# RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-4) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

## Class 8: RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-4) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8

The document RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-4) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 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

#### Answer 10: First, we have to find the area of the square lawn, which the total cost divided by the cost of levelling and turfing per square metre:

The length of one side of the square is equal to the square root of the area. We will use the long division method to find it as shown below:

Length of one side of the square = 73 m
The circumference of the square is 73 ×× 4 = 292 m
Total cost of fencing the lawn at Rs. 5 per metre = 292 ×× 5 = Rs. 1460

#### 23000 is 4 (704 − 700) less than 482. Hence, 4 must be added to 2300 to get a perfect square.

The document RD Sharma Solutions for Class 8 Math Chapter 3 - Squares and Square Roots (Part-4) Notes | Study RD Sharma Solutions for Class 8 Mathematics - Class 8 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
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## RD Sharma Solutions for Class 8 Mathematics

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