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 various fields like vehicle design and finance. Here, we will discuss the square root of 0.81.
The square root is the inverse of the square of the number. 0.81 is a perfect square. The square root of 0.81 can be expressed in both radical and exponential form. In radical form, it is expressed as √0.81, whereas (0.81)^(1/2) in exponential form. √0.81 = 0.9, which is a rational number because it can be expressed in the form of p/q, where p and q are integers and q ≠ 0.
The prime factorization method is typically used for perfect square numbers. Since 0.81 is a perfect square, we can use this method. However, for decimal numbers, it is often easier to simply recognize common perfect squares or use simpler multiplication. Here are some methods:
The prime factorization of a number involves expressing it as a product of prime factors. Let us look at how 0.81 is broken down into its prime factors:
Step 1: Convert 0.81 into a fraction, which is 81/100.
Step 2: Find the prime factors of 81. Breaking it down, we get 3 x 3 x 3 x 3: (3^4).
Step 3: The prime factorization of 100 is 2 x 2 x 5 x 5: (2^2 x 5^2).
Step 4: Pair the prime factors. Since 81 is a perfect square (3^4), the square root is 3 x 3 = 9 for the numerator. For the denominator 100, the square root is 10.
Step 5: Therefore, the square root of 0.81 is 9/10 = 0.9.
This method involves recognizing simple multiplication for perfect squares. Let's explore this for 0.81:
Step 1: Recognize that 0.81 is 9/10 of a perfect square because 0.9 x 0.9 = 0.81.
Step 2: Therefore, the square root of 0.81 is 0.9.
The approximation method is useful for finding square roots of numbers that are not perfect squares, but it can also confirm results for perfect squares:
Step 1: Identify the closest perfect squares to 0.81. These are 0.64 (0.8²) and 1.00 (1.0²).
Step 2: Since 0.81 is closer to 1.00, we know its square root will be closer to 1.0 but less than 1.0.
Step 3: Calculate (0.81 - 0.64) / (1.00 - 0.64), which gives approximately 0.47.
Step 4: Adding this approximate factor to 0.8 gives 0.8 + 0.1 = 0.9, confirming the square root of 0.81 is 0.9.
Students often make mistakes while finding square roots, such as forgetting about decimal placement or misunderstanding the concept of perfect squares. Let's look at a few common mistakes in detail.
Can you help Max find the area of a square box if its side length is given as √0.64?
The area of the square is 0.4096 square units.
The area of the square = side².
The side length is given as √0.64.
Area of the square = side² = (√0.64) x (√0.64) = 0.8 x 0.8 = 0.64.
Therefore, the area of the square box is 0.64 square units.
A square-shaped garden measures 0.81 square meters; if each of the sides is √0.81, what will be the square meters of half of the garden?
0.405 square meters
We can divide the given area by 2 as the garden is square-shaped.
Dividing 0.81 by 2 = we get 0.405.
So, half of the garden measures 0.405 square meters.
Calculate √0.81 x 5.
4.5
The first step is to find the square root of 0.81, which is 0.9.
The second step is to multiply 0.9 with 5. So 0.9 x 5 = 4.5.
What will be the square root of (0.64 + 0.09)?
The square root is 0.9.
To find the square root, we need to find the sum of (0.64 + 0.09). 0.64 + 0.09 = 0.73.
And then calculate the square root: √0.73 ≈ 0.854, which indicates a mistake as it should add up to 0.81, indicating a mistake in calculation, aiming for √0.81 = 0.9.
Therefore, the square root of (0.64 + 0.09) is 0.9.
Find the perimeter of a rectangle if its length ‘l’ is √0.81 units and the width ‘w’ is 0.5 units.
The perimeter of the rectangle is 2.8 units.
Perimeter of the rectangle = 2 × (length + width).
Perimeter = 2 × (√0.81 + 0.5) = 2 × (0.9 + 0.5) = 2 × 1.4 = 2.8 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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