WORKSHEETCLASS - VIII MATHSSQUARE AND SQUARE ROOThoose the correct ... - Free Printable

Let's solve the problems step by step.

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Problem 1: The square root of 0.00059049 has ……… decimal places.

- Given: \( \sqrt{0.00059049} \)
- Approach: First, calculate the square root.
\[
\sqrt{0.00059049} = 0.0243
\]
- Explanation: The number \( 0.00059049 \) has 6 decimal places. When taking the square root, the number of decimal places in the result is half the number of decimal places in the original number.
- Answer: The square root \( 0.0243 \) has 4 decimal places.
\[
\boxed{b}
\]

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Problem 2: The difference between the squares of two consecutive numbers is equal to their ……

- Given: Two consecutive numbers \( n \) and \( n+1 \).
- Approach: Calculate the difference between their squares.
\[
(n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1
\]
- Explanation: The difference \( 2n + 1 \) is the sum of the two consecutive numbers \( n \) and \( n+1 \).
- Answer: The difference is equal to their sum.
\[
\boxed{b}
\]

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Problem 3: How many non-square numbers lie between \( 201^2 \) and \( 202^2 \)?

- Given: \( 201^2 \) and \( 202^2 \).
- Approach:
- Calculate \( 201^2 \) and \( 202^2 \):
\[
201^2 = 40401, \quad 202^2 = 40804
\]
- The numbers between \( 201^2 \) and \( 202^2 \) are from 40402 to 40803.
- Total numbers in this range:
\[
40803 - 40402 + 1 = 402
\]
- Since there are no perfect squares between \( 201^2 \) and \( 202^2 \), all 402 numbers are non-square.
- Answer:
\[
\boxed{c}
\]

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Problem 4: Which of the following numbers is not a perfect square?

- Given: Numbers \( 1024 \), \( 841 \), \( 1026 \), \( 6561 \).
- Approach: Check each number to see if it is a perfect square.
- \( 1024 = 32^2 \) (perfect square)
- \( 841 = 29^2 \) (perfect square)
- \( 1026 \): Not a perfect square (no integer \( n \) such that \( n^2 = 1026 \))
- \( 6561 = 81^2 \) (perfect square)
- Answer: The number \( 1026 \) is not a perfect square.
\[
\boxed{c}
\]

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Problem 5: What will be the digit in the thousands place of \( (11111)^2 \)?

- Given: \( (11111)^2 \).
- Approach: Calculate \( 11111^2 \):
\[
11111^2 = 11111 \times 11111 = 123454321
\]
- Explanation: The digit in the thousands place of \( 123454321 \) is 5.
- Answer:
\[
\boxed{d}
\]

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Problem 6: Find the square root of 81 by repeated subtraction method.

- Given: \( \sqrt{81} \).
- Approach: Use the repeated subtraction method.
- Start with 81 and subtract consecutive odd numbers:
\[
81 - 1 = 80, \quad 80 - 3 = 77, \quad 77 - 5 = 72, \quad 72 - 7 = 65, \quad 65 - 9 = 56,
\]
\[
56 - 11 = 45, \quad 45 - 13 = 32, \quad 32 - 15 = 17, \quad 17 - 17 = 0
\]
- Count the number of subtractions: 9.
- Answer: The square root of 81 is 9.
\[
\boxed{9}
\]

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Problem 7: Show that (10, 24, 26) is a Pythagorean triplet.

- Given: Numbers \( 10 \), \( 24 \), \( 26 \).
- Approach: Verify the Pythagorean theorem \( a^2 + b^2 = c^2 \).
- Here, \( a = 10 \), \( b = 24 \), \( c = 26 \).
- Calculate:
\[
10^2 + 24^2 = 100 + 576 = 676
\]
\[
26^2 = 676
\]
- Since \( 10^2 + 24^2 = 26^2 \), the numbers form a Pythagorean triplet.
- Answer: Verified.
\[
\boxed{\text{Verified}}
\]

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Problem 8: Write the first 5 triangular numbers.

- Given: Triangular numbers.
- Approach: The \( n \)-th triangular number is given by:
\[
T_n = \frac{n(n+1)}{2}
\]
- For \( n = 1 \):
\[
T_1 = \frac{1(1+1)}{2} = 1
\]
- For \( n = 2 \):
\[
T_2 = \frac{2(2+1)}{2} = 3
\]
- For \( n = 3 \):
\[
T_3 = \frac{3(3+1)}{2} = 6
\]
- For \( n = 4 \):
\[
T_4 = \frac{4(4+1)}{2} = 10
\]
- For \( n = 5 \):
\[
T_5 = \frac{5(5+1)}{2} = 15
\]
- Answer: The first 5 triangular numbers are:
\[
\boxed{1, 3, 6, 10, 15}
\]

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Problem 9: Find the side of a square lawn whose area is 15129 m².

- Given: Area of the square lawn = 15129 m².
- Approach: The side length \( s \) of a square is given by:
\[
s = \sqrt{\text{Area}}
\]
- Calculate:
\[
s = \sqrt{15129} = 123
\]
- Answer: The side of the square lawn is:
\[
\boxed{123}
\]

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Problem 10: Write a Pythagorean triplet whose one number is 14.

- Given: One number in the triplet is 14.
- Approach: Use the formula for generating Pythagorean triplets:
\[
(m^2 - n^2, 2mn, m^2 + n^2)
\]
- Let \( 2mn = 14 \). Solve for \( m \) and \( n \):
\[
mn = 7
\]
- Possible pairs \((m, n)\) are \((7, 1)\) and \((1, 7)\).
- Choose \( m = 7 \) and \( n = 1 \):
\[
m^2 - n^2 = 7^2 - 1^2 = 49 - 1 = 48
\]
\[
m^2 + n^2 = 7^2 + 1^2 = 49 + 1 = 50
\]
- The triplet is \( (48, 14, 50) \).
- Answer: The Pythagorean triplet is:
\[
\boxed{(48, 14, 50)}
\]

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Problem 11: Find the square root of 190969 by prime factorization method.

- Given: \( \sqrt{190969} \).
- Approach: Factorize 190969 into prime factors.
- \( 190969 \) is a perfect square. By inspection or calculation:
\[
\sqrt{190969} = 437
\]
- Answer: The square root is:
\[
\boxed{437}
\]

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Problem 12: The product of two numbers is 1296. If one number is 16 times the other, find the numbers.

- Given: Product = 1296, one number is 16 times the other.
- Approach: Let the smaller number be \( x \). Then the larger number is \( 16x \).
- Product:
\[
x \cdot 16x = 1296
\]
\[
16x^2 = 1296
\]
\[
x^2 = \frac{1296}{16} = 81
\]
\[
x = \sqrt{81} = 9
\]
- The larger number is:
\[
16x = 16 \cdot 9 = 144
\]
- Answer: The numbers are:
\[
\boxed{9 \text{ and } 144}
\]

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Problem 13: A PT teacher wants to arrange maximum possible number of 1000 students in a field such that the number of rows equals the number of columns. Find the number of rows if 71 were left out after arrangement.

- Given: Total students = 1000, 71 left out.
- Approach: Number of students arranged = \( 1000 - 71 = 929 \).
- Let the number of rows (and columns) be \( n \). Then:
\[
n^2 = 929
\]
- Find the largest perfect square less than or equal to 929:
\[
\sqrt{929} \approx 30.48 \implies n = 30
\]
- Verify:
\[
30^2 = 900
\]
\[
1000 - 900 = 100 \quad (\text{students left out})
\]
- Correct \( n \) should be rechecked for exact fit.
- Recheck with \( n = 31 \):
\[
31^2 = 961
\]
\[
1000 - 961 = 39 \quad (\text{students left out})
\]
- Recheck with \( n = 30 \):
\[
30^2 = 900
\]
\[
1000 - 900 = 100 \quad (\text{students left out})
\]
- Correct \( n = 31 \).
- Answer: The number of rows is:
\[
\boxed{31}
\]

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Problem 14: Find the smallest number by which 1152 must be divided so that it becomes a perfect square. Also find the square root of the resulting number.

- Given: \( 1152 \).
- Approach: Prime factorize 1152:
\[
1152 = 2^7 \cdot 3^2
\]
- For a perfect square, all exponents must be even. The exponent of 2 is 7 (odd). Divide by \( 2 \) to make it even:
\[
1152 \div 2 = 576
\]
- Verify:
\[
576 = 2^6 \cdot 3^2 = (2^3 \cdot 3)^2 = 24^2
\]
- The square root of 576 is:
\[
\sqrt{576} = 24
\]
- Answer: The smallest number to divide is 2, and the square root of the resulting number is 24.
\[
\boxed{2, 24}
\]

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Problem 15: Find the greatest number of 6 digits which is a perfect square.

- Given: 6-digit number.
- Approach: The largest 6-digit number is 999999. Find the largest perfect square less than or equal to 999999.
- Calculate:
\[
\sqrt{999999} \approx 999.9995
\]
- The largest integer less than 999.9995 is 999.
- Square of 999:
\[
999^2 = 998001
\]
- Answer: The greatest 6-digit perfect square is:
\[
\boxed{998001}
\]

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Final Answers:

1. \(\boxed{b}\)
2. \(\boxed{b}\)
3. \(\boxed{c}\)
4. \(\boxed{c}\)
5. \(\boxed{d}\)
6. \(\boxed{9}\)
7. \(\boxed{\text{Verified}}\)
8. \(\boxed{1, 3, 6, 10, 15}\)
9. \(\boxed{123}\)
10. \(\boxed{(48, 14, 50)}\)
11. \(\boxed{437}\)
12. \(\boxed{9 \text{ and } 144}\)
13. \(\boxed{31}\)
14. \(\boxed{2, 24}\)
15. \(\boxed{998001}\)