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प्रश्न
Find the squares of the following numbers using column method. Verify the result by finding the square using the usual multiplication:
37
उत्तर
Here, a = 3, b = 7
Step 1. Make 3 columns and write the values of a2, 2 x a x b, and b2 in these columns.
| Column I | Column II | Column III |
| a2 | 2 x a x b | b2 |
| 9 | 42 | 49 |
Step 2. Underline the unit digit of b2 (in Column III) and add its tens digit, if any, with 2 x a x b (in Column II).
| Column I | Column II | Column III |
| a2 | 2 x a x b | b2 |
| 9 | 42 + 4 | 49 |
| 46 |
Step 3. Underline the unit digit in Column II and add the number formed by the tens and other digits, if any, with a2 in Column I.
| Column I | Column II | Column III |
| a2 | 2 x a x b | b2 |
| 9 + 4 | 42 + 4 | 49 |
| 13 | 46 |
Step 4. Underline the number in Column I.
| Column I | Column II | Column III |
| a2 | 2 x a x b | b2 |
| 9 + 4 | 42 + 4 | 49 |
| 13 | 46 |
Step 5. Write the underlined digits at the bottom of each column to obtain the square of the given number.
In this case, we have:
372 = 1369
Using multiplication:
37
37
259
111
1369
This matches with the result obtained using the column method.
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संबंधित प्रश्न
Write a Pythagorean triplet whose one member is 14.
Which of the following triplets are pythagorean?
(8, 15, 17)
Which of the following triplet pythagorean?
(12, 35, 38)
Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
\[ 1^2 + 2^2 = \frac{1}{6}\left[ 2 \times \left( 2 + 1 \right) \times \left( 2 \times 2 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :
12 + 22 + 32 + 42 + ... + 102
Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
\[ 1^2 + 2^2 = \frac{1}{6}\left[ 2 \times \left( 2 + 1 \right) \times \left( 2 \times 2 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :
52 + 62 + 72 + 82 + 92 + 102 + 112 + 122
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