Last updated on May 26th, 2025
If a number is multiplied by itself, the result is a square. The inverse of squaring a number is finding its square root. The square root is used in fields such as engineering, finance, and physics. Here, we will discuss the square root of 1.2.
The square root is the inverse of squaring a number. 1.2 is not a perfect square. The square root of 1.2 is expressed in both radical and exponential form. In the radical form, it is expressed as √1.2, whereas 1.2^(1/2) in the exponential form. √1.2 ≈ 1.095445, 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, for non-perfect square numbers, we use methods like the long division method and approximation method. Let us now learn the following methods:
The long division method is particularly used for non-perfect square numbers. In this method, we should check the closest perfect square number for the given number. Let us now learn how to find the square root using the long division method, step by step:
Step 1: Group the numbers from right to left. In the case of 1.2, consider it as 1.20.
Step 2: Find a number n whose square is less than or equal to 1. Here, n = 1 because 1 × 1 = 1. Subtracting, we get a remainder of 0.
Step 3: Bring down 20, making it the new dividend. Double the previous quotient (1) to get the new divisor, which is 2.
Step 4: Find n such that 2n × n ≤ 20. Here, n = 4, because 24 × 4 = 96.
Step 5: Subtract 96 from 120 to get 24. Bring down two zeros to make it 2400.
Step 6: Double the entire quotient (10.4) to get 208. Find the new n such that 208n × n ≤ 2400.
Step 7: Continue the steps until you achieve the desired precision. So, √1.2 ≈ 1.095445.
The approximation method is another method for finding square roots, and it is an easy method to find the square root of a given number. Let's learn how to find the square root of 1.2 using the approximation method.
Step 1: Identify the closest perfect squares around 1.2. These are 1 (1^2) and 1.44 (1.2^2). Thus, √1.2 is between 1 and 1.2.
Step 2: Apply the approximation formula: (Given number - smaller perfect square) / (larger perfect square - smaller perfect square). For 1.2, (1.2 - 1) / (1.44 - 1) = 0.2 / 0.44 ≈ 0.454545.
Step 3: Add this to the smaller perfect square's root: 1 + 0.454545 = 1.454545. Thus, √1.2 ≈ 1.095445, which is more accurate than the approximation.
Students often make mistakes when finding square roots, such as forgetting about the negative square root or misapplying methods. Let's look at a few of these mistakes in detail.
Can you help Max find the area of a square box if its side length is given as √1.2?
The area of the square is approximately 1.2 square units.
The area of the square = side^2.
The side length is given as √1.2.
Area of the square = (√1.2)^2 = 1.2.
Therefore, the area of the square box is approximately 1.2 square units.
A square-shaped building measuring 1.2 square feet is built; if each of the sides is √1.2, what will be the square feet of half of the building?
0.6 square feet
We can divide the given area by 2 since the building is square-shaped.
Dividing 1.2 by 2 = 0.6.
So, half of the building measures 0.6 square feet.
Calculate √1.2 × 5.
Approximately 5.477225.
First, find the square root of 1.2, which is approximately 1.095445.
Then multiply by 5. So, 1.095445 × 5 ≈ 5.477225.
What will be the square root of (0.8 + 0.4)?
The square root is approximately 1.095445.
To find the square root, sum (0.8 + 0.4) = 1.2, and then find √1.2, which is approximately 1.095445.
Find the perimeter of the rectangle if its length ‘l’ is √1.2 units and the width ‘w’ is 1 unit.
The perimeter of the rectangle is approximately 4.19089 units.
Perimeter of the rectangle = 2 × (length + width)
Perimeter = 2 × (√1.2 + 1) = 2 × (1.095445 + 1) = 2 × 2.095445 ≈ 4.19089 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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