CBSE Class 8 Mathematics Square And Square Roots Worksheet Set 04

CH-6 SQUARE AND SQUARE ROOTS

Question 1. Which of the following numbers are perfect squares?
(a) 3364
(b) 4489
(c) 4358
Answer: To check if a number is a perfect square, we can find its square root or check its units digit.
(a) For 3364: The units digit is 4. Let us test by estimation or prime factorization. \(58 \times 58 = 3364\). Thus, 3364 is a perfect square.
(b) For 4489: The units digit is 9. Let us test: \(67 \times 67 = 4489\). Thus, 4489 is a perfect square.
(c) For 4358: The units digit is 8. Since no square of a natural number can end in 2, 3, 7, or 8, 4358 is not a perfect square.
In simple words: Numbers that can be expressed as the product of an integer with itself are perfect squares. Numbers ending in 8 can never be perfect squares.

Exam Tip: Remember the property that numbers ending in 2, 3, 7, or 8 are never perfect squares.

Question 2. Using the property of squares, find the value of the following.
(a) \(24^2 - 23^2\)
(b) \(50^2 - 49^2\)
(c) \(105^2 - 104^2\)
Answer: Using the algebraic identity \(a^2 - b^2 = (a + b)(a - b)\):
(a) \(24^2 - 23^2 = (24 + 23)(24 - 23) = (47)(1) = 47\)
(b) \(50^2 - 49^2 = (50 + 49)(50 - 49) = (99)(1) = 99\)
(c) \(105^2 - 104^2 = (105 + 104)(105 - 104) = (209)(1) = 209\)
In simple words: The difference of squares of two consecutive numbers is simply equal to their sum.

Exam Tip: For consecutive integers \(n+1\) and \(n\), the difference of their squares is always \((n+1) + n = 2n + 1\).

Question 3. Without adding find the sum.
(a) \(1 + 3 + 5 + 7 + 9\)
(b) \(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23\)
Answer: The sum of the first \(n\) odd natural numbers is given by \(n^2\).
(a) Here, there are 5 consecutive odd numbers starting from 1.
Sum = \(5^2 = 25\)
(b) Here, there are 12 consecutive odd numbers starting from 1 up to 23.
Sum = \(12^2 = 144\)
In simple words: Count how many odd numbers are added together and square that count.

Exam Tip: This property only works when the series starts strictly from 1 and consists of consecutive odd integers.

Question 4. Find the square root of the following by prime factorisation method.
(a) 1024
(b) 8836
(c) 8464
Answer: (a) Prime factorization of 1024:
\(1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{10}\)
Group in pairs: \(\sqrt{1024} = 2 \times 2 \times 2 \times 2 \times 2 = 32\)
(b) Prime factorization of 8836:
\(8836 = 2 \times 2 \times 47 \times 47\)
Group in pairs: \(\sqrt{8836} = 2 \times 47 = 94\)
(c) Prime factorization of 8464:
\(8464 = 2 \times 2 \times 2 \times 2 \times 23 \times 23\)
Group in pairs: \(\sqrt{8464} = 2 \times 2 \times 23 = 92\)
In simple words: Break the number into prime factors, make pairs of identical factors, and take one from each pair to find the square root.

Exam Tip: Always double-check your prime factors by multiplying them back together before extracting pairs.

Question 5. Find the smallest number by which 2475 must be multiplied to get a perfect square.
Answer: First, find the prime factorization of 2475:
\(2475 = 3 \times 3 \times 5 \times 5 \times 11 = 3^2 \times 5^2 \times 11\)
Pairing the prime factors, we see that 3 and 5 have pairs, but 11 is left without a pair.
To make 2475 a perfect square, it must be multiplied by 11.
In simple words: Find the prime factors and multiply by the factor that does not form a complete pair.

Exam Tip: The same method applies if the question asks for the smallest number to divide by to make a perfect square.

Question 6. 4096 soldiers are arranged in an auditorium in such a manner that there are as many soldiers in a row as there are rows in the auditorium. How many rows are there in the auditorium?
Answer: Let the number of rows be \(x\).
Then the number of soldiers in each row is also \(x\).
Total number of soldiers = \(x \times x = x^2\)
Given total soldiers = 4096
\(x^2 = 4096\)
\(x = \sqrt{4096}\)
Finding the square root of 4096 by prime factorization:
\(4096 = 2^{12}\)
\(x = 2^6 = 64\)
Number of rows = 64
In simple words: When rows and columns are equal, find the square root of the total count.

Exam Tip: Word problems involving square arrangements always require finding the square root of the given total.

Question 7. Find the least perfect square exactly divisible by each one of the numbers 4, 5, 10.
Answer: First, find the LCM of 4, 5, and 10.
Prime factors: \(4 = 2^2\), \(5 = 5\), \(10 = 2 \times 5\)
LCM = \(2^2 \times 5 = 4 \times 5 = 20\)
Now, find the prime factorization of 20:
\(20 = 2 \times 2 \times 5 = 2^2 \times 5\)
To make 20 a perfect square, it must be multiplied by 5 (since 5 lacks a pair).
Least perfect square = \(20 \times 5 = 100\)
In simple words: Find the LCM first, then multiply its prime factors to make all of them form pairs.

Exam Tip: The LCM is the smallest number divisible by all given numbers. Making it a square gives the least square number divisible by them.

Question 8. Find the least square number exactly divisible by each one of the numbers 6, 9, 10, 15 and 20.
Answer: First, find the LCM of 6, 9, 10, 15, and 20.
Prime factorizations:
\(6 = 2 \times 3\)
\(9 = 3^2\)
\(10 = 2 \times 5\)
\(15 = 3 \times 5\)
\(20 = 2^2 \times 5\)
LCM = \(2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\)
Now, find the prime factorization of 180:
\(180 = 2^2 \times 3^2 \times 5\)
Here, 5 is unpaired. To make it a perfect square, multiply by 5.
Least square number = \(180 \times 5 = 900\)
In simple words: Find the LCM of the numbers and multiply by any factor needed to complete all pairs.

Exam Tip: Always check that your LCM is divisible by all the individual numbers before making it a square.

Question 9. Find the least number which must be added to 4215 to make it a perfect square.
Answer: We use the long division method to find the square root of 4215:
\(64^2 = 4096\)
\(65^2 = 4225\)
Since \(64^2 < 4215 < 65^2\), the next perfect square after 4215 is \(65^2 = 4225\).
Least number to be added = \(4225 - 4215 = 10\)
In simple words: Find the next greater square number and subtract the given number from it.

Exam Tip: Use long division or estimation of nearby squares for addition and subtraction problems.

Question 10. Find the least number which should be subtracted from 984 to make it a perfect square. Also find the square root of this perfect square.
Answer: Using long division or trial for 984:
\(31^2 = 961\)
\(32^2 = 1024\)
Since \(961 < 984 < 1024\), the greatest perfect square less than 984 is 961 (which is \(31^2\)).
Least number to be subtracted = \(984 - 961 = 23\)
Square root of the new perfect square (\(\sqrt{961}\)) = 31
In simple words: Find the largest square smaller than the number, subtract to find what to remove, and take its square root.

Exam Tip: Subtraction questions rely on finding the square just below the given number.

Question 11. The area of the square field is \(8464\text{ m}^2\). A man takes 3 rounds of this field. Find the distance covered by him.
Answer: Area of square field = \(8464\text{ m}^2\)
Side of the square field (\(s\)) = \(\sqrt{8464}\)
Using prime factorization or long division, \(\sqrt{8464} = 92\text{ m}\)
Perimeter of the field = \(4 \times \text{side} = 4 \times 92 = 368\text{ m}\)
Distance covered in 3 rounds = \(3 \times \text{Perimeter} = 3 \times 368 = 1104\text{ m}\)
In simple words: Find the side length from the area, calculate the perimeter, and multiply by the number of rounds.

Exam Tip: One round around a field equals its total perimeter.

Question 12. Find the square root of the following decimal numbers.
(a) 72.25
(b) 86.49
(c) 75.69
(d) 9.61
(e) 0.16
Answer: (a) \(\sqrt{72.25} = 8.5\) (since \(8.5 \times 8.5 = 72.25\))
(b) \(\sqrt{86.49} = 9.3\) (since \(9.3 \times 9.3 = 86.49\))
(c) \(\sqrt{75.69} = 8.7\) (since \(8.7 \times 8.7 = 75.69\))
(d) \(\sqrt{9.61} = 3.1\) (since \(3.1 \times 3.1 = 9.61\))
(e) \(\sqrt{0.16} = 0.4\) (since \(0.4 \times 0.4 = 0.16\))
In simple words: Ignore the decimal point temporarily, find the square root of the whole number, and place the decimal point after one digit.

Exam Tip: A decimal number with two decimal places will have a square root with one decimal place.

Question 13. PQRS is a rhombus. \(PQ = 12\text{ cm}\), \(QS = 16\text{ cm}\). Find the side of the rhombus.
Answer: In a rhombus, all four sides are equal.
Given: \(PQ = 12\text{ cm}\).
Since PQ is already one of the sides of the rhombus PQRS, the side of the rhombus is 12 cm.
*(Note: The diagonal length \(QS = 16\text{ cm}\) is extra information not needed to find the side since the side length is directly given in the problem statement).*
In simple words: All sides of a rhombus are equal in length, so if one side is given as 12 cm, every side is 12 cm.

Question 14. Express 81 as the sum of 9 odd numbers.
Answer: We know that the sum of the first \(n\) odd numbers is \(n^2\).
Here, \(n = 9\), so \(9^2 = 81\).
The first 9 odd numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17.
Sum = \(1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81\)
In simple words: Write down the first 9 consecutive odd numbers starting from 1.

Exam Tip: The number of terms must equal the square root of the given number.

Question 15. How many numbers lie between the squares of the following numbers?
(a) 15 and 16
(b) 56 and 57
Answer: For any natural number \(n\) and \(n+1\), the number of non-square numbers lying between \(n^2\) and \((n+1)^2\) is \(2n\).
(a) For 15 and 16, here \(n = 15\):
Number of numbers = \(2 \times 15 = 30\)
(b) For 56 and 57, here \(n = 56\):
Number of numbers = \(2 \times 56 = 112\)
In simple words: Multiply the smaller number by 2 to find how many numbers lie between their squares.

Exam Tip: Always use the shortcut formula \(2n\) where \(n\) is the smaller of the two consecutive numbers.

Question 16. Express the following as the sum of two consecutive integers.
(a) \(21^2\)
(b) \(13^2\)
(c) \(19^2\)
Answer: For any odd number \(n^2\), it can be expressed as the sum of two consecutive integers using the formula \(\frac{n^2 - 1}{2}\) and \(\frac{n^2 + 1}{2}\).
(a) For \(21^2 = 441\):
First integer = \(\frac{441 - 1}{2} = \frac{440}{2} = 220\)
Second integer = \(\frac{441 + 1}{2} = \frac{442}{2} = 221\)
Check: \(220 + 221 = 441 = 21^2\)
(b) For \(13^2 = 169\):
First integer = \(\frac{169 - 1}{2} = \frac{168}{2} = 84\)
Second integer = \(\frac{169 + 1}{2} = \frac{170}{2} = 85\)
Check: \(84 + 85 = 169 = 13^2\)
(c) For \(19^2 = 361\):
First integer = \(\frac{361 - 1}{2} = \frac{360}{2} = 180\)
Second integer = \(\frac{361 + 1}{2} = \frac{362}{2} = 181\)
Check: \(180 + 181 = 361 = 19^2\)
In simple words: Square the number and divide the difference and sum by 2 to get the two consecutive integers.

Exam Tip: This rule works exclusively for the squares of odd numbers.