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Extra questions and detailed answers from the CBSE 8th grade Mathematics chapter on Squares and Squares Roots
Updated:
May 8, 2023
The chapter on squares and square roots is divided into three parts. The first part deals with the concept of squares of numbers. The second part deals with the properties and patterns followed by square numbers. The third section deals with square roots, which is the inverse operation of finding squares.
In the chapter, students learn how to find the squares of numbers by repeated addition, how to recognize odd and even square numbers, and how to find square roots by factorization. Students also learn how to estimate square roots and how to use squares and square roots in real-life situations.
Overall, this chapter is essential for building a solid foundation in mathematics and lays the groundwork for more complex topics in the future.
Here are some key notes on CBSE Class 8 Mathematics - Squares and Squares Roots:
1. What is a perfect square?
Answer: A perfect square is a number that is produced when an integer multiplies itself. For example, 4 x 4 = 16, so 16 is a perfect square.
2. What is the square of 8?
Answer: The square of number 8 is 64. (8 x 8 = 64)
3. What is the smallest odd perfect square?
Answer: The smallest odd perfect square is 1.
4. Is 50 a square number?
Answer: No, 50 is not a square number.
5. What is a radical sign?
Answer: The radical sign (√) is a symbol used to indicate roots. For example, √9 indicates the square root of 9.
6. What is the square of 1?
Answer: The square of 1 is 1.
7. What is the difference between a square number and a perfect square?
Answer: A square number is any number that is produced when a number multiplies itself, whereas a perfect square is a number that is produced when an integer multiplies itself.
8. What are some examples of odd perfect squares?
Answer: Some examples of odd perfect squares are 1, 9, 25, 49, 81, 121, 169, 225, 361, 441, and so on.
9. What are some examples of even perfect squares?
Answer: Some examples of even perfect squares are 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, and so on.
10. What is a cube number?
Answer: A cube number is a number that is produced when a number multiplies itself three times. For example, 3 x 3 x 3 = 27, so 27 is a cube number.
11. What is the method of repeated addition?
Answer: The method of repeated addition is a method used to find the square of a number. It involves adding the number to itself repeatedly according to the value of the number. For example, to find the square of 5 using the method of repeated addition, we need to add 5 to itself five times, resulting in 25.
12. What is the value of √16?
Answer: The value of √16 is 4, because 4 x 4 = 16.
13. Is 49 a perfect square? If yes, find its square root.
Answer: Yes, 49 is a perfect square. Its square root is 7, because 7 x 7 = 49.
14. What is the value of √81?
Answer: The value of √81 is 9, because 9 x 9 = 81.
15. What is the importance of squares and square roots in real-life situations?
Answer: Squares and square roots are used in various real-life situations such as in construction, measurement, engineering, and design.
16. What is the value of √25?
Answer: The value of √25 is 5, because 5 x 5 = 25.
17. What is the square of the smallest prime number?
Answer: The smallest prime number is 2, and its square is 4.
18. How can we prove that a number is a perfect square?
Answer: We can prove that a number is a perfect square if its prime factorization contains only even exponents.
19. Find the square of 10.
Answer: The square of 10 is 100.
20. What is the square of the largest single-digit number?
Answer: The largest single-digit number is 9, and its square is 81.
1. Express 13 as the sum of three consecutive odd integers. Also, find the sum of the squares of these three integers.
Answer: Let x be the first of these odd integers. Then, the three consecutive odd integers are x, x+2, and x+4.
We have x + (x+2) + (x+4) = 13. Solving this equation, we get x = 3.
So, the three consecutive odd integers are 3, 5, and 7.
Now, the sum of the squares of these three integers is 3^2 + 5^2 + 7^2 = 83.
2. Determine the smallest number which must be added to 1030 so that the resulting number is a perfect square.
Answer: We start by finding the next perfect square greater than 1030, which is 32^2 = 1024.
Therefore, the number that must be added to 1030 to get a perfect square is 1024 - 1030 + 1 = -5 + 1 = -4.
So, to get a perfect square, we must add 4 to 1030, which gives us 1034.
3. Determine the value of x and y in the equation x^2 - y^2 = 336.
Answer: We begin by factorizing the left-hand side of the given equation as (x+y) (x-y) = 336.
Now, we need to find two factors of 336 whose sum or difference is even.
We can write 336 as 2 x 2 x 2 x 2 x 3 x 7.
Possible factors whose sum or difference is even are (2 x 168), (4 x 84), (8 x 42), (16 x 21).
Among these, the only pair whose product gives us 336 is (16 x 21).
Therefore, we have x+y = 42, and x-y = 16. Solving these equations simultaneously, we get x = 29 and y = 13.
4. Find the number of digits in the square root of 124416.
Answer: To find the number of digits in the square root of 124416, we first find the value of the square root, which is 352.
Now, the number of digits in 352 is 3. Therefore, the square root of 124416 has three digits.
5. If a number leaves a remainder of 2 when divided by 4, what can be said about its square?
Answer: A number that leaves a remainder of 2 when divided by 4 can be written as 4k + 2, where k is any integer.
Now, when we square this number, we get (4k + 2)^2 = 16k^2 + 16k + 4 = 4(4k^2 + 4k + 1) + 0.
So, the square of any number that leaves a remainder of 2 when divided by 4 will also leave a remainder of 2 when divided by 4.
6. Find the smallest natural number which when increased by 17 is a perfect square and when decreased by 17 is also a perfect square.
Answer: Let x be the required natural number. Then, the given conditions can be written as (x+17) = a^2 and (x-17) = b^2, where a and b are integers.
Subtracting the second equation from the first, we get 34 = a^2 - b^2 = (a+b)(a-b).
Now, 34 can only be factored as (1 x 34) or (2 x 17).
Solving the resulting two pairs of equations, we get x=187 and x=288 as the possible solutions.
Therefore, the smallest natural number which satisfies the given conditions is 187.
7. Find the square of the smallest odd prime greater than 5.
Answer: The smallest odd prime greater than 5 is 7.
Therefore, the square of this number is 7^2 = 49, which is the square of the smallest odd prime greater than 5.
8. A natural number between 2000 and 3000 leaves a remainder of 4 when divided by 9, and a remainder of 3 when divided by 7. Find the number.
Answer: Let the required number be x. Then, we have x = 9a + 4 and x = 7b + 3, where a and b are integers.
Solving these equations simultaneously, we get b = 9k + 4 and a = 14k + 8, where k is any integer.
Now, we need to find the smallest value of k that satisfies the given conditions, and also gives a value of x between 2000 and 3000.
Checking values of k, we find that k = 2 satisfies both conditions, and gives x = 2005 (which is between 2000 and 3000).
Therefore, the required natural number is 2005.
9. Find the two consecutive positive integers whose sum is a perfect square, and whose difference is a perfect cube.
Answer: Let the smaller of the two consecutive positive integers be x. Then, the larger integer is x+1.
We are given that (x) + (x+1) = k^2 for some positive integer k, and (x+1) - (x) = n^3 for some positive integer n.
Simplifying the second equation, we get n^3 = 1, which has only one integer solution, n = 1.
Therefore, the difference between the two consecutive integers is 1, and their sum is k^2.
Solving these equations simultaneously, we get x = (k^2 - 1)/2 and x+1 = (k^2 + 1)/2.
Now, we need to find the smallest value of k that gives us a value of x between 1 and 10^6.
Checking values of k, we find that k = 15 satisfies both conditions, and gives x = 112 and x+1 = 113.
Therefore, the two consecutive positive integers are 112 and 113.
10. Find the smallest positive integer value of x for which 150x is a perfect cube.
Answer: We can write 150x as (2 x 3 x 5^2) x x.
For 150x to be a perfect cube, x must be multiplied by 2^2 and an odd number such that the product is a perfect cube.
Now, let x = 3^2 x y, where y is any positive integer.
Substituting this value of x in the expression for 150x, we get 150x = 3^2 x 2^2 x 5^2 x y.
For this expression to be a perfect cube, y must be multiplied by 5^2 to give a perfect cube.
Therefore, let y = 5^2 x z, where z is any positive integer.
Substituting these values of x and y in the expression for 150x, we get 150x = 3^2 x 2^2 x 5^3 x z^3.
For this expression to be a perfect cube, z must be multiplied by 3^3 to give a perfect cube.
Therefore, let z = 3^3.
Substituting this value of z in the expression for 150x, we get 150x = 3^2 x 2^2 x 5^3 x 3^3.
Simplifying this expression, we get x = 3.
Therefore, the smallest positive integer value of x for which 150x is a perfect cube is 3.
11. Find the largest number i if 8642+i is a perfect square.
Answer: We can write 8642 + i as 8642 + i = 82 × 105 + i.
For 8642 + i to be a perfect square, we need to find the largest perfect square that is less than 8642 + i.
The largest perfect square less than 8642 + i is 81 × 100 = 8100.
Subtracting 8100 from both sides, we get
8642 + i - 8100 = i + 542
Now, we need to find the largest value of i such that i + 542 is a perfect square.
Let i + 542 = k^2 where k is a positive integer.
Then, we have i = k^2 - 542.
Since i must be the largest possible value, we need to find the largest value of k such that k^2 - 542 is less than or equal to 0.
Solving for k, we get k = 24.
Therefore, the largest number i such that 8642+i is a perfect square is i = (24)^2 - 542 = 2.
12. Find the smallest five-digit number which is a perfect square and is divisible by 10.
Answer: A perfect square that is divisible by 10 must have 0 at its units place.
The smallest possible five-digit number that ends with 0 is 10,000.
Squaring 10,000, we get 10,000,000, which is a seven-digit number.
The next smallest five-digit number that ends with 0 is 20,000.
Squaring 20,000, we get 400,000,000, which is an eight-digit number.
Therefore, no five-digit number that is a perfect square and is divisible by 10 exists.
13. Find the smallest number by which 392 must be multiplied so that the product is a perfect square.
Answer: We can write 392 as the product of its prime factors: 392 = 2^3 x 7^2.
For the product of 392 and the unknown number to be a perfect square, we need to multiply 392 by the square of a factor of 2 and by the square of a factor of 7.
Therefore, we need to find the smallest number by which 392 must be multiplied such that it becomes 2^4 x 7^2.
This is equivalent to multiplying 392 by 2 and then multiplying the result by 2 again.
Therefore, the smallest number by which 392 must be multiplied so that the product is a perfect square is 2 x 2 = 4.
14. Find the value of k for which the roots of the quadratic equation kx^2 - 7x + 1 = 0 are both integers.
Answer: Let the two roots of the quadratic equation kx^2 - 7x + 1 = 0 be p and q, where p and q are integers.
By Vieta's formulas, we have p + q = 7/k and pq = 1/k.
Since p and q are integers, their product, pq, must also be an integer.
Therefore, k must divide 1. This means that k can only be 1 or -1.
Substituting k = -1 into the quadratic equation, we get -x^2 - 7x + 1 = 0.
Using the quadratic formula, we get
x = (-(-7) ± √((-7)^2 - 4(-1)(1))) / (2(-1))
= (7 ± √29) / 2
Both of these solutions are not integers.
Therefore, k cannot be -1.
Substituting k = 1 into the quadratic equation, we get x^2 - 7x + 1 = 0.
Using the quadratic formula, we get
x = (7 ± √45) / 2
x = (7 ± 3√5) / 2
Both of these solutions are not integers.
Therefore, there is no value of k for which the roots of the given quadratic equation are both integers.