The document RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-7) Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.

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**We can round it off to four decimal places, i.e. 3.4651. **

3.31662

** and ****evaluate each of the following:**

(i)

Given: âˆš3 = 1.732

(ii) 28.867

**(i) **

**(ii) **

**(iii) **

**(iv) **

**(v) **

7

15

=1.732Ã—2.236=3.873

**=1.414Ã—6.083 **

**=8.602 **

** **

** =1.414Ã—6.403 **

**=9.055 **

**=1.414Ã—3Ã—3.317 **

**= 14.070 **

**=2Ã—3Ã—1.732Ã—2.2361 **

**=23.24 **

**=1.7321Ã—5.385Ã—10 **

**=93.27 **

=11Ã—5.3851

=59.235

=13 Ã—6.4031

=83.239

**=1.732 Ã—5Ã—7Ã—2.646 **

**=160.41 **

**=2Ã—2Ã—1.414Ã—6.4031 **

**=36.222 **

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

**Substituting the values: **

**= 2Ã—2Ã—11.4455 (using the table to find âˆš**2)

= 64.75

4955 is equal to 5 Ã— 991, which means that

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:

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

We have:

Their difference is 0.071.

Thus, for the difference of 1 (50 âˆ’ 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 Ã— 0.0701 = 0.03905

7+0.03905=7.03905

Finally, we have:

Ã—10=7.03905Ã—10=70.3905

** **(using the square root table to find âˆš11)

** =0.829 **

(using the square root table to find

0.581

**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: **

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

**We have: **

**Their difference is 0.414.Thus, for the difference of 1 (2 âˆ’ 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 Ã— 0.414 = 0.00414 **

âˆ´ 1+0.00414=1.00414

Ã—10=1.00414Ã—10=10.0414This value is really close to the one from the key answer.

13.21

**Their difference is 0.136.Thus, for the difference of 1 (14 âˆ’ 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:0.136Ã—0.21=0.02856**

From the square root table, we have:

Their difference is 0.107.

Thus, for the difference of 1 (22 âˆ’ 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:

0.107Ã—0.97=0.104

4.583+0.104â‰ˆ4.687

=1.414Ã—2.236Ã—3.317 (Using the square root table to find all the square roots)

=10.488

=1.414Ã—1.732Ã—2.236Ã—6.083 (Using the table to find all the square roots )=33.312

**Their difference is 0.1474.Thus, for the difference of 1 (12 âˆ’ 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 Ã— 0.1474 = 0.0162 **

3.3166+0.0162=3.328â‰ˆ3.333

âˆ´

=5Ã—3.605

=18.030

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

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 m

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

**âˆ´ **

**=2Ã—2Ã—5 **

=129.60 m

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