We will learn how to solve real-life problems using approximate values of square roots.
How to Use Approximate Values
Depending on the problem, use approximate values such as √2 ≈ 1.414, √3 ≈ 1.732, and √5 ≈ 2.236.
📘 Example ①
Find the length of the diagonal of a square with side length 1 m (to the second decimal place).
Solution: Diagonal = √(1² + 1²) = √2 ≈ 1.41 m
Find the length of the diagonal of a square with side length 1 m (to the second decimal place).
Solution: Diagonal = √(1² + 1²) = √2 ≈ 1.41 m
📘 Example ②
Find the approximate value of 3√2 to the second decimal place (√2 ≈ 1.414).
Solution: 3 × 1.414 = 4.242 ≈ 4.24
Find the approximate value of 3√2 to the second decimal place (√2 ≈ 1.414).
Solution: 3 × 1.414 = 4.242 ≈ 4.24
💡 Key Points
- Simplify the expression before substituting an approximate value
- Carefully calculate the product of the coefficient and the approximate value
- Check the number of decimal places specified in the problem and round the answer
Practice Problems
- Find the approximate value of 2√3 to the second decimal place (√3 ≈ 1.732).
- Find the length of the diagonal of a square with side length 5 m (to the second decimal place, √2 ≈ 1.414).
Answers and Explanations
- Answer: 2 × 1.732 = 3.464 ≈ 3.46
- Answer: Diagonal = 5√2 ≈ 5 × 1.414 = 7.07 m