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 the field of vehicle design, finance, etc. Here, we will discuss the square root of 805.
The square root is the inverse of the square of the number. 805 is not a perfect square. The square root of 805 is expressed in both radical and exponential form. In the radical form, it is expressed as √805, whereas (805)^(1/2) in the exponential form. √805 ≈ 28.37252, 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.
For perfect square numbers, the prime factorization method is used. However, for non-perfect square numbers like 805, the long-division method and approximation method 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: Group the digits of 805 from right to left as 05 and 8.
Step 2: Find the greatest number whose square is less than or equal to the first group. For 8, that number is 2, because 2 x 2 = 4. The remainder is 8 - 4 = 4.
Step 3: Bring down the next group, making the new dividend 405. Double the quotient (2), and use it as part of the new divisor.
Step 4: Find a digit, say n, such that 4n x n is less than or equal to 405. Here, n is 8 because 48 x 8 = 384.
Step 5: Subtract 384 from 405 to get 21, and the quotient is 28.
Step 6: Since the remainder is less than the divisor, add a decimal point and bring down pairs of zeroes.
Step 7: Continue the process to get more decimal places as needed.
So the square root of √805 ≈ 28.37.
The approximation method is another method for finding square roots. It is an easy method to find the square root of a given number. Let us learn how to find the square root of 805 using the approximation method.
Step 1: Identify the two closest perfect squares that 805 falls between. These are 784 (28^2) and 841 (29^2). Hence, √805 is between 28 and 29.
Step 2: Use the formula: (Given number - smaller perfect square) / (Larger perfect square - smaller perfect square). (805 - 784) / (841 - 784) ≈ 21 / 57 ≈ 0.368 Step 3: Add this decimal to the smaller square root: 28 + 0.368 ≈ 28.368.
So, the square root of 805 is approximately 28.368.
Students often make mistakes while finding the square root, such as forgetting about the negative square root or skipping steps in the long division method. 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 √805?
The area of the square is approximately 805 square units.
The area of the square = side^2.
The side length is given as √805.
Area of the square = (√805)^2 = 805.
Therefore, the area of the square box is approximately 805 square units.
A square-shaped building measuring 805 square feet is built; if each of the sides is √805, what will be the square feet of half of the building?
402.5 square feet
We can divide the given area by 2 as the building is square-shaped.
Dividing 805 by 2 gives us 402.5.
So half of the building measures 402.5 square feet.
Calculate √805 x 5.
Approximately 141.86
The first step is to find the square root of 805, which is approximately 28.37.
The second step is to multiply 28.37 by 5.
So, 28.37 x 5 ≈ 141.86.
What will be the square root of (805 + 36)?
The square root is approximately 29.
To find the square root, we need to find the sum of (805 + 36). 805 + 36 = 841, and √841 = 29.
Therefore, the square root of (805 + 36) is ±29.
Find the perimeter of the rectangle if its length ‘l’ is √805 units and the width ‘w’ is 38 units.
The perimeter of the rectangle is approximately 132.74 units.
Perimeter of the rectangle = 2 × (length + width).
Perimeter = 2 × (√805 + 38) ≈ 2 × (28.37 + 38) = 2 × 66.37 ≈ 132.74 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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