RD Sharma Solutions Class 8 Squares And Square Roots Exercise 3.2

RD Sharma Class 8 Solutions Chapter 3 Ex 3.2 PDF Free Download

RD Sharma Solutions Class 8 Chapter 3 Exercise 3.2

1) The following number are not perfect squares. Give reason:

(i) 1547

(ii) 45743

(iii)8948

(iv) 333333

Solution:

If a number ends with 2, 3, 7 or 8, it cannot be a perfect square.

(i) Given number: 1547

The last digit of a number is 7. Therefore, 1547 cannot be a perfect square.

(ii) Given number: 45743

The last digit of a number is 3. Therefore, 45743 cannot be a perfect square.

(iii) Given number: 8948

The last digit of a number is 8. Therefore, 8948 cannot be a perfect square.

(iv) Given number: 333333

The last digit of a number is 3. Therefore, 333333 cannot be a perfect square.

2.) Show that the following numbers are not  perfect squares:

(i) 9327

(ii) 4058

(iii) 22453

(iv)743522

Solution:

If a number ends with 2, 3, 7 or 8, it cannot be a perfect square.

(i) Given number: 9327

The last digit of a number is 7. Therefore, 9327 cannot be a perfect square.

(ii) Given number: 4058

The last digit of a number is 8. Therefore, 4058 cannot be a perfect square.

(iii) Given number: 22453

The last digit of a number is 3. Therefore, 22453 cannot be a perfect square.

(iv) Given number: 743522

The last digit of a number is 2. Therefore, 743522 cannot be a perfect square.

 

3.) The square of which of the following numbers would be an odd number?

(i) 731

 (ii)3456

(iii) 5559

 (iv) 42008

Solution:

We know that the square of an odd number is always an odd number.

(i) Given number: 731

Since the number 731 is an odd number, the square of a number will be an odd number.

(ii)Given number: 3456

Since the number 3456 is an even number, the square of a number will not be an odd number.

(iii) Given number: 5559

Since the number 5559 is an odd number, the square of a number will be an odd number.

(iv) Given number: 42008

Since the number 42008 is an even number, the square of a number will not be an odd number.

Therefore, the squares of the numbers 731 and 5559 will be the odd numbers.

4.) What will be the unit digit of the squares of the following numbers?

(i) 52

(ii) 977

 (iii) 4583

(iv) 78367

 (v) 52698

(vi) 99880

(vii) 12796

(viii) 55555

 (ix) 53924

Solution:

The unit’s digit of a number is the last digit of the number.

So, to find the unit digits of the squares of the numbers are as follows:

(i) Given number: 52

The last digit of a number is 2

Therefore, the square of the unit digit is 22, which is equal to 4.

So, the unit digit is 4.

(ii) Given number: 977

The last digit of a number is 7

Therefore, the square of the unit digit is 72, which is equal to 49

So, the unit digit is 9.

(iii) Given number: 4583

The last digit of a number is 3

Therefore, the square of the unit digit is 32, which is equal to 9

So, the unit digit is 9.

(iv) Given number: 78367

The last digit of a number is 7

Therefore, the square of the unit digit is 72, which is equal to 49

So, the unit digit is 9.

(v) Given number: 52698

The last digit of a number is 8

Therefore, the square of the unit digit is 82, which is equal to 64

So, the unit digit is 4.

(vi) Given number: 99880

The last digit of a number is 0

Therefore, the square of the unit digit is 02, which is equal to 0

So, the unit digit is 0.

(vii) Given number: 12796

The last digit of a number is 6

Therefore, the square of the unit digit is 62, which is equal to 36

So, the unit digit is 6.

(viii) Given number: 55555

The last digit of a number is 5

Therefore, the square of the unit digit is 52, which is equal to 25

So, the unit digit is 5.

(ix) Given number: 53924

The last digit of a number is 4

Therefore, the square of the unit digit is 42, which is equal to 16

So, the unit digit is 6.

5.) Observe the following pattern:

1 + 3 = 22

1+ 3 + 5 = 32

1+ 3 +5 + 7= 42 and write the value of  1+ 3 + 5 + 7 + 9 + … up to n terms.

Solution:

From the above-given pattern of numbers, it is observed that the sum of the first n positive odd numbers is equal to the square of the n-th positive number.

Now substitute into the formula, we get

1 + 3 + 5 + 7 +….+ n = n2

Therefore, the required value of the given pattern is 1 + 3 + 5 + 7 +….+ n = n2

6.) Observe the following pattern:

22 – 12 = 2 +1

32 – 22 = 3 +2

42 – 32 = 4+ 3

52 – 42 = 5+ 4

and find the value of

i) 1002 – 992

(ii) 1112 – 1092

(iii) 992 – 962

Solution:

From the above-given pattern, it is observed that the difference between the squares of two consecutive numbers is the sum of the numbers itself. So, the formula for the given condition is as follows:

(n+1)2 — (n)2 = (n+1) +  n

So with the help of the formula, find the values,

(i) 1002 – 992

Substitute in the above formula, we get

= (99 + 1) + 99

Therefore, 1002 – 992= 199

(ii) 1112 – 1092

= 1112– 1102+ 1102 – 1092

= (111 + 110) + (110 + 109)

Therefore, 1112 – 1092= 440

(iii) 992 – 962

= 992 – 982 + 982 – 972 + 972 – 962

= 99 + 98 + 98 + 97 + 97 + 96

Therefore, 992 – 962= 585

7.) Which of the following triplets is Pythagorean?

(i) (8, 15, 17)

(ii) (18, 80, 82)

(iii) (14, 48, 51)

(iv) (10, 24, 51)

(v) (16, 63, 65)

(vi) (12, 35, 38)

Solution:

We know that, Pythagorean triplets (a, b, c) states that he sum of the squares of the two smallest numbers is equal to the square of the biggest number.

(i) Given triplet: (8, 15, 17)

The smallest two numbers are 8 and 15. Therefore, the sum of their squares is:

82 + 152 = 289 = 172

Thus, (8, 15, 17) is a Pythagorean triplet.

(ii) Given triplet: (18, 80, 82)

The smallest  two numbers are 18 and 80.Therefore, the sum of their squares is: 182 + 802= 6724 = 822

Thus, (18, 80, 82) is a Pythagorean triplet.

(iii) Given triplet: (14, 48, 51)

The smallest two numbers are 14 and 48. Therefore, the sum of their squares is:

142 + 482 = 2500, this is not equal to 512 = 2601

Thus, (14, 48, 51) is not a Pythagorean triplet.

(iv) Given triplet: (10, 24, 51)

The smallest two numbers are 10 and 24.Therefore, the sum of their squares is:

102 + 242 = 676 = 262

Thus, (10, 24, 26) is a Pythagorean triplet.

(v) Given triplet: (16, 63, 65)

The smallest two numbers are 16 and 63. Therefore, the sum of their squares is:

162 + 632 = 4225 = 652

Thus, (16, 63, 65) is a Pythagorean triplet.

(vi) Given triplet: (12, 35, 38)

The smallest two numbers are 12 and 35. Therefore, thesum of their squares is:

122 + 352 = 1369, which is not equal to 382 = 1444

Thus, (12, 35, 38) is not a Pythagorean triplet.

Hence, Only (i), (ii), (iv) and (v) are Pythagorean triplets.

8.) Observe the following pattern:

(1 x 2) + (2 x 3) = \(\frac{2\times 3\times 4}{3}\)

(1 x 2) + (2 x 3) + (3 x 4) = \(\frac{3\times 4\times 5}{3}\)

(1 x 2) + (2 x3 ) + (3 x 4) + ( 4x 5) = \(\frac{4\times 5\times 6}{3}\) and find the value of (1 x 2) + (2 x 3) + (3 x 4) + ( 4 x5) + (5 x 6)

Solution:

From the given pattern, it is observed taht the RHS of the three equalities is a fraction whose numerator is the multiplication of three consecutive numbers and whose denominator is 3.

When the biggest number (factor) on the LHS is 3, the multiplication of the three numbers on the RHS starts with 2.

when the biggest number (factor) on the LHS is 4, the multiplication of the three numbers on the RHS starts with 3.

when the biggest number (factor) on the LHS is 5, the multiplication of the three numbers on the RHS starts with 4.

Using the above-given pattern, (1 x 2) + (2 x 3) + (3 x 4) + (4 x 5) + (5 x 6) has 6 as the largest number. Therefore, the multiplication of the three numbers on the RHS will starts with 5.

Therefore,

1 x 2+ 2 x3 + 3 x 4 + 4 x 5 + 5 x 6 = \(\frac{5\times 6\times 7}{3} = 70\)

9.) Observe the following pattern:

1 = \(\frac{1}{2}\) {1 x ( 1+ 1)}

1 +2 = \(\frac{1}{2}\) {2 x {2 +2)}

1 + 2 +3 = \(\frac{1}{2}\) {3 x (3 + 1)}

1 + 2 + 3 +4 =\(\frac{1}{2}\) {4 x (4+ 1)} and find the values of each of the following :

(i) 1+ 2 + 3 + 4 + 5 … + 50

(ii) 31 + 32+ … + 50

Solution:

From the given data, observe the numbers on the RHS of the given three equalities.

In 1st  equality, the biggest number on the LHS is 1,

so it has 1, 1 and 1 as the three numbers.

The second equality, the biggest number on the LHS is 2,

so it has 2, 2 and 1 as the three numbers

The third equality, the biggest number on the LHS is 3,

so it has 3, 3 and 1 as the three numbers.

The fourth equality, the biggest number on the LHS is 4,

so it has 4, 4 and 1 as the three numbers.

Hence, if the biggest number on the LHS is n, the three numbers on the RHS will be n, n and 1.

So, with the help of this property, we can calculate the sums for (i) and (ii) as follows:

(i) 1 + 2 + 3 +…..  + 50

= \(\frac{1}{2}\) x 50 x (50 + 1)

1 + 2 + 3 +…..  + 50 = 1275 —-(1)

(ii) The sum can be expressed as the difference of the two sums as follows:

31 + 32 +…. + 50

= (1 + 2 + 3 +…. + 50) – ( 1 + 2 + 3 +  +30) —-(2)

substitute the value of (1) in (2), we get

Now, to find the ( 1 + 2 + 3 +… +30)

1 + 2+…. +30 = \(\frac{1}{2}\)  (30 x (30 + 1)) = 465 —(3)

substitute the value of (1) and (3) in (2), we get

= 1275 — 465

Therefore, 31 + 32 + …+ 50=810

11.) Which of the following numbers are squares of even numbers: 121, 225, 256, 324, 1296, 6561, 5476, 4489, 373758

Solution:

When the last digit of a number is odd, the square of a number cannot be an even number. So leave the odd numbers 121, 225, 6561 and 4489.

So, for those even numbers like 256, 324, 1296, 5476 and 373758, use the prime factorisation method to make it as a pair of equal factors.

For the given number,

(i) 256 =2x2x2x2x2x2x2x2 = (2 x 2) x (2 x 2) x (2 x 2) x (2 x 2)

Here, all the factors are paired. So, 256 is a perfect square of an even number.

(ii) 324 =2x2x3x3x3x3 = (2 x 2) x (3 x 3) x (3 x 3)

Here, all the factors are paired. So, 324 is a perfect square of an even number.

(iii) 1296 =2x2x2x2x3x3x3x3 = (2 x 2) x (2 x 2) x (3 x 3) x (3 x 3)

Here, all the factors are paired. So, 1296 is a perfect square of an even number.

(iv) 5476 = 2 x 2 x 37 x 37 = (2 x 2) x (37 x 37)

Here, all the factors are paired. So, 5476 is a perfect square of an even number.

(v) 373758 =2 x 3 x 7 x 11 x 809

Here, the factors are not paired. Each factor appears only once. Hence, 373758 is not a perfect square of an even number.

Therefore, the numbers that are the squares of even numbers are 256, 324, 1296 and 5476.

12.) By just examining the units digit, can you tell which of the following cannot be whole squares?

(i) 1026

(ii) 1028

(iii) 1024

(iv) 1022

(v) 1023

(vi) 1027

Solution:

If the unit’s digit of a number is 2, 3, 7 or 8, the number cannot be a whole square.

(i) Given number: 1026

The last digit of a number is 6. Therefore, 1026 can be a perfect square.

(ii) Given number: 1028

The last digit of a number is 8. Therefore, 1028 cannot be a perfect square.

(iii) Given number: 1024

The last digit of a number is 4. Therefore, 1024 can be a perfect square.

(iv) Given number: 1022

The last digit of a number is 2. Therefore, 1022 cannot be a perfect square.

(v) Given number: 1023

The last digit of a number is 3. Therefore, 1023 cannot be a perfect square.

(vi) Given number: 1027

The last digit of a number is 7. Therefore, 1027 cannot be a perfect square.

Therefore, it is observed that the numbers that cannot be a whole square are: 1028, 1022, 1023 and 1027

13.) Write five numbers which you cannot decide whether they are squares.

Solution:

It is known that if a number having the unit digit as 2, 3, 7 or 8, those numbers cannot be a perfect square.

On the other hand, if a number having the unit digit as 1, 4, 5, 6, 9 or 0, those numbers might be a perfect square. But in some cases, it is necessary to check whether the number is a perfect square or not.

But we cannot quickly decide whether the following numbers are squares of a number by applying the above two conditions for certain cases like,

1111, 1444, 1555, 1666, 1999

14.) Write five numbers which you cannot decide whether they are square just be looking at the units digit.

Solution:

It is known that if a number having the unit digit as 2, 3, 7 or 8, those numbers cannot be a perfect square.

On the other hand, if a number having the unit digit as 1, 4, 5, 6, 9 or 0, those numbers might be a perfect square. But in some cases, it is necessary to check whether the number is a perfect square or not.

But we cannot quickly decide whether the following numbers are squares of a number by applying the above two conditions for certain cases like,

1111, 1001, 1555, 1666 and 1999

15.) Write True (T) and false (F) for the following statements.

(i) The number of digits in a square number is even.

Soln: False

consider an example:

if 100 is the square of a number 10, but the number of digits in 100 is three, which is not an even number.

(ii) The square of a prime number is prime.

Soln: False

Suppose, if p is a prime number, then the square of a prime number becomes p2.

It has at least three factors such as 1, p and p2.

Since phas more than two factors, it is not a prime number.

(iii) The sum of two square numbers is a square number.

Soln: False

Consider, 1 is the square of a number (1 = 12). But the sum of a square number i.e., 1 + 1 = 2, is not a square number.

(iv) The difference of two square numbers is a square number.

Soln: False

Consider an example,

4 and 1 are squares of  a number (4 = 22, 1 = 12). But the difference of a square number i.e., 4 — 1 = 3,  is not a square number.

(v) The product of two square numbers is a square number.

Soln: True

Assume that, if a2 and bare two squares, their product is a2 x b2 = (ab)2, which is a square number.

(vi) No square number is negative.

Soln: True

It is known that the square of a negative number will be positive because negative times negative is positive. example (-2)= (-2) x (-2) = 4

(vii) There is no square number between 50 and 60

Soln: True

There are no square numbers between 50 and 60 because 72 = 49 and 82 = 64.

Since the numbers 7 and 8 are consecutive numbers

(viii) There are fourteen square number up to 200.

Soln: True

It is observed that there are 14 square numbers below 200.

Because 142 is equal to 196, that is below 200.

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