 100
(v) ab – a – b.
This expression is a trinomial because it contains three terms: ab, -a, and -b
(vi) 5 – 3t.
This expression is a binomial because it contains two terms : 5 and – 31.
(vii) 4p²q – 4pq².
This expression is a binomial because it contains two terms: 4p²q and – 4pq².
(viii) 7mn.
This expression is a monomial because it contains only one term : 7mn.
(ix) z² – 3z + 8.
This expression is a trinomial because it contains three terms : z², – 3z and 8.
(x) a² + b².
This expression is a binomial because it contains two terms: a² and b².
(xi) z² + z.
This expression is a binomial because it contains two terms : z² and z.
(xii) 1 + x + x².
This expression is a trinomial because it contains three terms: 1, x, and x².

Question 6.
State whether a given pair of terms is of like or unlike terms :
(i) 1, 100
Solution:
Like

(ii) -7x, \(\frac{5}{2}\)x
Solution:
Like

(iii) – 29x, – 29y
Solution:
Unlike

(iv)14xy, 42yx
Solution:
Like

(v) 4m²p, 4mp²
Answer:
Unlike

(vi) 12xz, 12x²z²
Solution:
Unlike

Question 7.
Identify like terms in the following :

Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1, 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 11.1 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 11.1.

• Perimeter and Area Class 7 Ex 11.2
• Perimeter and Area Class 7 Ex 11.3
• Perimeter and Area Class 7 Ex 11.4
• Perimeter and Area Class 7 MCQ

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1

Question 1.
The length and the breadth of a rectangular piece of land is 500 m and 300 m respectively. Find
(i) its area
(ii) the cost of the land, if 1 m² of the land cost ₹ 10,000.
Solution:

Here, length = 500 m, breadth = 300 m
(i) Area = length × breadth = (500 × 300) m² = 1,50,000 m²
(ii) Cost of land at the rate of ₹ 10,000 per 1 m² = ₹ (10,000 × 1,50,000) = ₹ 1,50,00,00,000.

Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:

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:

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:

Question 5.
The area of a square park is the same as 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:


Hence, the breadth of the rectangular park is 40 m.

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:

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 (figure). Find the cost of whitewashing the wall, if the rate of whitewashing the wall is ₹ 20 per m².
Solution:

We hope the NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.1, 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 10.1 are part of NCERT Solutions for Class 7 Maths. Here we have given NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1.

• Practical Geometry Class 7 Ex 10.2
• Practical Geometry Class 7 Ex 10.3
• Practical Geometry Class 7 Ex 10.4
• Practical Geometry Class 7 Ex 10.5

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1

Question 1.
Draw a line, say, AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
Solution:
Steps of Construction

1. Draw a line AB.
2. Take a point C outside it.
3. Take any point D on AB.
4. Join C to D.
   
5. with D as centre and a convenient radius, draw an arc cutting AB at F and CD at E.
6. Now with C as centre and the same radius as in step 5, draw an arc GH cutting CD at I.
7. Place the pointed tip of the compasses at F and adjust the opening so that the pencil tip is at E.
8. With the same opening as in step 7 and with I as centre, draw an arc cutting the arc GH at J.
9. Now join CJ to draw a line ‘KL’. Then KL is the required line.

Question 2.
Draw a line l. Draw a perpendicular to l at any point on l. On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to l.
Solution:
Steps of Construction

1. Draw a line l.
2. Take any point A on line l.
   
3. Construct an angle of 90° at point A of line l and draw a line AL perpendicular to line l.
4. Mark a point X on AL such that AX = 4 cm.
5. At X construct an angle of 90° and draw a line XC perpendicular to line AL.
6. Then line XC (line m) is the required line through X such that m || l.

Question 3.
Let l be a line and P be a point not on l. Through P, draw a line m parallel to l. Now join P to any point Q on l. Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet l at S. What shape do the two sets of parallel lines enclose?
Solution:
Steps of Construction

1. Draw a line l and take a point P not on it.
2. Take any point Q on l.
3. Join Q to P.
   
4. Draw a line m parallel to line l, as shown in the figure. Then line m || line l.
5. Join P to any point Q on l.
6. Choose any point R on m.
7. Join R to Q.
8. Through R, draw a line n parallel to the line PQ.
9. Let the line n meet the line l at S.
10. Then, the shape enclosed by the two sets of parallel lines is a parallelogram.

We hope the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.1 helps you. If you have any query regarding NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry EX 10.1, drop a comment below and we will get back to you at the earliest.