MATHEMATICS  PLEDGE

Mathematics is my world. 

All maths problems are my friends and relatives. 

I love my maths and I am proud of it's speed and accuracy.

 I shall always try to solve all the problems

 I shall learn principles, formulas and all sums with happiness and treat every problem with respect. 

I shall mind all the chapters.

To my maths and my teachers I pledge my devotion. In solving the sums and finding the answers alone lies my happiness.

Jai Maths

Math Pledge

I pledge to honor mathematics,

The universal language of logic and reasoning.

I shall explore its wonders with curiosity,

And use its power to solve problems for the benefit of all.

I shall strive to learn and apply its principles,

To contribute to the progress of science, technology, and society.

I will respect the legacy of great mathematicians,

And draw inspiration from their dedication and brilliance.

To mathematics, the foundation of innovation,

I pledge my commitment, creativity, and perseverance.

Mathematics Pledge

I pledge to embrace the beauty of mathematics,
To explore its logic and patterns with curiosity.
I will strive to understand its principles,
And apply them to solve problems in everyday life.
I will respect the contributions of great mathematicians,
And draw inspiration from their dedication and genius.
With persistence and determination,
I will conquer challenges, both in mathematics and beyond.
I vow to promote the value of mathematics,
As a tool for progress, creativity, and discovery.
Today and every day,
I honor the legacy of mathematics in building a better world.

Class – 7 CH-13 EXPONENTS AND POWERS MATHS NCERT SOLUTIONS

 

Class – 7 CH-13 EXPONENTS AND POWERS 

MATHS NCERT SOLUTIONS

 Exercise 13.1 

   

Question 1:   

Find the value of: 

(i) 26 (ii) 93  (iii) 112 (iv) 54 

  SOLUTION 1:  

(i) 26 = 2 x 2 x 2 x 2 x 2 x 2 = 64 

(ii) 93 = 9 x 9 x 9 = 729 

(iii) 112 = 11 x 11 = 121 

(iv) 54= 5 x 5 x 5 x 5 = 625 

 

 

 

Question 2:   

Express the following in exponential form: 

(i) 6 x 6 x 6 x 6 (ii) tt  

(iii) b b b b    (iv) 5 x 5 x 7 x 7 x 7 

(v) 2 2  a a  (vi) a a a c c c c d       

  SOLUTION 2:  

(i) 6 x 6 x 6 x 6 = 64 

(ii) t t  t2  

(iii) b b b b    b4  

(iv) 5 x 5 x 7 x 7 x 7 = 52 x 73 

(v) 2    2 a a 22 a2  

(vi) a a a c c c c d         a3 c4 d 

 

 

 

Question 3:   

Express each of the following numbers using exponential notation: 

(i) 512  (ii) 343 (iii) 729 (iv) 3125 

  SOLUTION 3:  

(i) 512  

 

 

 

   

  

 

 

 

 

 

 

 

 

 

 

 

 

(ii) 

  

 

 

 

 

(iii) 

  

 

 

 

 

 

 

 

 

(iv) 

 

 

 

 

 

 

 

  2 512 

2 256 

2 128 

2 64 

2 32 

2 16 

2 8 

2 4 

2 2 

  1 

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 29 

343  

7 343 

7 49 

7 7 

  1 

343 = 7 x 7 x 7 = 73 

729  

3 729 

3 243 

3 81 

3 27 

3 9 

3 3 

  1 

729 = 3 x 3 x 3 x 3 x 3 x 3 = 36 

3125  

5 3125 

5 625 

5 125 

5 25 

5 5 

  1 

3125 = 5 x 5 x 5 x 5 x 5 

 

   

Question 4:   

Identify the greater number, wherever possible, in each of the following: 

(i) 43 and 34 (ii) 53 or 35 

(iii) 28 or 82 (iv) 1002 or 2100 

(v) 210 or 102   SOLUTION 4:  

(i) 43 = 4 x 4 x 4 = 64 

34 = 3 x 3 x 3 x 3 = 81 

Since 64 < 81 

Thus, 34 is greater than 43. 

 

(ii) 53 = 5 x 5 x 5 = 125 

35 = 3 x 3 x 3 x 3 x 3 = 243 

Since,  125 < 243 

Thus, 34 is greater than 53. 

 

(iii) 28 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256 

  82 = 8 x 8 = 64 

  Since, 256 > 64 

Thus, 28 is greater than 82. 

 

(iv) 1002 = 100 x 100 = 10,000 

  2100 = 2 x 2 x 2 x 2 x 2 x …..14 times x ……… x 2 = 16,384 x ….. x 2 

  Since, 10,000 < 16,384 x ……. x 2    Thus, 2100 is greater than 1002. 

 

(v) 210 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1,024 

  102 = 10 x 10 = 100 

  Since, 1,024 > 100 

  Thus, 210 > 102 

 

 

 

 

Question 5:   

Express each of the following as product of powers of their prime factors: 

(i) 648  (ii) 405 (iii) 540 (iv) 3,600 

  SOLUTION 5:  

(i) 648 = 23 x 34 

2 648 

2 324 

2 162 

3 81 

3 27 

3 9 

3 3 

  1 

(ii) 405 = 5 x 34 

5 405 

3 81 

3 27 

3 9 

3 3 

  1 

(iii) 540 = 22 x 33 x 5 

2 540 

2 270 

3 135 

3 45 

3 15 

5 5 

  1 

 

 

4

   

(iv) 3,600 = 24 x 32 x 52 

 

 

 

 

 

 

 

 

 

  

Question 6:   

Simplify: 

(i) 2 x 103 (ii) 

(iii) 23 x 5  (iv) 

(v) 0 x 102    (vi) (vii) 24 x 32     (viii) 

  SOLUTION 6:  

(i) 2 x 103  = 2 x 10 x 10 x 10   

(ii) 72 x 22  = 7 x 7 x 2 x 2   

(iii) 23 x 5    = 2 x 2 x 2 x 5   

(iv) 3 x 44    = 3 x 4 x 4 x 4 x 4   

(v) 0 x 102  = 0 x 10 x 10     

(vi) 53 x 33  = 5 x 5 x 3 x 3 x 3   

(vii) 24 x 32  = 2 x 2 x 2 x 2 x 3 x 3  

(viii) 32 x 104  = 3 x 3 x 10 x 10 x 10 x 10  

 

Question 7:   

Simplify: 

(i) 43    (ii) 

(iii) 32  52    (iv) 

  SOLUTION 7:  

(i) 43        4  4  4 64  

 

  2 3600 

2 1800 

2 900 

2 450 

3 225 

3 75 

5 25 

5 5 

  1 

72 x 22 3 x 44 

52 x 33 

32 x 104 

= 2,000 

= 196 

= 40 

= 768 

= 0 

= 675 

= 144 

= 90,000 

  3  23  

  23  103 

 

(ii)   3  23          3  2  2  2 24  

(iii) 32  52          3  3  5  5 225  

(iv) 23  103         2  2  2  10  10  10 

 

 

Question 8:   

Compare the following numbers: 

(i) 2.7 x 1012; 1.5 x 108   (ii) 4 x 1014; 3 x 1017   SOLUTION 8:  

(i) 2.7 x 1012 and 1.5 x 108 

On comparing the exponents of base 10, 

 2.7 x 1012 > 1.5 x 108 

 

(ii) 4 x 1014 and 3 x 1017 

  On comparing the exponents of base 10, 

  4 x 1014 < 3 x 1017  

6

   

 Exercise 13.2  

Question 1:   

Using laws of exponents, simplify and write the SOLUTION in exponential form: 

(i) 32 x 34 x 38 (ii) 615  610 

(iii) a3a2  (iv) 7x 72  

(v) (52)2  53 (vi) 25 x 55 

(vii) a b4  4  (viii) (34)3 

(ix) (220  215) x 23 (x) 8 8t  2 

  SOLUTION 1:  

(i) 3 3 3 32  4 8 2 4 8   314   a a am n  m n   

(ii) 615  610 615 10  65     a a am  n  m n  

(iii) a a a3 2 3 2 a5   a a am n  m n  

(iv) 7x  72 7x2     a a am n  m n  

(v) 523  5 53 2 3   5 5 53 6 3     amn am n  

            = 56 3 53     a a am  n  m n  

(vi) 2555 2 5 5 105   a bm m a b m 

(vii) a b4 4 a b 4   a bm m a b m 

(viii) 343 = 34 3  312   amn am n  

(ix) 220 21523 = 220 15 23     a a am  n  m n  

  = 2 2 25 3 5 3 28     a a am n  m n  

(x) 8t  82 8t2     a a am  n  m n  

 

 

 

 

 

Question 2:   

Simplify and express each of the following in exponential form: 

(i) 2 3 433 32  4   (ii) 5235457  



4 53  (iv) 3 7 11 2 8  

(iii) 25 

21 11

(v) 37 0  30 40  

(vi) 2

3 34  3

(vii) 20  30 40  (viii) 30 2050  

28a5 a5  8

(ix) 43a3  (x)  a3 a  

(xi) 4455a ba b5 28 3  (xii)  3 22 2 

  SOLUTION 2:  

2 3 4 2 3 23 4 3 4 2 23 2 34

(i) 3 32  3 2 5  3 2 5     a a am n  m n  

3 2

      = 20 33 = 1 3 3 33  

 

3 n

      =   54 25 5 34 3   a a am  n  m n  2 35

(ii) 52 5457 565457   am am n  

     = 56 4 57  510 57   a a am n  m n  

     = 510 7  53     a a am  n  m n  

 

(iii) 25 54  3 524   5 5 53 8 3     amn am n  

    = 58 3 55   a a am  n  m n  

 

3 7 11 3 7 11

(iv)  21 112 3 8  3 7 11  2 38  31 1 72 1 118 3     a a am  n  m n  

          = 30  71 115 = 7 11 5  

 

 

 

   

 

(v) 

  

 

(vi) 

 

(vii) 

 

(viii) 

 

(ix) 

  

  

 

(x) 

  

 

(xi) 

  

 

(xii) 

  

 

 

 

 

   

37 37 37

   3 3 34  3 4 3 37

= 37 7  30 1   

20      30 40 1 1 1 3   

20      30 40 1 1 1 1   

30 2050      1 1 1 2 1 2   

28a5 28a5 28a5

3 a3  223a3 26a3   4 

= 28 6 a5 2  22 a2     

= 2a2   

 a5  8 a5 3 a8  a2 a8   

 a3 a 

     = a2 8  a10   

45a b8 3 45 5 a8 5 b3 2   40 a3 b  

 45a b5 2

    = 1  a b ab3 3   

2 23 2 23 1 2 242   

   = 24 2 28 

   

 

 

 

 

 

 

 

 

 

 

 

  



























 a a am n  m n   a a am  n  m n  

 a0 1 

 a0 1 

 a0 1 

 amn am n  

 a a am  n  m n   a bm m a b m 

 a a am  n  m n   a a am n  m n  

 a a am  n  m n   a0 1 

 a a am n  m n  

 

Question 3:   

Say true or false and justify your SOLUTION:

(i) 10 x 1011 = 10011 (ii) 23 > 52 

(iii) 23 x 32 = 65 (iv) 30 = (1000)0 

  SOLUTION 3:  

(i) 10 10 11 10011    

L.H.S. 101 11 = 1012    and R.H.S. 10211 1022  

Since, L.H.S.  R.H.S. 

Therefore, it is false. 

 

(ii) 23  52  

  L.H.S. 2 83     and R.H.S. 52  25  

  Since, L.H.S. is not greater than R.H.S. 

  Therefore, it is false. 

 

(iii) 23 32 65  

  L.H.S. 23   32 8 9 72  and R.H.S. 65  7,776  

Since, L.H.S.  R.H.S. 

Therefore, it is false. 

 

(iv) 30 10000  

  L.H.S. 30 1    and R.H.S. 10000 = 1 

Since, L.H.S. = R.H.S. 

Therefore, it is true. 

 

 

Question 4:   

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

(i) 108 x 192    (ii) 270 (iii) 729 x 64    (iv) 768 

  SOLUTION 4:  

(i) 108 x 192  

 

 

   

  

  

 

 

 

 

 

 

 

 

  

  

 

 

 

 

 

 

 

(ii) 

  

 

 

 

 

 

 

 

(iii) 

   

   

108 x 192 

   

   

270 

270  

729 x 64  

729 x 64  2 192 

2 96 

2 48 

2 24 

2 12 

2 6 

3 3 

  1 

= 2233263  

= 22 6 33 1  

= 28 34 

2 108 

2 54 

3 27 

3 9 

3 3 

  1 

 

2 270 

3 135 

3 45 

3 15 

5 5 

  1 

= 2 35 5  

= 36 26 

2 64 

2 32 

2 16 

2 8 

 

2 4 

2 2 

  1 

 

3 729 

3 243 

3 81 

3 27 

3 9 

3 3 

  1 

 

 

 

 

 

 

 

 

 

(iv) 768  

2 768 

2 384 

2 192 

2 96 

2 48 

2 24 

2 12 

2 6 

3 3 

  1 

  768 = 28 3  

 

 

 

 

 

 

 

 

 

 

 

Question 5:   

Simplify: 

252 73

(i) 3   

8 7

25 5

(ii)  3 2t4t8  

10 

3 10 255

(iii)  7 5 5  

5 6

 

 

  SOLUTION 5:  

25273 25 2 73

(i) 837  2337  

2 710  3

=  2 79   

= 210 9 73 1  2 72  

= 2 x 49  

= 98 

 

25 5 2 t8 5 52 2 t8

(ii) 103t4  5 2 3t4  

52 2 t8 4

=  2 33 3   

54 t4

  =  2 53 3  

54 3 t4

=  23   

5t4

=    

8

 

3 10 255 5 3 2 5 55   5 2

(iii) 5 67  5 = 57  2 35   

3 2 5 55  5 5 2

 

= 5 2 37  5 5   

3 2 55 5 5 2

 

  = 5 2 37  5 5  

3 2 55 5 7

 

= 5 2 37  5 5 

  = 25 5 35 5 55 5  

  = 20  30 50  

  = 1 x 1 x 1 

  = 1 

 

7

 Exercise 13.3  

Question 1:   

Write the following numbers in the expanded form: 

279404,  3006194, 2806196, 120719, 20068 

  SOLUTION 1:  

(i) 2,79,404  = 2,00,000 + 70,000 + 9,000 + 400 + 00 + 4 

 = 2 x 100000 + 7 x 10000 + 9 x 1000 + 4 x 100 + 0 x 10 + 4 x 1 

 = 2 10          5 7 104 9 103 4 102 0 101 4 100  (ii) 30,06,194 = 30,00,000 + 0 + 0 + 6,000 + 100 + 90 + 4 

= 3 x 1000000 + 0 x 100000 + 0 x 10000 + 6 x 1000 + 1 x 100 + 9 x 10 + 4 x 1 

  = 3 10            6 0 105 0 104 6 103 1 102 9 10 4 100  

(iii) 28,06,196 = 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6 

= 2 x 1000000 + 8 x 100000 + 0 x 10000 + 6 x 1000 + 1 x 100 + 9 x 10 + 6 x 1 

  = 2 10            6 8 105 0 104 6 103 1 102 9 10 6 100  

(iv) 1,20,719 = 1,00,000 + 20,000 + 0 + 700 + 10 + 9 

 = 1 x 100000 + 2 x 10000 + 0 x 1000 + 7 x 100 + 1 x 10 + 9 x 1 

= 1 10          5 2 104 0 103 7 102 1 101 9 100  

(v) 20,068 = 20,000 + 00 + 00 + 60 + 8 

 = 2 x 10000 + 0 x 1000 + 0 x 100 + 6 x 10 + 8 x 1 

      = 2 10        4 0 103 0 102 6 101 8 100 

 

Question 2:   

Find the number from each of the following expanded forms: 

(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100 

(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100 

(c) 3 x 104 + 7 x 102 + 5 x 100 

(d) 9 x 105 + 2 x 102 + 3 x 101   SOLUTION 2:  

(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100 

= 8 x 10000 + 6 x 1000 + 0 x 100 + 4 x 10 + 5 x 1 

= 80000 + 6000 + 0 + 40 + 5 

= 86,045 

(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100 

= 4 x 100000 + 0 x 10000 + 5 x 1000 + 3 x 100 + 0 x 10 + 2 x 1 

= 400000 + 0 + 5000 + 3000 + 0 + 2 

= 4,05,302 

 

(c) 3 x 104 + 7 x 102 + 5 x 100 

  = 3 x 10000 + 0 x 1000 + 7 x 100 + 0 x 10 + 5 x 1 

  = 30000 + 0 + 700 + 0 + 5 

  = 30,705 

(d) 9 x 105 + 2 x 102 + 3 x 101 

  = 9 x 100000 + 0 x 10000 + 0 x 1000 + 2 x 100 + 3 x 10 + 0 x 1 

  = 900000 + 0 + 0 + 200 + 30 + 0 

  = 9,00,230 

Question 3:   

Express the following numbers in standard form: 

(i) 5,00,00,000 (ii) 70,00,000 

(iii) 3,18,65,00,000 (iv) 3,90,878 

(v) 39087.8 (vi) 3908.78 

  SOLUTION 3:  

(i) 5,00,00,000    = 5 x 1,00,00,000 = 5 10 7  

(ii) 70,00,000 = 7 x 10,00,000 = 7 10 6  

(iii) 3,18,65,00,000  = 31865 x 100000  

= 3.1865 x 10000 x 100000 = 3.1865 10 9  

(iv) 3,90,878 = 3.90878 x 100000 = 3.90878 10 5  

(v) 39087.8 = 3.90878 x 10000 = 3.90878 10 4  

(vi) 3908.78  = 3.90878 x 1000 = 3.90878 10 3 

 

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. 

(b) Speed of light in vacuum is 300,000,000 m/s. 

(c) Diameter of Earth id 1,27,56,000 m. 

(d) Diameter of the Sun is 1,400,000,000 m. 

(e) In a galaxy there are on an average 100,000,000,0000 stars. 

(f) The universe is estimated to be about 12,000,000,000 years old. 

(g) 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. 

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm. 

(i) The Earth has 1,353,000,000 cubic km of sea water. 

(j) The population of India was about 1,027,000,000 in march, 2001. 

  SOLUTION 4:  

(a) The distance between Earth and Moon  = 384,000,000 m 

= 384 x 1000000 m  

= 3.84 x 100 x 1000000  

= 3.84 10 8 m 

 

(b) Speed of light in vacuum = 300,000,000 m/s 

  = 3 x 100000000 m/s 

   

  = 3 10 8 m/s 

(c) Diameter of the Earth = 1,27,56,000 m 

  = 12756 x 1000 m  

= 1.2756 x 10000 x 1000 m 

   

  = 1.2756 10 7 m 

(d) Diameter of the Sun = 1,400,000,000 m 

  = 14 x 100,000,000 m  

= 1.4 x 10 x 100,000,000 m 

   

  = 1.4 10 9 m 

(e) Average of Stars = 100,000,000,000 

  = 1 x 100,000,000,000 

   

  = 1 10 11  

(f) Years of Universe = 12,000,000,000 years 

  = 12 x 1000,000,000 years 

  = 1.2 x 10 x 1000,000,000 years 

  = 1.2 10 10 years 

 

(g) Distance of the Sun from the centre of the Milky Way Galaxy  

= 300,000,000,000,000,000,000 m 

  = 3 x 100,000,000,000,000,000,000 m 

  = 3 10 20 m 

 

(h) Number of molecules in a drop of water weighing 1.8 gm  

= 60,230,000,000,000,000,000,000  

  = 6023 x 10,000,000,000,000,000,000  

  = 6.023 x 1000 x 10,000,000,000,000,000,000  

  = 6.023 10 22  

 

  

   

 

(i) The Earth has Sea water 

   

   

   

 

(j) The population of India 

   

   

   

  = 1,353,000,000 km3 

= 1,353 x 1000000 km3 

 = 1.353 x 1000 x 1000,000 km3  = 1.353 10 9 km3 

= 1,027,000,000 

= 1027 x 1000000 

= 1.027 x 1000 x 1000000 

= 1.027 10 9 

 

D

Class – 7 CH-15 VISUALIZING SOLID SHAPES MATHS NCERT SOLUTIONS

 

Class – 7 CH-15 VISUALIZING SOLID SHAPES 

MATHS NCERT SOLUTIONS

 Exercise 15.1 

      

Question 1:   

Identify the nets which can be used to make cubes (cut out copies of the nets and try it): 

  

  SOLUTION 1:  

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. 

Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box. 

  

Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7. 

  SOLUTION 2:  

  

Question 3:   

Can this be a net for a die? Explain your SOLUTION. 

 

  SOLUTION 3:  

No, this cannot be a net for a die. 

Because one pair of opposite faces will have 1 and 4 on them and another pair of opposite faces will have 3 and 6 on them whose total is not equal to 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 faces are there in the net here? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.) 

  

  SOLUTION 4:  

There three faces are given: 

  

 

 

Question 5:   

Match the nets with appropriate solids: 

 

 SOLUTION 5:  

              Solids                Their nets 

  

 

 Exercise 15.2  

Question 1:   

Use isometric dot paper and make an isometric sketch for each one of the given shapes: 

 

  

 

  

   

Question 2:   

The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid. 

  SOLUTION 2:  

The dimensions of given cuboid are 5 cm, 3 cm and 2 cm: 

  

Three different isometric sketches are: 

   

Question 3:   

Three cubes each with 2 cm edge are placed side by side to form a cuboid. Sketch an oblique or isometric sketch of this cuboid. 

  SOLUTION 3:  

Oblique sketch: 

  

Isometric sketch 

   

Question 4:   

Make an oblique sketch for each one of the given isometric shapes: 

  

  SOLUTION 4:  

Oblique sketches: 

  

Question 5:   

Give (i) an oblique sketch and (ii) an isometric sketch for each of the following: (a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique?) (b) A cube with an edge 4 cm long. 

  SOLUTION 5:  

(a) A cuboid of dimension 5 cm, 3 cm and 2 cm. 

 

(i) Oblique sketch (ii) Isometric sketch 

  

 

(b) A cube with an edge 4 cm long. 

(i) Oblique sketch        (ii) Isometric sketch 

 

 

 

 

Question 6:   

An isometric sheet is attached at the end of the book. You could try to make on it some cubes or cuboids of dimensions specified by your friend.   SOLUTION 6:  

Cubes and cuboids shapes on isometric sheet given below: 

  

You can also draw more shapes of cubes and cuboids.  

 

 Exercise 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 1:  

 

  

1

  

 

 

 

 

2

 

 Exercise 15.4  

Question 1:   

A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then SOLUTION these questions). 

A book 

 

          (i)          (ii)           (iii)   SOLUTION 1:  

S.No. Object Shadow Shape’s name 

(i) A ball   Circle 

(ii) A cylindrical pipe   Line 

(iii) A book   Rectangle 

 

 

 

 

Question 2:   

Here are the shadows of some 3-D objects, when seen under the lamp of the overhead projector. Identify the solid (s) that match each shadow. (There may be multiple SOLUTIONs for these!) 

 

  SOLUTION 2:  

S. No. Shadow Shape’s Name 3-D objects 

(i)   Circle Chapatti, Football, Disc, Plate etc. 

(ii)   Square Die, Square paper sheet, cubical magic box, Chalk box etc. 

(iii)   Triangle Ice-cream cone, Birthday cap, etc. 

(iv)   Rectangle Geometry box, Book, Table etc. 

 

Question 3:   

Examine if the following are true statements: 

(i) 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 3:  

(i) True  

(ii) False 

 

 

 

Class – 7 CH-14 SYMMETRY MATHS NCERT SOLUTIONS

 

Class – 7 CH-14 SYMMETRY 

MATHS NCERT SOLUTIONS

 Exercise 14.1 

       

Question 1:   

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

 

 

S.No. Punched holed figures The axes of symmetry 

(a)   (rectangle) 

(b)     

 

(c)     

(d)     

(e)     

(f)     

(g)     

(h)     

(i)     

(j)     

(k)     

(l)     

 

 

Question 2:   

Express the following in exponential form: 

 

 

 

  SOLUTION 2:  

S.No. Line(s) of symmetry Other holes on figures 

(a)     

(b)     

(c)     

(d)     

(e)     

 

 

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? 

 

  

   (a) (b) (c) (d)     (e)       (f) 

  SOLUTION 3:  

S.No. Question figures Complete figures Names of the figure 

(a)     Square 

(b)     Triangle 

(c)     Rhombus 

(d)     

Circle 

 

(e)     Pentagon 

(f)     Octagon 

 

  

Question 4:   

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

 

 

 

 

  SOLUTION 4:  

S.No. Problem Figures Lines of symmetry 

(a)     

(b)     

(c)     

(d)     

(e)     

(f)     

(g)     

(h)     

 

 

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 5:  

SOLUTION figures are: 

  

Yes, there is more than one way. 

Yes, this figure will be symmetric about both the diagonals. 

 

 

 

Question 6:   

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

  

  (a)    (b)       (c)  (d) 

  SOLUTION 6:  

   

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 7:  

S.No. Figure’s name Diagram with symmetry Number of lines 

(a) Equilateral triangle   3 

(b) Isosceles triangle   1 

(c) Scalene triangle   0 

(d) Square   4 

(e) Rectangle   2 

(f) Rhombus   2 

(g) Parallelogram   0 

(h) Quadrilateral   0 

(i) Regular Hexagon   6 

(j) Circle   Infinite 

 

 

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 8:  

(a) Vertical mirror – A, H, I, M, O, T, U, V, W, X and Y 

            mirror                mirror 

                             

(b) Horizontal mirror – B, C, D, E, H, I, O and X 

   

 

(c) Both horizontal and vertical mirror – H, I, O and X  

Question 9:   

Give three examples of shapes with no line of symmetry. 

  SOLUTION 9:  

The three examples are: 

 Quadrilateral 

 Scalene triangle 

 Parallelogram 

 

Question 10:   

What other name can you give to the line of symmetry of: (a) an isosceles triangle? (b) a circle?   SOLUTION 10:  

(a) The line of symmetry of an isosceles triangle is median or altitude. 

(b) The line of symmetry of a circle is diameter. 

 

 Exercise 14.2  

Question 1:   

Which of the following figures have rotational symmetry of order more than 1: 

 

  

(d) (e) (f) 

Rotational symmetry of order more than 1 are a,b,d,e and  f  because in these figures, a complete turn, more than 1 number of times, an object looks exactly the same. 

 

 

Question 2:   

Give the order the rotational symmetry for each figure: 

    

       (f)       (g)            (h) 

S.No. Problem figures Rotational figures Order of rotational symmetry 

(a)     2 

(b)     2 

(c)     3 

(d)     4 

(e)     4 

(f)    

  5 

(g)     6 

(h)     3 

 

 

 

 Exercise 14.3  

Question 1:   

Name any two figures that have both line symmetry and rotational symmetry. 

  SOLUTION 1:  Circle and Square. 

 

 

Question 2:   

Draw, wherever possible, a rough sketch of: 

(i) a triangle with both line and rotational symmetries 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 a 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 2:  

(i) An equilateral triangle has both line and rotational symmetries of order more than 1. 

 

 

 

Line symmetry:  

 

 

 

 

Rotational symmetry: 

   

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

 

 

  Line symmetry:  

 

 

 

 

 

   

 

 

Rotational symmetry: 

 

 

 

 

(iii) It is not possible because order of rotational symmetry is more than 1 of a figure, most acertain the line of symmetry. 

(iv) A trapezium which has equal non-parallel sides, a quadrilateral with line symmetry but not a rotational symmetry of order more than 1. 

 

 

Line symmetry: 

 

 

 

   

 

 

 

Rotational symmetry: 

 

 

 

 

 

Question 3:   

In a figure has two or more lines of symmetry, should it have rotational symmetry of order more than 1?   SOLUTION 3:  

Yes, because every line through the centre forms a line of symmetry and it has rotational symmetry around the centre for every angle. 

 

 

Question 4:   

Fill in the blanks: 

Shape Centre of Rotation Order of Rotation Angle of Rotation 

Square  

Rectangle  

Rhombus  

Equilateral triangle  

Regular hexagon  

Circle  

Semi-circle  

  SOLUTION 4:  

Shape Centre of Rotation Order of Rotation Angle of Rotation 

Square Intersecting point of diagonals. 4 90  

Rectangle Intersecting point of diagonals. 2 180   

Rhombus Intersecting point of diagonals. 2 180   

Equilateral triangle Intersecting point of medians. 3 120   

Regular hexagon Intersecting point of diagonals. 6 60   

Circle Centre infinite At every point 

Semi-circle Mid-point of diameter 1 360   

Question 5:   

Name the quadrilateral which has both line and rotational symmetry of order more than 

1. 

  SOLUTION 5:  

Square has both line and rotational symmetry of order more than 1. 

 

 

Line symmetry: 

 

 

Rotational symmetry: 

  

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 6:  

Other angles will be 120 ,180 ,240 ,300 ,360    . 

 

 

For 60 rotation:  

It will rotate six times. 

 

  

  

 

For 120 rotation:  

It will rotate three times. 

  

For 180 rotation:  

It will rotate two times. 

  

 

For 360 rotation:  

It will rotate one time. 

  

Question 7:   

Can we have a rotational symmetry of order more than 1 whose angle of rotation is: 

 

(i) 45  (ii) 17 ? 

  SOLUTION 7:  

(i) If the angle of rotation is 45, then symmetry of order is possible and would be 8 rotations. 

(ii) If the angle of rotational is 17 , then symmetry of order is not possible because 360 is not complete divided by 17 .