Last updated on May 26th, 2025
If a number is multiplied by itself, the result is a square. The inverse of the square is a square root. The square root is used in fields like vehicle design, finance, etc. Here, we will discuss the square root of 0.91.
The square root is the inverse of the square of a number. 0.91 is not a perfect square. The square root of 0.91 is expressed in both radical and exponential forms. In the radical form, it is expressed as √0.91, whereas (0.91)^(1/2) in the exponential form. √0.91 ≈ 0.9539, 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, the prime factorization method is not used for non-perfect square numbers, where long-division and approximation methods are used. 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: To begin with, pair the numbers starting from the decimal point and moving to the right. For 0.91, we pair as 91.
Step 2: Now, find a number whose square is less than or equal to 91. This number is 9 because 9 x 9 = 81.
Step 3: Now, subtract 81 from 91, giving a remainder of 10. Bring down two zeros to make it 1000.
Step 4: Double the divisor (9) to get 18. We need to find a digit to add to 18 to make a new divisor, which when multiplied by the same digit gives a product less than or equal to 1000.
Step 5: The digit is 5 because 185 x 5 = 925.
Step 6: Subtract 925 from 1000 to get 75. Bring down two more zeros.
Step 7: Continue this process to refine the square root to more decimal places. So the square root of √0.91 ≈ 0.9539.
The approximation method is another method for finding square roots. It is an easy method to estimate the square root of a given number. Now let us learn how to find the square root of 0.91 using the approximation method.
Step 1: Identify the closest perfect squares to 0.91. The perfect squares are 0.81 (0.9^2) and 1 (1^2).
Step 2: Apply the formula to estimate: (Given number - smaller perfect square) / (Larger perfect square - smaller perfect square). So, (0.91 - 0.81) / (1 - 0.81) ≈ 0.526. Add this decimal to the smaller square root: 0.9 + 0.0526 ≈ 0.9539. Therefore, the square root of 0.91 is approximately 0.9539.
Students often make mistakes while finding square roots, such as forgetting about the negative square root or skipping steps in long division methods. Let's examine a few common mistakes in detail.
Can you help Max find the area of a square box if its side length is given as √0.81?
The area of the square is 0.6561 square units.
The area of a square = side^2. The side length is given as √0.81. Area of the square = (√0.81)^2 = 0.9 x 0.9 = 0.81. Therefore, the area of the square box is 0.81 square units.
A square-shaped building measuring 0.91 square meters is built; if each of the sides is √0.91, what will be the square meters of half of the building?
0.455 square meters
We can divide the given area by 2 since the building is square-shaped.
Dividing 0.91 by 2 = 0.455 So half of the building measures 0.455 square meters.
Calculate √0.91 x 5.
4.7695
The first step is to find the square root of 0.91, which is approximately 0.9539.
The second step is to multiply 0.9539 by 5. So, 0.9539 x 5 ≈ 4.7695.
What will be the square root of (0.49 + 0.42)?
The square root is approximately 0.9539.
To find the square root, we need to find the sum of (0.49 + 0.42). 0.49 + 0.42 = 0.91, and then √0.91 ≈ 0.9539.
Therefore, the square root of (0.49 + 0.42) is approximately ±0.9539.
Find the perimeter of the rectangle if its length ‘l’ is √0.81 units and the width ‘w’ is 1 unit.
The perimeter of the rectangle is 3.8 units.
Perimeter of the rectangle = 2 × (length + width)
Perimeter = 2 × (√0.81 + 1) = 2 × (0.9 + 1) = 2 × 1.9 = 3.8 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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