GSEB Class 8 Maths Solutions Chapter 6 Square and Square Roots InText Questions

Try These (Page 90)

 

Question 1. Find the perfect square numbers between the following ranges:
1. 30 and 40
2. 50 and 60.
Answer:
1. To find the perfect square number between 30 and 40, we list squares of whole numbers:
\( 1 \times 1 = 1 \)
\( 2 \times 2 = 4 \)
\( 3 \times 3 = 9 \)
\( 4 \times 4 = 16 \)
\( 5 \times 5 = 25 \)
\( 6 \times 6 = 36 \)
\( 7 \times 7 = 49 \)
From this list, 36 is the only perfect square number that falls between 30 and 40.

2. To find the perfect square number between 50 and 60, we check the squares of whole numbers.
We know \( 7 \times 7 = 49 \) and \( 8 \times 8 = 64 \).
This shows that there is no perfect square number that exists between 49 and 64.
Therefore, there is no perfect square number between 50 and 60.
In simple words: For each range, list the squares of numbers. The square that fits between the given numbers is the answer. If no square fits, then there is no perfect square in that range.

Exam Tip: Remember that perfect squares are the result of multiplying a whole number by itself. Quickly list squares around the given range to find the answer.

 

Try These (Page 90)

 

Question 1. Can we say whether the following numbers are perfect squares? How do we know? Write five numbers which you can decide by looking at their ones digit that they are not square numbers?
1. 1057
2. 23453
3. 7928
4. 222222
5. 1069
6. 2061
Answer:
A number is a perfect square only if its units digit is 0, 1, 4, 5, 6, or 9. If a number ends with 2, 3, 7, or 8, it cannot be a perfect square.

1. For 1057:
The ending digit is 7. Since 7 is not a possible units digit for a perfect square, 1057 cannot be a square number.

2. For 23453:
The ending digit is 3. As 3 is not among the allowed units digits for perfect squares, 23453 cannot be a square number.

3. For 7928:
The ending digit is 8. Because 8 is not a valid units digit for a perfect square, 7928 cannot be a square number.

4. For 222222:
The ending digit is 2. Since 2 is not a units digit for any perfect square, 222222 cannot be a square number.

5. For 1069:
The ending digit is 9. A number ending in 9 might be a perfect square (e.g., \( 3^2=9 \), \( 7^2=49 \), \( 13^2=169 \)). We need to check further.
We find that \( 31 \times 31 = 961 \).
Also, \( 32 \times 32 = 1024 \).
And \( 33 \times 33 = 1089 \).
There is no natural number between 1024 and 1089 that is a square number. Therefore, 1069 cannot be a square number.

6. For 2061:
The ending digit is 1. A number ending in 1 might be a perfect square (e.g., \( 1^2=1 \), \( 9^2=81 \), \( 11^2=121 \)). We need to check further.
We find that \( 45 \times 45 = 2025 \).
Also, \( 46 \times 46 = 2116 \).
There is no natural number between 2025 and 2116 that is a square number. Therefore, 2061 is not a square number.

Five numbers that are not square numbers, identifiable by their ones digit, are those ending in 2, 3, 7, or 8. Examples include:
1234 (ends in 4, but is not a perfect square)
4312 (ends in 2)
5678 (ends in 8)
87543 (ends in 3)
1002007 (ends in 7)
In simple words: To check if a number is a perfect square, first look at its last digit. If it ends in 2, 3, 7, or 8, it is definitely not a perfect square. If it ends in 0, 1, 4, 5, 6, or 9, it might be, so you need to check further.

Exam Tip: Memorize the list of possible unit digits for perfect squares (0, 1, 4, 5, 6, 9) and impossible unit digits (2, 3, 7, 8). This is a quick trick to eliminate many non-square numbers.

 

Question 2. Write five numbers which you cannot decide just by looking at their units digit (or ones place) whether they are square numbers or not?
Answer:
Any natural number that ends with 0, 1, 4, 5, 6, or 9 may or may not be a square number. For these numbers, simply checking the units digit is not enough to determine if they are perfect squares.
Five such numbers are:
56790 (ends in 0)
3671 (ends in 1)
2454 (ends in 4)
76555 (ends in 5)
69209 (ends in 9)
In simple words: Numbers that end in 0, 1, 4, 5, 6, or 9 could be perfect squares, or they might not be. You need to do more calculations to find out. The last digit alone does not give the full answer.

Exam Tip: Be aware that while some unit digits rule out perfect squares, others only indicate a possibility. Further calculation (like finding the square root or checking numbers around it) is needed for numbers ending in 0, 1, 4, 5, 6, or 9.

 

Property 2.

If a number has 1 or 9 in the units place, then its square ends in 1.
For example:
\( (1)^2 = 1 \)
\( (9)^2 = 81 \)
\( (11)^2 = 121 \)
\( (19)^2 = 361 \)

 

Try These (Page 91)

 

Question 1. Which of \( 123^2 \), \( 77^2 \), \( 82^2 \), \( 161^2 \), \( 109^2 \) would end with digit 1?
Answer:
The squares of numbers that end in 1 or 9 will have 1 as their units digit.
From the given numbers:
- \( 123^2 \) ends with \( 3^2 = 9 \)
- \( 77^2 \) ends with \( 7^2 = 49 \), so it ends with 9.
- \( 82^2 \) ends with \( 2^2 = 4 \)
- \( 161^2 \) ends with \( 1^2 = 1 \)
- \( 109^2 \) ends with \( 9^2 = 81 \), so it ends with 1.
Therefore, the squares of 161 and 109 would end in 1.
In simple words: To find if a number's square ends in 1, you only need to look at its last digit. If the original number ends in 1 or 9, then its square will end in 1.

Exam Tip: Focus only on the unit digit of the base number to determine the unit digit of its square. This is a crucial shortcut for such questions.

 

Property 3.

When a square number ends in 6, then the number whose square it is, will have 4 or 6 in its unit place.

 

Try These (Page 91)

 

Question 1. Which of the following numbers would have digit 6 at unit place?
1. \( 19^2 \)
2. \( 24^2 \)
3. \( 26^2 \)
4. \( 36^2 \)
5. \( 34^2 \)
Answer:
A number's square will have 6 in its units place if the original number itself ends in 4 or 6.

1. For \( 19^2 \): The unit digit is 9. \( 9^2 = 81 \). So, \( 19^2 \) would not have a unit digit of 6; it would end in 1.

2. For \( 24^2 \): The unit digit is 4. \( 4^2 = 16 \). So, \( 24^2 \) would have 6 as its unit digit.

3. For \( 26^2 \): The unit digit is 6. \( 6^2 = 36 \). So, \( 26^2 \) would have 6 as its unit digit.

4. For \( 36^2 \): The unit digit is 6. \( 6^2 = 36 \). So, \( 36^2 \) would end in 6.

5. For \( 34^2 \): The unit digit is 4. \( 4^2 = 16 \). So, \( 34^2 \) would have 6 as its unit digit.
In simple words: Look at the last digit of each number. If that digit is 4 or 6, then when you square the number, its last digit will be 6. Otherwise, it won't be 6.

Exam Tip: To quickly find the unit digit of a squared number, just square its unit digit. For example, to find the unit digit of \( 34^2 \), just square 4 (which is 16), and the unit digit is 6.

 

Try These (Page 92)

 

Question 1. What will be the "ones digit" in the square of the following numbers?
1. 1234
2. 26387
3. 52698
4. 99880
5. 21222
6. 9106
Answer:
To determine the ones digit of a number's square, we only need to consider the ones digit of the original number and then square that digit.

1. For 1234:
The ones digit is 4. Squaring 4 gives \( 4^2 = 16 \). So, the ones digit of \( (1234)^2 \) will be 6.

2. For 26387:
The ones digit is 7. Squaring 7 gives \( 7^2 = 49 \). So, the ones digit of \( (26387)^2 \) will be 9.

3. For 52698:
The ones digit is 8. Squaring 8 gives \( 8^2 = 64 \). So, the ones digit of \( (52698)^2 \) will be 4.

4. For 99880:
The ones digit is 0. Squaring 0 gives \( 0^2 = 0 \). So, the ones digit of \( (99880)^2 \) will be 0.

5. For 21222:
The ones digit is 2. Squaring 2 gives \( 2^2 = 4 \). So, the ones digit of \( (21222)^2 \) will be 4.

6. For 9106:
The ones digit is 6. Squaring 6 gives \( 6^2 = 36 \). So, the ones digit of \( (9106)^2 \) will be 6.
In simple words: To find the last digit of a squared number, just square the last digit of the original number. The last digit of that result will be your answer.

Exam Tip: This method works because only the unit digit affects the unit digit of its square. This is a time-saving technique for calculations.

 

Property 4.

A square number can only have an even number of zeros at the end.

 

Property 5.

The squares of odd numbers are odd and the squares of even numbers are even.

 

Try These (Page 92)

 

Question 1. The square of which of the following numbers would be an odd number/an even number? Why?
1. 727
2. 158
3. 269
4. 1980
Answer:
Based on Property 5, the square of an odd number is always odd, and the square of an even number is always even.

1. For 727:
Since 727 is an odd number, its square \( (727)^2 \) will also be an odd number.

2. For 158:
Since 158 is an even number, its square \( (158)^2 \) will also be an even number.

3. For 269:
Since 269 is an odd number, its square \( (269)^2 \) will also be an odd number.

4. For 1980:
Since 1980 is an even number, its square \( (1980)^2 \) will also be an even number.
In simple words: An easy rule to remember is that if you square an odd number, you get an odd result. If you square an even number, you get an even result. Just check if the original number is odd or even.

Exam Tip: This property is fundamental. Simply classify the base number as odd or even to determine the nature of its square, rather than calculating the full square.

 

Question 2. Find the number of zeros in the square of the following numbers:
1. 60
2. 400
Answer:
According to Property 4, a perfect square can only have an even number of zeros at the end. When a number with zeros at the end is squared, the number of zeros in its square is double the number of zeros in the original number.

1. For 60:
The number 60 has 1 zero at the end.
Therefore, its square \( (60)^2 \) will have \( 1 \times 2 = 2 \) zeros at the end. \( (60)^2 = 3600 \).

2. For 400:
The number 400 has 2 zeros at the end.
Therefore, its square \( (400)^2 \) will have \( 2 \times 2 = 4 \) zeros at the end. \( (400)^2 = 160000 \).
In simple words: When you square a number that ends in zeros, simply double the count of zeros it has. For instance, one zero becomes two zeros, and two zeros become four zeros.

Exam Tip: Always double the count of trailing zeros when squaring a number ending in zeros. Ensure the final number of zeros is always even.

 

Property 6.

The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers.

 

Property 7.

There are \( 2n \) non-perfect square numbers between the squares of the numbers \( n \) and \( n + 1 \).

 

Try These (Page 94)

 

Question 1. How many natural numbers lie between \( 9^2 \) and \( 10^2 \)? Between \( 11^2 \) and \( 12^2 \)?
Answer:
According to Property 7, the number of non-perfect square numbers between the squares of two consecutive natural numbers \( n \) and \( n+1 \) is \( 2n \).

(a) For between \( 9^2 \) and \( 10^2 \):
Here, \( n = 9 \).
The number of natural numbers between \( 9^2 \) and \( 10^2 \) is \( 2 \times n \), which is \( 2 \times 9 = 18 \).

(b) For between \( 11^2 \) and \( 12^2 \):
Here, \( n = 11 \).
The number of natural numbers between \( 11^2 \) and \( 12^2 \) is \( 2 \times n \), which is \( 2 \times 11 = 22 \).
In simple words: To find how many numbers are between two consecutive squares (like \( n^2 \) and \( (n+1)^2 \)), just multiply the smaller number (\( n \)) by 2. This gives you the count of all the numbers that are not perfect squares in that range.

Exam Tip: Always identify the smaller number \( n \) in the consecutive pair \( n^2 \) and \( (n+1)^2 \). The formula \( 2n \) only counts the non-square numbers *between* the squares, not including the squares themselves.

 

Question 2. How many non-square numbers lie between the following pairs of numbers:
1. \( 100^2 \) and \( 101^2 \)
2. \( 90^2 \) and \( 91^2 \)
3. \( 1000^2 \) and \( 1001^2 \)
Answer:
Using the rule from Property 7, the number of non-perfect square numbers between \( n^2 \) and \( (n+1)^2 \) is \( 2n \).

1. For between \( 100^2 \) and \( 101^2 \):
Here, \( n = 100 \).
The count of non-square numbers is \( 2 \times n = 2 \times 100 = 200 \).
Therefore, 200 non-square numbers lie between \( 100^2 \) and \( 101^2 \).

2. For between \( 90^2 \) and \( 91^2 \):
Here, \( n = 90 \).
The count of non-square numbers is \( 2 \times n = 2 \times 90 = 180 \).
Therefore, 180 non-square numbers lie between \( 90^2 \) and \( 91^2 \).

3. For between \( 1000^2 \) and \( 1001^2 \):
Here, \( n = 1000 \).
The count of non-square numbers is \( 2 \times n = 2 \times 1000 = 2000 \).
Therefore, 2000 non-square numbers lie between \( 1000^2 \) and \( 1001^2 \).
In simple words: To quickly find how many non-square numbers are between two squares of consecutive numbers, just double the smaller number. So, for \( 100^2 \) and \( 101^2 \), you double 100 to get 200 numbers.

Exam Tip: This formula \( 2n \) is efficient for quickly determining the quantity of non-square numbers. Be careful to correctly identify \( n \) as the base of the smaller square.

 

Property 8.

The sum of the first \( n \) odd natural numbers is \( n^2 \).
If a number is a square number, it must be the sum of the successive odd numbers starting from 1.

 

Try These (Page 94)

 

Question 1. Find whether each of the following numbers is a perfect square or not?
1. 121
2. 55
3. 81
4. 49
5. 69
Answer:
To determine if a number is a perfect square using the repeated subtraction of odd numbers method, we subtract consecutive odd numbers (starting from 1) from the given number. If we reach 0, the number is a perfect square. The number of steps taken indicates its square root.

1. For 121:
\( 121 - 1 = 120 \)
\( 120 - 3 = 117 \)
\( 117 - 5 = 112 \)
\( 112 - 7 = 105 \)
\( 105 - 9 = 96 \)
\( 96 - 11 = 85 \)
\( 85 - 13 = 72 \)
\( 72 - 15 = 57 \)
\( 57 - 17 = 40 \)
\( 40 - 19 = 21 \)
\( 21 - 21 = 0 \)
Since we obtained 0 after subtracting 11 successive odd numbers, 121 is a perfect square, and its square root is \( \sqrt{121} = 11 \).

2. For 55:
\( 55 - 1 = 54 \)
\( 54 - 3 = 51 \)
\( 51 - 5 = 46 \)
\( 46 - 7 = 39 \)
\( 39 - 9 = 30 \)
\( 30 - 11 = 19 \)
\( 19 - 13 = 6 \)
\( 6 - 15 = -9 \)
Since we did not reach 0 (and went into negative numbers), 55 is not a perfect square.

3. For 81:
\( 81 - 1 = 80 \)
\( 80 - 3 = 77 \)
\( 77 - 5 = 72 \)
\( 72 - 7 = 65 \)
\( 65 - 9 = 56 \)
\( 56 - 11 = 45 \)
\( 45 - 13 = 32 \)
\( 32 - 15 = 17 \)
\( 17 - 17 = 0 \)
Since we obtained 0 after subtracting 9 successive odd numbers, 81 is a perfect square, and its square root is \( \sqrt{81} = 9 \).

4. For 49:
\( 49 - 1 = 48 \)
\( 48 - 3 = 45 \)
\( 45 - 5 = 40 \)
\( 40 - 7 = 33 \)
\( 33 - 9 = 24 \)
\( 24 - 11 = 13 \)
\( 13 - 13 = 0 \)
Since we obtained 0 after subtracting 7 successive odd numbers, 49 is a perfect square, and its square root is \( \sqrt{49} = 7 \).

5. For 69:
\( 69 - 1 = 68 \)
\( 68 - 3 = 65 \)
\( 65 - 5 = 60 \)
\( 60 - 7 = 53 \)
\( 53 - 9 = 44 \)
\( 44 - 11 = 33 \)
\( 33 - 13 = 20 \)
\( 20 - 15 = 5 \)
\( 5 - 17 = -12 \)
Since we did not reach 0 (and went into negative numbers), 69 is not a perfect square.
In simple words: To find if a number is a perfect square, keep taking away odd numbers in order (1, 3, 5, etc.). If you end up at zero, it's a perfect square. The count of odd numbers you subtracted is its square root. If you go past zero into negative numbers, it's not a perfect square.

Exam Tip: This method provides a clear way to verify perfect squares and find their roots. Be meticulous in your subtraction steps to avoid errors. Ensure you subtract all odd numbers in sequence.

 

Property 9.

The square of an odd number can be expressed as the sum of two consecutive natural numbers.

 

Try These (Page 95)

 

Question 1. Express the following as the sum of two consecutive integers.
1. \( 21^2 \)
2. \( 13^2 \)
3. \( 11^2 \)
4. \( 19^2 \)
Answer:
According to Property 9, the square of any odd number \( n \) can be written as the sum of two consecutive integers using the formulas: \( \frac{n^2 - 1}{2} \) and \( \frac{n^2 + 1}{2} \).

1. For \( n = 21 \):
\( n^2 = 21^2 = 441 \)
The first integer is \( \frac{21^2 - 1}{2} = \frac{441 - 1}{2} = \frac{440}{2} = 220 \).
The second integer is \( \frac{21^2 + 1}{2} = \frac{441 + 1}{2} = \frac{442}{2} = 221 \).
So, \( 21^2 = 220 + 221 = 441 \).

2. For \( n = 13 \):
\( n^2 = 13^2 = 169 \)
The first integer is \( \frac{13^2 - 1}{2} = \frac{169 - 1}{2} = \frac{168}{2} = 84 \).
The second integer is \( \frac{13^2 + 1}{2} = \frac{169 + 1}{2} = \frac{170}{2} = 85 \).
So, \( 13^2 = 84 + 85 = 169 \).

3. For \( n = 11 \):
\( n^2 = 11^2 = 121 \)
The first integer is \( \frac{11^2 - 1}{2} = \frac{121 - 1}{2} = \frac{120}{2} = 60 \).
The second integer is \( \frac{11^2 + 1}{2} = \frac{121 + 1}{2} = \frac{122}{2} = 61 \).
So, \( 11^2 = 60 + 61 = 121 \).

4. For \( n = 19 \):
\( n^2 = 19^2 = 361 \)
The first integer is \( \frac{19^2 - 1}{2} = \frac{361 - 1}{2} = \frac{360}{2} = 180 \).
The second integer is \( \frac{19^2 + 1}{2} = \frac{361 + 1}{2} = \frac{362}{2} = 181 \).
So, \( 19^2 = 180 + 181 = 361 \).
In simple words: To write an odd number's square as the sum of two numbers, use two easy formulas. For a number \( n \), the first number is \( n^2 \) minus 1, then divided by 2. The second number is \( n^2 \) plus 1, then divided by 2. Add these two results to get \( n^2 \).

Exam Tip: This pattern only works for the squares of odd numbers. Remember the two formulas for consecutive integers: \( (n^2-1)/2 \) and \( (n^2+1)/2 \).

 

Question 2. Do you think the reverse is also true, i.e., is the sum of any two consecutive positive integers a perfect square of a number? Give example to support your answer?
Answer:
No, the reverse is not always true. The sum of any two consecutive positive integers is not necessarily a perfect square.

For example:
- If we take the consecutive integers 5 and 6, their sum is \( 5 + 6 = 11 \). However, 11 is not a perfect square.
- If we take 21 and 22, their sum is \( 21 + 22 = 43 \). But 43 is not a perfect square.
- Consider 2 and 3, their sum is \( 2 + 3 = 5 \), which is not a perfect square.
This shows that while every square of an odd number can be expressed as a sum of two consecutive integers, not every sum of two consecutive integers is a perfect square.
In simple words: No, if you add any two numbers that come right after each other, the total is not always a perfect square. For example, 5 plus 6 equals 11, and 11 is not a perfect square.

Exam Tip: Always test such statements with counterexamples. Even one example that disproves the statement is enough to show it is not universally true.

 

Property 10.

The difference between the squares of two consecutive natural numbers is equal to the sum of the two numbers?
Examples:
\( 9^2 - 8^2 = 81 - 64 = 17 = 9 + 8 \)
\( 10^2 - 9^2 = 100 - 81 = 19 = 10 + 9 \)
\( 15^2 - 14^2 = 225 - 196 = 29 = 15 + 14 \)
\( 101^2 - 100^2 = 10201 - 10000 = 201 = 101 + 100 \)

 

Property 11.

If \( (n + 1) \) and \( (n - 1) \) are two consecutive even or odd natural numbers, then \( (n + 1) \times (n - 1) = n^2 - 1 \).
For example,
\( 10 \times 12 = (11 - 1) \times (11 + 1) = 11^2 - 1 \)
\( 11 \times 13 = (12 - 1) \times (12 + 1) = 12^2 - 1 \)
\( 25 \times 27 = (26 - 1) \times (26 + 1) = 26^2 - 1 \)

 

Try These (Page 95)

 

Question 1. Write the square of the following numbers, making use of the above pattern?
1. \( 111111^2 \)
2. \( 11111111^2 \)
Answer:
This pattern describes how powers of numbers consisting solely of the digit '1' can be calculated. For \( n \) number of 1s, the square will be \( 123...n...(n-1)...321 \).

1. For \( (111111)^2 \):
The number 111111 has six '1's. So, its square will follow the pattern up to 6 and then back down.
\( (111111)^2 = 12345654321 \)

2. For \( (11111111)^2 \):
The number 11111111 has eight '1's. So, its square will follow the pattern up to 8 and then back down.
\( (11111111)^2 = 1234567654321 \)
In simple words: When you square a number made only of ones, like 11 or 111, you get a special pattern. If it has six ones, the answer goes up to 6 and then back down to 1. If it has eight ones, it goes up to 8 and then back down to 1.

Exam Tip: This is a special pattern for repetitive unit '1's. The digits in the square increase from 1 to the number of '1's, then decrease back to 1.

 

Question 2. Find the square of the following numbers using the above pattern?
1. \( 6666667^2 \)
2. \( 66666667^2 \)
Answer:
The numbers given are of the form '6' repeated several times followed by '7'. The pattern for such numbers is not explicitly stated but is implicitly related to \( (2/3 \times 11...1)^2 \). However, by observing the results in the source, we can infer the pattern directly for these specific types of numbers.

1. For \( (6666667)^2 \):
This number has six '6's followed by a '7'. The pattern for its square appears to be '4' repeated \( (n-1) \) times, then an '8' repeated \( (n-1) \) times, followed by '89', where \( n \) is the count of repeating digits (here 6). Let's verify with the provided solution.
\( (6666667)^2 = 44444448888889 \)
(This means six '4's, then six '8's, then 89, which is 6 '4's + (6-1) '8's + '89' for the total 7 digits in the base number (6666667). This pattern is \( (n) \times 4 \) and \( (n-1) \times 8 \). It's more complex, but the solution provides direct mapping for this type of problem.)

2. For \( (66666667)^2 \):
This number has seven '6's followed by a '7'.
\( (66666667)^2 = 4444444488888889 \)
(This follows the pattern of seven '4's, then seven '8's, then 89.)
In simple words: For numbers like 666...67, the square follows a pattern. If there are 'n' sixes before the 7, the answer will start with 'n' fours, then 'n' eights, and then '89' at the very end.

Exam Tip: For specific patterns like those involving repeating digits, carefully observe the number of repetitions and how they translate into the digits of the square. These questions often test pattern recognition rather than direct computation.

 

Try These (Page 97)

 

Question 1. Find the square of the following numbers containing 5 in unit place?
1. 15
2. 95
3. 105
4. 205
Answer:
For a number ending in 5, its square can be calculated quickly. If the number is of the form \( (10x + 5) \), its square is \( x(x+1) \times 100 + 25 \), where \( x \) is the digit(s) before 5.

1. For 15:
Here, \( x = 1 \).
\( (15)^2 = 1 \times (1 + 1) \times 100 + 25 \)
\( = 1 \times 2 \times 100 + 25 \)
\( = 200 + 25 \)
\( = 225 \)

2. For 95:
Here, \( x = 9 \).
\( (95)^2 = 9 \times (9 + 1) \times 100 + 25 \)
\( = 9 \times 10 \times 100 + 25 \)
\( = 9000 + 25 \)
\( = 9025 \)

3. For 105:
Here, \( x = 10 \).
\( (105)^2 = 10 \times (10 + 1) \times 100 + 25 \)
\( = 10 \times 11 \times 100 + 25 \)
\( = 11000 + 25 \)
\( = 11025 \)

4. For 205:
Here, \( x = 20 \).
\( (205)^2 = 20 \times (20 + 1) \times 100 + 25 \)
\( = 20 \times 21 \times 100 + 25 \)
\( = 42000 + 25 \)
\( = 42025 \)
In simple words: To square a number that ends in 5, take the part of the number before the 5. Multiply that part by one more than itself. Then, put "25" at the end of that result. For example, for 15, take 1, multiply by 2 (which is 1+1) to get 2, then add 25 to get 225.

Exam Tip: This shortcut for squaring numbers ending in 5 is very useful for mental math and quick calculations. Remember the pattern: \( (\text{first digits}) \times (\text{first digits} + 1) \) followed by 25.

 

Try These (Page 99)

 

Question 1.
1. \( 11^2 = 121 \). What is the square root of 121?
2. \( 14^2 = 196 \). What is the square root of 196?
Answer:
The square root of a number is the value that, when multiplied by itself, gives the original number.

1. Given \( 11^2 = 121 \). This means that 11 multiplied by itself is 121.
Therefore, the square root of 121 is 11.

2. Given \( 14^2 = 196 \). This means that 14 multiplied by itself is 196.
Therefore, the square root of 196 is 14.
In simple words: The square root is the number you multiply by itself to get the original number. So if \( 11 \times 11 = 121 \), then 11 is the square root of 121.

Exam Tip: Understand that squaring and finding the square root are inverse operations. If \( a^2 = b \), then \( \sqrt{b} = a \).

 

Try These (Page 99)

 

Question 1.
\( (-1)^2 = 1 \). Is -1, a square root of 1?
\( (-2)^2 = 4 \). Is -2, a square root of 4?
\( (-9)^2 = 81 \). Is -9, a square root of 81?
Answer:
1. We know that \( (-1) \times (-1) = 1 \). So, \( (-1)^2 = 1 \). This indicates that -1 is indeed a square root of 1. A positive number generally has two square roots: one positive and one negative.

2. Similarly, \( (-2) \times (-2) = 4 \). So, \( (-2)^2 = 4 \). This means -2 is a square root of 4.

3. And \( (-9) \times (-9) = 81 \). So, \( (-9)^2 = 81 \). This confirms -9 is a square root of 81.

However, in mathematics, when the square root symbol \( \sqrt{} \) is used, it usually refers to the principal (non-negative) square root. Therefore, when we write \( \sqrt{16} \), it means the positive square root, which is 4, and not -4. Similarly, \( \sqrt{25} \) means the positive square root of 25, which is 5.
In simple words: Yes, a negative number multiplied by itself gives a positive number. So, -1 is a square root of 1, -2 is a square root of 4, and -9 is a square root of 81. But when we use the \( \sqrt{} \) symbol, we usually mean the positive square root.

Exam Tip: Remember that every positive number has both a positive and a negative square root. However, the radical symbol \( \sqrt{} \) specifically denotes the principal (positive) square root. Be clear about the context when answering questions about square roots.

 

Try These (Page 100)

 

Question 1. By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square, then find its square root?
1. 121
2. 55
3. 36
4. 49
5. 90
Answer:
To determine if a number is a perfect square using repeated subtraction, we consecutively subtract odd numbers (1, 3, 5, 7...) from the given number. If the final result is 0, the number is a perfect square, and the count of subtraction steps gives its square root. If the result is not 0 (e.g., negative), it is not a perfect square.

1. For 121:
\( 121 - 1 = 120 \)
\( 120 - 3 = 117 \)
\( 117 - 5 = 112 \)
\( 112 - 7 = 105 \)
\( 105 - 9 = 96 \)
\( 96 - 11 = 85 \)
\( 85 - 13 = 72 \)
\( 72 - 15 = 57 \)
\( 57 - 17 = 40 \)
\( 40 - 19 = 21 \)
\( 21 - 21 = 0 \)
We obtained 0 after subtracting 11 successive odd numbers. Therefore, 121 is a perfect square, and its square root is \( \sqrt{121} = 11 \).

2. For 55:
\( 55 - 1 = 54 \)
\( 54 - 3 = 51 \)
\( 51 - 5 = 46 \)
\( 46 - 7 = 39 \)
\( 39 - 9 = 30 \)
\( 30 - 11 = 19 \)
\( 19 - 13 = 6 \)
\( 6 - 15 = -9 \)
Since we reached a negative number and did not obtain 0, 55 is not a perfect square.

3. For 36:
\( 36 - 1 = 35 \)
\( 35 - 3 = 32 \)
\( 32 - 5 = 27 \)
\( 27 - 7 = 20 \)
\( 20 - 9 = 11 \)
\( 11 - 11 = 0 \)
We obtained 0 after subtracting 6 successive odd numbers. Therefore, 36 is a perfect square, and its square root is \( \sqrt{36} = 6 \).

4. For 49:
\( 49 - 1 = 48 \)
\( 48 - 3 = 45 \)
\( 45 - 5 = 40 \)
\( 40 - 7 = 33 \)
\( 33 - 9 = 24 \)
\( 24 - 11 = 13 \)
\( 13 - 13 = 0 \)
We obtained 0 after subtracting 7 successive odd numbers. Therefore, 49 is a perfect square, and its square root is \( \sqrt{49} = 7 \).

5. For 90:
\( 90 - 1 = 89 \)
\( 89 - 3 = 86 \)
\( 86 - 5 = 81 \)
\( 81 - 7 = 74 \)
\( 74 - 9 = 65 \)
\( 65 - 11 = 54 \)
\( 54 - 13 = 41 \)
\( 41 - 15 = 26 \)
\( 26 - 17 = 9 \)
\( 9 - 19 = -10 \)
Since we reached a negative number and did not obtain 0, 90 is not a perfect square.
In simple words: To check if a number is a perfect square, keep subtracting odd numbers (1, 3, 5, etc.) in order. If you reach zero, it's a perfect square, and the number of steps is its square root. If you go below zero, it's not a perfect square.

Exam Tip: This method is effective for smaller numbers. Ensure you subtract odd numbers sequentially. Any skipped odd number or incorrect subtraction will lead to an erroneous result.

 

Try These (Page 103)

 

Question 1. Can we say that if a perfect square is of n-digits, then its square root will have \( \frac{n}{2} \) digits if n is even or \( \frac{(n+1)}{2} \) if n is odd?
Answer:
Yes, it is true that this rule helps determine the number of digits in the square root of a perfect square.

Examples:
1. Consider 529:
The number 529 has 3 digits, so \( n = 3 \) (which is an odd number).
Using the formula for an odd number of digits: \( \frac{n+1}{2} = \frac{3+1}{2} = \frac{4}{2} = 2 \) digits.
The square root of 529 is 23, which indeed has 2 digits.

2. Consider 1296:
The number 1296 has 4 digits, so \( n = 4 \) (which is an even number).
Using the formula for an even number of digits: \( \frac{n}{2} = \frac{4}{2} = 2 \) digits.
The square root of 1296 is 36, which indeed has 2 digits.
In simple words: Yes, this rule is correct. If a perfect square has an even number of digits, divide that number by two to get the number of digits in its square root. If it has an odd number of digits, add one to that number and then divide by two to get the number of digits in its square root.

Exam Tip: This rule is very helpful for quickly estimating the number of digits in a square root without actually calculating it. Remember the two separate formulas for even and odd digit counts.

 

Try These (Page 105)

 

Question 1. Without calculating square roots, find the number of digits in the square root of the following numbers?
1. 25600
2. 100000000
3. 36864
Answer:
We can use the rule: if a perfect square has \( n \) digits, its square root will have \( \frac{n}{2} \) digits if \( n \) is even, or \( \frac{(n+1)}{2} \) digits if \( n \) is odd.

1. For 25600:
This number has 5 digits, so \( n = 5 \) (an odd number).
The number of digits in its square root will be \( \frac{n+1}{2} = \frac{5+1}{2} = \frac{6}{2} = 3 \) digits.

2. For 100000000:
This number has 9 digits, so \( n = 9 \) (an odd number).
The number of digits in its square root will be \( \frac{n+1}{2} = \frac{9+1}{2} = \frac{10}{2} = 5 \) digits.

3. For 36864:
This number has 5 digits, so \( n = 5 \) (an odd number).
The number of digits in its square root will be \( \frac{n+1}{2} = \frac{5+1}{2} = \frac{6}{2} = 3 \) digits.
In simple words: To find out how many digits a square root will have without calculating it, count the digits in the original number. If it's an even count, divide by two. If it's an odd count, add one to it, then divide by two. That result tells you how many digits are in the square root.

Exam Tip: Double-check whether \( n \) is odd or even before applying the corresponding formula. This helps in quick and accurate estimation of the number of digits.

 

Try These (Page 107)

 

Question 1. Estimate the value of the following to the nearest whole number?
1. \( \sqrt{80} \)
2. \( \sqrt{1000} \)
3. \( \sqrt{350} \)
4. \( \sqrt{500} \)
Answer:
To estimate the value of a square root to the nearest whole number, we find the two consecutive perfect squares that bracket the given number.

1. For \( \sqrt{80} \):
We know the perfect squares around 80:
\( 8^2 = 64 \)
\( 9^2 = 81 \)
Since 80 lies between 64 and 81 (i.e., \( 64 < 80 < 81 \)), this means \( 8^2 < 80 < 9^2 \).
Therefore, \( 8 < \sqrt{80} < 9 \).
Since 80 is much closer to 81 than to 64, \( \sqrt{80} \) is approximately 9.

2. For \( \sqrt{1000} \):
We know the perfect squares around 1000:
\( 30^2 = 900 \)
\( 31^2 = 961 \)
\( 32^2 = 1024 \)
Since 1000 lies between 961 and 1024 (i.e., \( 961 < 1000 < 1024 \)), this means \( 31^2 < 1000 < 32^2 \).
Therefore, \( 31 < \sqrt{1000} < 32 \).
Since 1000 is closer to 1024 than to 961, \( \sqrt{1000} \) is approximately 32.

3. For \( \sqrt{350} \):
We know the perfect squares around 350:
\( 18^2 = 324 \)
\( 19^2 = 361 \)
Since 350 lies between 324 and 361 (i.e., \( 324 < 350 < 361 \)), this means \( 18^2 < 350 < 19^2 \).
Therefore, \( 18 < \sqrt{350} < 19 \).
Since 350 is closer to 361 (difference 11) than to 324 (difference 26), \( \sqrt{350} \) is approximately 19.

4. For \( \sqrt{500} \):
We know the perfect squares around 500:
\( 22^2 = 484 \)
\( 23^2 = 529 \)
Since 500 lies between 484 and 529 (i.e., \( 484 < 500 < 529 \)), this means \( 22^2 < 500 < 23^2 \).
Therefore, \( 22 < \sqrt{500} < 23 \).
Since 500 is closer to 484 (difference 16) than to 529 (difference 29), \( \sqrt{500} \) is approximately 22.
In simple words: To estimate a square root, find the two perfect squares it sits between. The square root will be between the roots of those two squares. Then, see which perfect square it is closest to, and that will give you the closest whole number estimate for the square root.

Exam Tip: A good estimation involves identifying the two closest perfect squares. Always check which perfect square the given number is nearer to, as this determines the nearest whole number estimate for its square root.

 

Question 1. By repeated subtraction of odd numbers starting from 1, find whether the following numbers are perfect squares or not? If the number is a perfect square, then find its square root?
Answer:
4. For 49:
1. \( 49 - 1 = 48 \)
2. \( 48 - 3 = 45 \)
3. \( 45 - 5 = 40 \)
4. \( 40 - 7 = 33 \)
5. \( 33 - 9 = 24 \)
6. \( 24 - 11 = 13 \)
7. \( 13 - 13 = 0 \)
Since we achieved 0 after subtracting 7 consecutive odd numbers starting from 1, 49 is a perfect square. The square root of 49 is 7, as it took 7 steps.
5. For 90:
1. \( 90 - 1 = 89 \)
2. \( 89 - 3 = 86 \)
3. \( 86 - 5 = 81 \)
4. \( 81 - 7 = 74 \)
5. \( 74 - 9 = 65 \)
6. \( 65 - 11 = 54 \)
7. \( 54 - 13 = 41 \)
8. \( 41 - 15 = 26 \)
9. \( 26 - 17 = 9 \)
10. \( 9 - 19 = -10 \)
We did not reach 0 by subtracting successive odd numbers. Therefore, 90 is not a perfect square.
In simple words: To check if a number is a perfect square, keep subtracting odd numbers (1, 3, 5, etc.) from it. If you reach zero, it's a perfect square, and the number of subtractions is its square root. If you can't reach zero, it's not.

Exam Tip: Remember to subtract consecutive odd numbers (1, 3, 5, 7...) in sequence. If at any point the remainder becomes negative before reaching zero, the number is not a perfect square.

 

Try These (Page 103)

 

Question 1. Can we say that if a perfect square is of n-digits, then its square root will have \( \frac{n}{2} \) digits if n is even or \( \frac{(n+1)}{2} \) if n is odd?
Answer: Yes, it is true that the number of digits in the square root of a perfect square follows this rule:

Examples:
1. For 529 (which is a perfect square):
\( n = 3 \) (an odd number)
So, the number of digits in its square root is \( \frac{n+1}{2} = \frac{3+1}{2} = \frac{4}{2} = 2 \).
The actual square root of 529 is 23, which indeed has 2 digits.
2. For 1296 (which is a perfect square):
\( n = 4 \) (an even number)
So, the number of digits in its square root is \( \frac{n}{2} = \frac{4}{2} = 2 \).
The actual square root of 1296 is 36, which also has 2 digits.
In simple words: This rule helps us quickly find out how many digits the square root of a large number will have without actually calculating the square root. You just need to know if the original number has an even or odd number of digits.

Exam Tip: This property is useful for quickly verifying answers or estimating the magnitude of square roots, especially for larger numbers. Make sure to apply the correct formula for 'n' being even or odd.

 

Try These (Page 105)

 

Question 1. Without calculating square roots, find the number of digits in the square root of the following numbers?
1. 25600
2. 100000000
3. 36864
Answer:
1. For 25600:
The number of digits is \( n = 5 \) (an odd number).
So, the number of digits in its square root will be \( \frac{n+1}{2} = \frac{5+1}{2} = \frac{6}{2} = 3 \) digits.
2. For 100000000:
The number of digits is \( n = 9 \) (an odd number).
So, the number of digits in its square root will be \( \frac{n+1}{2} = \frac{9+1}{2} = \frac{10}{2} = 5 \) digits.
3. For 36864:
The number of digits is \( n = 5 \) (an odd number).
So, the number of digits in its square root will be \( \frac{n+1}{2} = \frac{5+1}{2} = \frac{6}{2} = 3 \) digits.
In simple words: We count how many digits are in the number. If it's an odd count, we add 1 and divide by 2 to get the number of digits in the square root. If it's an even count, we just divide by 2.

Exam Tip: To avoid errors, carefully count the number of digits 'n' in the given number before applying the formula. This method provides a quick way to predict the size of the square root.

 

Try These (Page 107)

 

Question 1. Estimate the value of the following to the nearest whole number?
1. \( \sqrt{80} \)
2. \( \sqrt{1000} \)
3. \( \sqrt{350} \)
4. \( \sqrt{500} \)
Answer:
1. For \( \sqrt{80} \):
We know that \( 8^2 = 64 \) and \( 9^2 = 81 \).
Since 80 is between 64 and 81, this means \( 64 < 80 < 81 \).
Therefore, \( 8^2 < 80 < 9^2 \), which implies \( 8 < \sqrt{80} < 9 \).
Since 80 is very close to 81, \( \sqrt{80} \) is closer to 9.
2. For \( \sqrt{1000} \):
We know that \( 30^2 = 900 \), \( 31^2 = 961 \), and \( 32^2 = 1024 \).
Since 1000 is between 961 and 1024, this means \( 961 < 1000 < 1024 \).
Therefore, \( 31^2 < 1000 < 32^2 \), which implies \( 31 < \sqrt{1000} < 32 \).
Since 1000 is closer to 1024 than to 961 (difference of 24 vs 39), \( \sqrt{1000} \) is closer to 32.
3. For \( \sqrt{350} \):
We know that \( 18^2 = 324 \) and \( 19^2 = 361 \).
Since 350 is between 324 and 361, this means \( 324 < 350 < 361 \).
Therefore, \( 18^2 < 350 < 19^2 \), which implies \( 18 < \sqrt{350} < 19 \).
Since 350 is closer to 361 than to 324 (difference of 11 vs 26), \( \sqrt{350} \) is closer to 19.
4. For \( \sqrt{500} \):
We know that \( 22^2 = 484 \) and \( 23^2 = 529 \).
Since 500 is between 484 and 529, this means \( 484 < 500 < 529 \).
Therefore, \( 22^2 < 500 < 23^2 \), which implies \( 22 < \sqrt{500} < 23 \).
Since 500 is closer to 484 than to 529 (difference of 16 vs 29), \( \sqrt{500} \) is closer to 22.
In simple words: To estimate a square root, find the two whole numbers whose squares are just below and just above the number. The square root will be between these two numbers, and closer to the one its square is nearer to.

Exam Tip: When estimating square roots, it's helpful to know the squares of numbers up to at least 30 by heart. This speeds up finding the two closest perfect squares for estimation.