Class 8 Exam  >  Class 8 Notes  >  Mathematics (Maths) Class 8  >  RD Sharma Solutions: Exercise 3.9 - Squares And Square Roots

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8 PDF Download

Q.1. Using square root table, find the square root 7

Ans: From the table, we directly find that the square root of 7 is 2.646.


Q.2. Using square root table, find the square root 15

Ans: Using the table to find √3 and √5

√15 = √3 x √5      

= 1.732 × 2.236       

= 3.873


Q.3. Using square root table, find the square root 74

Ans: Using the table to find  √2 and √37

√74 = √2 x √37

= 1.414 x 6.083

= 8.602


Q.4. Using square root table, find the square root 82

Ans: Using the table to find √2– and √41

√82 = √2 × √41

= 1.414 × 6.403       

= 9.055


Q.5. Using square root table, find the square root 198

Ans: Using the table to find √2 and √11

√198 = √2 × √9 × 11       

= 1.414 × 3 × 3.317        

= 14.070


Q.6. Using square root table, find the square root 540

Ans: Using the table to find √3 and √5

√540 = √54 × √10         

=2 × 3√3 × √5

=2 × 3 × 1.732 × 2.2361         

=23.24


Q.7. Using square root table, find the square root 8700 

Ans: Using the table to find √3 and √29

√8700 = √3 × √29 × √100

= 1.7321 × 5.385 × 10          

= 93.27


Q.8. Using square root table, find the square root 3509

Ans: Using the table to find √29

√3509 = √121 × √29        

= 11 × 5.3851          

= 59.235


Q.9. Using square root table, find the square root 6929

Ans: Using the table to find √41

√6929 = √169 × √41          

= 13 × 6.4031          

= 83.239


Q.10. Using square root table, find the square root 25725

Ans: Using the table to find √3 and √7

√25725 = √3×5×5×7×7×7           

= √3 × 5 × 7 × 7

= 1.732 ×5×7×2.646           

= 160.41


Q.11. Using square root table, find the square root 1312

Ans: Using the table to find √2 and √41

√1312 = 2 × 2 × 2 × 2 × 2 × 41      

= 2 × 2 2 × 41          

= 2 × 2 × 1.414 × 6.4031          

= 36.222


Q.12. Using square root table, find the square root 4192

Ans: 

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

The square root of 131 is not listed in the table. Hence, we have to apply long division to find it.

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Substituting the values:

= 2×2×11.4455          (using the table to find √2)

= 64.75


Q.13. Using square root table, find the square root 4955

Ans: On prime factorisation:

4955 is equal to 5 × 991, which means thatExercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

The square root of 991 is not listed in the table; it lists the square roots of all the numbers below 100.

Hence, we have to manipulate the number such that we get the square root of a number less than 100. This can be done in the following manner:

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Now, we have to find the square root of 49.55.

We have: Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Their difference is 0.071.

Thus, for the difference of 1 (50 − 49), the difference in the values of the square roots is 0.071.

For the difference of 0.55, the difference in the values of the square roots is:

0.55 × 0.0701 = 0.03905

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Finally, we have:

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8


Q.14. Using square root table, find the square rootExercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Ans:                     

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

=Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

=Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8(using the square root table to find √11)

= 0.829


Q.15. Using square root table, find the square rootExercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Ans: 

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

= Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

= 0.581


Q.16. Using square root table, find the square rootExercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Ans: 

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

The square root of 101 is not listed in the table. This is because the table lists the square roots of all the numbers below 100.

Hence, we have to manipulate the number such that we get the square root of a number less than 100. This can be done in the following manner:

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Now, we have to find the square root of 1.01.

We have:

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Their difference is 0.414.

Thus, for the difference of 1 (2 − 1), the difference in the values of the square roots is 0.414.

For the difference of 0.01, the difference in the values of the square roots is:

0.01 × 0.414 = 0.00414

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Finally,Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8= 0.772 

This value is really close to the one from the key answer.


Q.17. Using square root table, find the square root 13.21

Ans: From the square root table, we have:

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Their difference is 0.136.

Thus, for the difference of 1 (14 − 13), the difference in the values of the square roots is 0.136.

For the difference of 0.21, the difference in the values of their square roots is:

0.136 × 0.21=0.02856

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8


Q.18. Using square root table, find the square root 21.97

Ans: We have to find √21.97

From the square root table, we have:

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Their difference is 0.107.

Thus, for the difference of 1 (22 − 21), the difference in the values of the square roots is 0.107.

For the difference of 0.97, the difference in the values of their square roots is:

0.107×0.97=0.104

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8


Q.19. Using square root table, find the square root 110

Ans: 

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

= 1.414 × 2.236 × 3.317        (Using the square root table to find all the square roots)       =10.488


Q.20. Using square root table, find the square root 1110

Ans: 

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

= 1.414 × 1.732 × 2.236 × 6.083       (Using the table to find all the square roots )         

= 33.312


Q.21. Using square root table, find the square root 11.11

Ans: We have:

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

Their difference is 0.1474.

Thus, for the difference of 1 (12 − 11), the difference in the values of the square roots is 0.1474.

For the difference of 0.11, the difference in the values of the square roots is:

0.11 × 0.1474 = 0.0162

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8


Q.22. The area of a square field is 325 m2. Find the approximate length of one side of the field.

Ans: The length of one side of the square field will be the square root of 325.

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

= 5 ×√13

= 5 × 3.605         

= 18.030

Hence, the length of one side of the field is 18.030 m.


Q.23. Find the length of a side of a square, whose area is equal to the area of a rectangle with sides 240 m and 70 m.

Ans: 

The area of the rectangle = 240 m × 70 m = 16800 m2

Given that the area of the square is equal to the area of the rectangle.

Hence, the area of the square will also be 16800 m2.

The length of one side of a square is the square root of its area.

Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

=Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

=Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8

= 129.60 m

Hence, the length of one side of the square is 129.60 m

The document Exercise 3.9 - Squares And Square Roots RD Sharma Solutions | Mathematics (Maths) Class 8 is a part of the Class 8 Course Mathematics (Maths) Class 8.
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