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Class 8 Exam > Class 8 Notes > RD Sharma Solutions for Class 8 Mathematics > RD Sharma Solutions - Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math

Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math RD Sharma Solutions | RD Sharma Solutions for Class 8 Mathematics PDF Download

QUESTION 7: BY WHAT NUMBER SHOULD EACH OF THE FOLLOWING NUMBERS BE MULTIPLIED TO GET A PERFECT SQUARE IN EACH CASE? ALSO, FIND THE NUMBER WHOSE SQUARE IS THE NEW NUMBER.
(I) 8820
(II) 3675
(III) 605
(IV) 2880
(V) 4056
(VI) 3468
(VII) 7776
ANSWER 7: Factorising each number.
(i) 8820 = 2 x 2 x 3 x 3 x 5 x 7 x 7

Grouping them into pairs of equal factors:
8820 = (2 x 2) x (3 x 3) x (7 x 7) x 5
The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 8820 must be multiplied by 5 for it to be a perfect square.
The new number would be (2 x 2) x (3 x 3) x (7 x 7) x (5 x 5).
Furthermore, we have:
(2 x 2) x (3 x 3) x (7 x 7) x (5 x 5) = (2 x 3 x 5 x 7) x (2 x 3 x 5 x 7)
Hence, the number whose square is the new number is:
2 x 3 x 5 x 7 = 210
(ii) 3675 = 3 x 5 x 5 x 7 x 7

Grouping them into pairs of equal factors:
3675 = (5 x 5) x (7 x 7) x 3
The factor, 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3675 must be multiplied by 3 for it to be a perfect square.
The new number would be (5 x 5) x (7 x 7) x (3 x 3).
Furthermore, we have:
(5 x 5) x (7 x 7) x (3 x 3) = (3 x 5 x 7) x (3 x 5 x 7)
Hence, the number whose square is the new number is:
3 x 5 x 7 = 105
(iii) 605 = 5 x 11 x 11

Grouping them into pairs of equal factors:
605 = 5 x (11 x 11)
The factor, 5 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 605 must be multiplied by 5 for it to be a perfect square.
The new number would be (5 x 5) x (11 x 11).
Furthermore, we have:
(5 x 5) x (11 x 11) = (5 x 11) x (5 x 11)
Hence, the number whose square is the new number is:
5 x 11 = 55
(iv) 2880 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5

Grouping them into pairs of equal factors:
2880 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 5
There is a 5 as the leftover. For a number to be a perfect square, each prime factor has to be paired. Hence, 2880 must be multiplied by 5 to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5).
Furthermore, we have:
(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (5 x 5) = (2 x 2 x 2 x 3 x 5) x (2 x 2 x 2 x 3 x 5)
Hence, the number whose square is the new number is:
2 x 2 x 2 x 3 x 5 = 120
(v) 4056 = 2 x 2 x 2 x 3 x 13 x 13

Grouping them into pairs of equal factors:
4056 = (2 x 2) x (13 x 13) x 2 x 3
The factors at the end, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 4056 must be multiplied by 6 (2 x 3) for it to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (3 x 3) x (13 x 13).
Furthermore, we have:
(2 x 2) x (2 x 2) x (3 x 3) x (13 x 13) = (2 x 2 x 3 x 13) x (2 x 2 x 3 x 13)
Hence, the number whose square is the new number is:
2 x 2 x 3 x 13 = 156
(vi) 3468 = 2 x 2 x 3 x 17 x 17

Grouping them into pairs of equal factors:
3468 = (2 x 2) x (17 x 17) x 3
The factor 3 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3468 must be multiplied by 3 for it to be a perfect square.
The new number would be (2 x 2) x (17 x 17) x (3 x 3).
Furthermore, we have:
(2 x 2) x (17 x 17) x (3 x 3) = (2 x 3 x 17) x (2 x 3 x 17)
Hence, the number whose square is the new number is:
2 x 3 x 17 = 102
(vii) 7776 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3

Grouping them into pairs of equal factors:
7776 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x 2 x 3
The factors, 2 and 3 at the end are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 7776 must be multiplied by 6 (2 x 3) for it to be a perfect square.
The new number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3).
Furthermore, we have:
(2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3) x (3 x 3) = (2 x 2 x 2 x 3 x 3 x 3) x (2 x 2 x 2 x 3 x 3 x 3)
Hence, the number whose square is the new number is:
2 x 2 x 2 x 3 x 3 x 3 = 216
Question 8: By what numbers should each of the following be divided to get a perfect square in each case? Also, find the number whose square is the new number.
(i) 16562
(ii) 3698
(iii) 5103
(iv) 3174
(v) 1575
Answer 8: Factorising each number.
(i) 16562 = 2 x 7 x 7 x 13 x 13

Grouping them into pairs of equal factors:
16562 = 2 x (7 x 7) x (13 x 13)
The factor, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 16562 must be divided by 2 for it to be a perfect square.
The new number would be (7 x 7) x (13 x 13).
Furthermore, we have:
(7 x 7) x (13 x 13) = (7 x 13) x (7 x 13)
Hence, the number whose square is the new number is:
7 x 13 = 91
(ii) 3698 = 2 x 43 x 43

Grouping them into pairs of equal factors:
3698 = 2 x (43 x 43)
The factor, 2 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3698 must be divided by 2 for it to be a perfect square.
The new number would be (43 x 43).
Hence, the number whose square is the new number is 43.
(iii) 5103 = 3 x 3 x 3 x 3 x 3 x 3 x 7

Grouping them into pairs of equal factors:
5103 = (3 x 3) x (3 x 3) x (3 x 3) x 7
The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 5103 must be divided by 7 for it to be a perfect square.
The new number would be (3 x 3) x (3 x 3) x (3 x 3).
Furthermore, we have:
(3 x 3) x (3 x 3) x (3 x 3) = (3 x 3 x 3) x (3 x 3 x 3)
Hence, the number whose square is the new number is:
3 x 3 x 3 = 27
(iv) 3174 = 2 x 3 x 23 x 23

Grouping them into pairs of equal factors:
3174 = 2 x 3 x (23 x 23)
The factors, 2 and 3 are not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 3174 must be divided by 6 (2 x 3) for it to be a perfect square.
The new number would be (23 x 23).
Hence, the number whose square is the new number is 23.
(v) 1575 = 3 x 3 x 5 x 5 x 7

Grouping them into pairs of equal factors:
1575 = (3 x 3) x (5 x 5) x 7
The factor, 7 is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 1575 must be divided by 7 for it to be a perfect square.
The new number would be (3 x 3) x (5 x 5).
Furthermore, we have:
(3 x 3) x (5 x 5) = (3 x 5) x (3 x 5)
Hence, the number whose square is the new number is:
3 x 5 = 15
QUESTION 9: FIND THE GREATEST NUMBER OF TWO DIGITS WHICH IS A PERFECT SQUARE.
ANSWER 9: WE KNOW THAT 102 IS EQUAL TO 100 AND 92 IS EQUAL TO 81.
SINCE 10 AND 9 ARE CONSECUTIVE NUMBERS, THERE IS NO PERFECT SQUARE BETWEEN 100 AND 81. SINCE 100 IS THE FIRST PERFECT SQUARE THAT HAS MORE THAN TWO DIGITS, 81 IS THE GREATEST TWO-DIGIT PERFECT SQUARE.

Question 10: Find the least number of three digits which is perfect square.
Answer 10: Let us make a list of the squares starting from 1.
1² = 1
2² = 4
3² = 9
4² = 16
5² = 25
6² = 36
7² = 49
8² = 64
9² = 81
10² = 100
The square of 10 has three digits. Hence, the least three-digit perfect square is 100.
QUESTION 11: FIND THE SMALLEST NUMBER BY WHICH 4851 MUST BE MULTIPLIED SO THAT THE PRODUCT BECOMES A PERFECT SUQARE.
ANSWER 11: Prime factorisation of 4851:
4851 = 3 x 3 x 7 x 7 x 11
Grouping them into pairs of equal factors:
4851 = (3 x 3) x (7 x 7) x 11
The factor, 11 is not paired. The smallest number by which 4851 must be multiplied such that the resulting number is a perfect square is 11.
QUESTION 12: FIND THE SMALLEST NUMBER BY WHICH 28812 MUST BE DIVIDED SO THAT THE QUOTIENT BECOMES A PERFECT SQUARE.
ANSWER 12: Prime factorisation of 28812:
28812 = 2 x 2 x 3 x 7 x 7 x 7 x 7
Grouping them into pairs of equal factors:
28812 = (2 x 2) x (7 x 7) x (7 x 7) x 3
The factor, 3 is not paired. Hence, the smallest number by which 28812 must be divided such that the resulting number is a perfect square is 3.
QUESTION 13: FIND THE SMALLEST NUMBER BY WHICH 1152 MUST BE DIVIDED SO THAT IT BECOMES A PERFECT SQUARE. ALSO, FIND THE NUMBER WHOSE SQUARE IS THE RESULTING NUMBER.
ANSWER 13:
Prime factorisation of 1152:
1152 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Grouping them into pairs of equal factors:
1152 = (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) x 2
The factor, 2 at the end is not paired. For a number to be a perfect square, each prime factor has to be paired. Hence, 1152 must be divided by 2 for it to be a perfect square.
The resulting number would be (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3).
Furthermore, we have:
(2 x 2) x (2 x 2) x (2 x 2) x (2 x 2) x (3 x 3) = (2 x 2 x 2 x 3) x (2 x 2 x 2 x 3)
Hence, the number whose square is the resulting number is:
2 x 2 x 2 x 3 = 24.

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FAQs on Chapter 3 - Squares and Square Roots (Ex-3.1) Part - 2, Class 8 Math RD Sharma Solutions - RD Sharma Solutions for Class 8 Mathematics

1. What are squares and square roots?

Ans. Squares are the result of multiplying a number by itself. For example, the square of 3 is 3 multiplied by 3, which equals 9. Square roots, on the other hand, are the numbers that, when multiplied by themselves, give the original number. So, the square root of 9 is 3, because 3 multiplied by 3 equals 9.

2. How do I find the square root of a number?

Ans. To find the square root of a number, you can use the long division method or the prime factorization method. In the long division method, you divide the number into groups of two digits from right to left, starting with the largest perfect square less than or equal to the leftmost group. Then, you find the square root of the leftmost group and subtract the square of that number from the leftmost group. This process is repeated until all groups are covered. The number you get after subtracting the squares is the remainder. To find the square root, you bring down the next group of digits and repeat the process. In the prime factorization method, you express the number as a product of its prime factors, and then take the square root of each factor.

3. What is the value of the square root of 0?

Ans. The square root of 0 is 0. This is because 0 multiplied by 0 equals 0. In general, the square root of any number multiplied by itself will give the original number. Since 0 multiplied by 0 is 0, the square root of 0 is 0.

4. Can the square root of a negative number be determined?

Ans. No, the square root of a negative number cannot be determined using real numbers. This is because when we multiply a number by itself, the result is always positive or zero. For example, the square of 3 is 9, and the square of -3 is also 9. However, we can use complex numbers to determine the square root of a negative number. Complex numbers involve the use of imaginary numbers, where the square root of -1 is denoted as "i". So, the square root of -9 can be represented as 3i or -3i.

5. What are some real-life applications of squares and square roots?

Ans. Squares and square roots have various applications in real-life situations. Some examples include: 1. Architecture: Square roots are used in construction and architecture to determine the measurements of structures, ensuring proper dimensions and stability. 2. Electrical Engineering: Square roots are used in electrical engineering to calculate voltage, current, and resistance in circuits. 3. Finance and Investment: Square roots are used in financial calculations, such as calculating compound interest, risk analysis, and stock market forecasting. 4. Physics: Square roots are used in physics to calculate distances, velocities, and accelerations, as well as in formulas related to force, energy, and power. 5. Medical Imaging: Square roots are used in medical imaging techniques, such as MRI and CT scans, to reconstruct images from the data collected by the machines.

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