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# RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-5) Class 8 Notes | EduRev

## RD Sharma Solutions for Class 8 Mathematics

Created by: Abhishek Kapoor

## Class 8 : RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-5) Class 8 Notes | EduRev

The document RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-5) Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
All you need of Class 8 at this link: Class 8

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

(xiii)

(xiv)

(xv)

#### Answer 1: (i) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

=3×7=21

=31

=
21/31
(ii)We know:

Now, let us compute the square roots of the numerator and the denominator separately.

2×3×3=18

(iii) By looking at the book's answer key, the fraction should be  We know:

(iv) We know:

(v) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

=2×7=14

(vi) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(vii) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(viii) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(ix) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(x) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(xi) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(xii) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(xiii) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(xiv) We know:

Now, let us compute the square roots of the numerator and the denominator separately.

(xv) We know:

Now let us compute the square roots of the numerator and the denominator separately.

7×7=49

3×5=15

∴∴

#### (iv) We have:

=

#### (v) We have:

= 30

#### Answer 3: The length of one side is the square root of the area of the field. Hence, we need to calculate the value of

We have

Now, to calculate the square root of the numerator and the denominator:

We know that:

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