Last updated on May 26th, 2025
If a number is multiplied by the same number, the result is a square. The inverse of the square is a square root. The square root is used in fields such as vehicle design, finance, and more. Here, we will discuss the square root of 1.3.
The square root is the inverse of the square of a number. 1.3 is not a perfect square. The square root of 1.3 can be expressed in both radical and exponential forms. In radical form, it is expressed as √1.3, whereas it is expressed as (1.3)^(1/2) in exponential form. √1.3 ≈ 1.140175, which is an irrational number because it cannot be expressed in the form of p/q, where p and q are integers and q ≠ 0.
The prime factorization method is used for perfect square numbers. However, this method is not applicable for non-perfect squares like 1.3, where methods such as long division and approximation are used. Let us learn the following methods:
The long division method is particularly used for non-perfect square numbers. Let us now learn how to find the square root using the long division method, step by step.
Step 1: Begin by grouping the numbers from right to left. For 1.3, consider it as 1.30.
Step 2: Find n whose square is less than or equal to 1. We can use n = 1 since 1 × 1 = 1. Subtract 1 from 1 to get a remainder of 0, and bring down 30 to make the new dividend 30.
Step 3: Double the divisor (1), which becomes 2, and use it as the new divisor.
Step 4: Find a digit 'd' such that 2d × d is less than or equal to 30. We find d = 1 because 21 × 1 = 21 is less than 30.
Step 5: Subtract 21 from 30 to get a remainder of 9 and bring down 00 to make the new dividend 900.
Step 6: The new divisor becomes 22 (by adding 1 to 21). Find a digit 'd' such that 22d × d is less than or equal to 900. We find d = 4 because 224 × 4 = 896.
Step 7: Continue this process to get the square root to the desired precision.
The square root of 1.3 is approximately 1.14.
The approximation method is an easy way to find the square root of a given number. Now let us learn how to find the square root of 1.3 using the approximation method.
Step 1: Identify the closest perfect squares to 1.3. The closest perfect squares are 1 (1^2) and 1.44 (1.2^2). Therefore, √1.3 lies between 1 and 1.2.
Step 2: Apply the approximation formula: (Given number - smaller perfect square) ÷ (Larger perfect square - smaller perfect square). Using the formula: (1.3 - 1) ÷ (1.44 - 1) ≈ 0.6818.
Step 3: Add this decimal to the smaller square root: 1 + 0.6818 ≈ 1.14.
Therefore, the square root of 1.3 is approximately 1.14.
Students often make mistakes while finding the square root, such as ignoring negative square roots, skipping steps in the long division method, etc. Let's discuss a few common mistakes in detail.
Can you help Max find the area of a square box if its side length is given as √1.3?
The area of the square is approximately 1.3 square units.
The area of a square = side².
The side length is given as √1.3.
Area of the square = (√1.3)² = 1.3.
Therefore, the area of the square box is approximately 1.3 square units.
A square-shaped plot measuring 1.3 square meters is built; if each of the sides is √1.3, what will be the area of half of the plot?
0.65 square meters
We can just divide the given area by 2 since the plot is square-shaped.
Dividing 1.3 by 2 = 0.65
Hence, half of the plot measures 0.65 square meters.
Calculate √1.3 × 5.
Approximately 5.701
The first step is to find the square root of 1.3, which is approximately 1.14.
The second step is to multiply 1.14 by 5. So, 1.14 × 5 ≈ 5.701.
What will be the square root of (1 + 0.3)?
Approximately 1.14
To find the square root, we calculate (1 + 0.3) = 1.3, and then √1.3 ≈ 1.14.
Therefore, the square root of (1 + 0.3) is approximately ±1.14.
Find the perimeter of a rectangle if its length 'l' is √1.3 units and the width 'w' is 3 units.
The perimeter of the rectangle is approximately 8.28 units.
Perimeter of the rectangle = 2 × (length + width).
Perimeter = 2 × (√1.3 + 3) ≈ 2 × (1.14 + 3) ≈ 2 × 4.14 = 8.28 units.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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