The addition of the squares of two numbers in squares is 221. What are those numbers?
10 and 11
Numbers: Finding the Pair Whose Squares Sum to 221
The question asks us to identify two numbers such that when each number is squared, and their squares are added together, the total sum is 221. This can be represented mathematically as \(a^2 + b^2 = 221\), where 'a' and 'b' are the two unknown numbers.
Understanding the Calculation of Squares
To find the numbers, we will check each pair provided in the options. For each pair, we will perform the following steps:
- Square the first number (\(a^2\)).
- Square the second number (\(b^2\)).
- Add the two squared results together (\(a^2 + b^2\)).
- Compare this sum with 221.
Step-by-Step Analysis of Options
1. Analyzing Numbers: 100 and 111
- Square of 100: \(100^2 = 100 \times 100 = 10,000\)
- Square of 111: \(111^2 = 111 \times 111 = 12,321\)
- Sum of squares: \(10,000 + 12,321 = 22,321\)
- The sum \(22,321\) is much greater than 221. Therefore, this option is not the correct pair of numbers.
2. Analyzing Numbers: 9 and 10
- Square of 9: \(9^2 = 9 \times 9 = 81\)
- Square of 10: \(10^2 = 10 \times 10 = 100\)
- Sum of squares: \(81 + 100 = 181\)
- The sum \(181\) is less than 221. Therefore, this option is not the correct pair of numbers.
3. Analyzing Numbers: 10 and 11
- Square of 10: \(10^2 = 10 \times 10 = 100\)
- Square of 11: \(11^2 = 11 \times 11 = 121\)
- Sum of squares: \(100 + 121 = 221\)
- The sum \(221\) exactly matches the required sum. This indicates that 10 and 11 are the correct numbers.
4. Analyzing Numbers: 11 and 12
- Square of 11: \(11^2 = 11 \times 11 = 121\)
- Square of 12: \(12^2 = 12 \times 12 = 144\)
- Sum of squares: \(121 + 144 = 265\)
- The sum \(265\) is greater than 221. Therefore, this option is not the correct pair of numbers.
Summary of Number Calculations
The table below summarizes the calculations for each pair of numbers to help visualize the results and confirm the correct option.
| Numbers | First Number Squared | Second Number Squared | Sum of Squares (\(a^2 + b^2\)) | Matches 221? |
|---|---|---|---|---|
| 100 and 111 | \(100^2 = 10,000\) | \(111^2 = 12,321\) | \(22,321\) | No |
| 9 and 10 | \(9^2 = 81\) | \(10^2 = 100\) | \(181\) | No |
| 10 and 11 | \(10^2 = 100\) | \(11^2 = 121\) | \(221\) | Yes |
| 11 and 12 | \(11^2 = 121\) | \(12^2 = 144\) | \(265\) | No |
Conclusion
From the analysis, the numbers whose squares add up to 221 are 10 and 11. This pair satisfies the condition given in the problem.
Important Questions from Square and Square Root
The sum of the squares of two positive integers is 306. If the square of the larger integer is 25 times the smaller integer, then the difference between the two integers is
- \(\sqrt{625.0025 }=\)
The least number which is a perfect square and is divisible by each of the numbers 4, 10 and 12 is :
Find the value of \(\sqrt{9604} \).
If (584)2 = 341056, then the value of square root of 34.1056 is:
