Class 8 Maths Chapter 3 Square Identities Questions and Answers Kerala Syllabus

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SCERT Class 8 Maths Chapter 3 Solutions Square Identities

Class 8 Kerala Syllabus Maths Solutions Chapter 3 Square Identities Questions and Answers

Square Identities Class 8 Questions and Answers Kerala Syllabus

Squares of Sums (Page No. 40)

Question 1.
Calculate the squares below in your head:
(i) 15²
(ii) 25²
(iii) 33²
(iv) \(\left(5 \frac{1}{2}\right)^2\)
(v) \(\left(10 \frac{1}{2}\right)^2\)
(vi) \(\left(25 \frac{1}{2}\right)^2\)
(vii) \(\left(5 \frac{1}{5}\right)^2\)
Answer:
(i) 15² = 10² + 5² + 2 × 10 × 5 = 225
(ii) 25² = 20² + 5² + 2 × 20 × 5 = 625
(iii) 33² = 30² + 3² + 2 × 30 × 3 = 1089
(iv) \(5 \frac{1^2}{2}=5^2+5+\frac{1}{2^2}=30 \frac{1}{4}\)
(v) \(10 \frac{1}{2}^2=10^2+10+\frac{1}{2^2}=110 \frac{1}{4}\)
(vi) \(25 \frac{1}{2}^2=25^2+25+\frac{1}{2^2}=650 \frac{1}{4}\)
(vii) \(5 \frac{1}{5}^2=5^2+2+\frac{1}{5^2}=27 \frac{1}{25}\)

Question 2.
In the general identity on the square of a sum of two numbers, what special case do we get by taking one of the numbers as 2? Do the calculations below in your head, using this:
(i) What is 22²?
(ii) What is 52²?
(iii) 25² = 625; What is 27²?
Answer:
(x + 1)² = x² + 2 × x × 2 + 2² = x² + 4x + 4
(i) 22² = (20 + 2)²
= 20² + 2 × 20 × 2 + 2²
= 400 + 80 + 4
= 484

(ii) 52² = (50 + 2)²
= 50² + 2 × 50 × 2 + 2²
= 2500 + 200 + 4
= 2704

(iii) 25² = 625
27² = (25 + 2)²
= 25² + 2 × 25 × 2 + 2²
= 625 + 100 + 4
= 729

Question 3.
43² = 1849
(i) What is 44²?
(ii) What is 46²?
Answer:
43² = 1849
(i) 44² = (43 + 1)²
= 43² + 2 × 43 × 1 + 1²
= 1849 + 86 + 1
= 1936

(ii) 46² = (43 + 3)²
= 43² + 2 × 43 × 3 + 3²
= 1849 + 258 + 9
= 2116

Textbook Page No. 45

Question 1.
Calculate the squares of some odd numbers and check the following statement. Explain why they are true:
(i) The square of any odd number is odd.
(ii) The square of any odd number leaves a remainder of 1 on division by 4.
(iii) The square of any odd number leaves a remainder of 1 on division by 8.
Answer:

(i) Let the odd number be, 2n + 1 where n = 0, 1, 2, 3,…..
(2n + 1)² = 4n² + 4n + 1 = 2(2n² + 2n) + 1
∴ The square of an odd number is of the form 2n + 1, which is an odd number.

(ii) (2n + 1)² = 4n² + 4n + 1 = 4(n² + n) + 1
That means the square of an odd number is 1 more than a multiple of 4.
∴ It leaves a remainder of 1 on division by 4.

(iii) (2n + 1)² = 4n(n + 1) + 1
Among the two consecutive numbers n and n + 1, one is always even.
So n(n + 1) is always even.
Thus, 4n(n + 1) is divisible by 8.
Hence, the square is of the form multiple of 8 + 1.
That means it leaves a remainder of 1 on division by 8.

Question 2.
Calculate the remainders on dividing the squares of some natural numbers by 4. Explain why the Statements below are true:
(i) The square of any natural number divided by 4 leaves a remainder of 0 or 1.
(ii) Any natural number that leaves a remainder of 2 or 3 on division by 4 is not a perfect square.
Answer:
The natural numbers on division by 4 are of the form 4n, 4n + 1, 4n + 2, 4n + 3, where n = 0, 1, 2,…
The possible remainders are 0, 1, 2, 3.
(i) Even natural numbers can be written as 2n
Squaring it gives (2n)² = 4n²
Here, the remainder is 0.
Odd natural numbers can be written as 2n + 1
Squaring it gives,
(2n + 1)² = 4n² + 4n + 1 = 4(n² + n) + 1
Here, the remainder is 1.
∴ The square of any natural number divided by 4 leaves a remainder of 0 or 1.

(ii) The square of any natural number divided by 4 leaves a remainder of 0 or 1.
∴ Any natural number leaving a remainder of 2 or 3 when divided by 4 is not a perfect square.

Question 3.
See this pattern:
3 = 2² – 1²
5 = 3² – 2²
7 = 4² – 3²
Check whether some other odd numbers can also be written as the difference of two perfect squares. Explain why all odd numbers greater than 1 can be written like this.
(Hint: Recall the general form of an odd number seen in class 7).
Answer:
All odd numbers greater than 1 can be written as the difference of two perfect squares.
Specifically, the difference of the squares of two consecutive natural numbers.
That is, (n + 1)² – n² = 2n + 1, always an odd number.
So, for any odd number n, it can be written as (n + 1)² – n².

Question 4.
Give reasons for the fact that the square of any number ending in 1 also ends in 1. What about numbers ending in 5? And numbers ending in 6?
Answer:
The last digit of a square is determined by the last digit of the original number.
The numbers ending in 1, 5, or 6 will have squares ending in 1, 5, or 6, respectively.
Because their last digits, when multiplied by themselves, result in the same digit.

Squares of Differences (Page No. 50)

Question 1.
Calculate the squares below in your head:
(i) 29²
(ii) 38²
(iii) 999²
(iv) \(\left(9 \frac{1}{2}\right)^2\)
(v) (9.7)²
Answer:
(i) 29² = (30 – 1)²
= 30² – 2 × 30 × 1 + 1²
= 900 – 60 + 1
= 841

(ii) 38² = (40 – 2)²
= 40² – 2 × 40 × 2 + 2²
= 1600 – 160 + 4
= 1444

(iii) 999² = (1000 – 1)²
= 1000000 – 2000 + 1
= 998001

(iv) \(\left(9 \frac{1}{2}\right)^2=9 \times 10+\frac{1}{4}=90 \frac{1}{4}\)
Or
\(\left(9 \frac{1}{2}\right)^2=\left(10-\frac{1}{2}\right)^2=10^2+\left(\frac{1}{2}\right)^2-10=90 \frac{1}{4}\)

(v) (9.7)² = (10 – 0.3)²
= 10² + (0.3)² – 2 × 10 × 0.3
= 100 + 0.09 – 6
= 94.09

Question 2.
See these computations:
3² – (2 × 3) = 3 = 2² – 1
4² – (2 × 4) = 8 = 3² – 1
5² – (2 × 5) = 15 = 4² – 1
Explain the general principle of these using algebra.
Answer:
3² – (2 × 3) = 3 = 2² – 1
4² – (2 × 4) = 8 = 3² – 1
5² – (2 × 5) = 15 = 4² – 1
…………………………………..
…………………………………..
n² – (2 × n) = (n – 1)² – 1

Textbook Page No. 53, 54

Question 1.
See these computations:
\(\left(\frac{1}{2}\right)^2+\left(1 \frac{1}{2}\right)^2=2 \frac{1}{2}\); 2 = 2 × 1²
\(\left(1 \frac{1}{2}\right)^2+\left(2 \frac{1}{2}\right)^2=8 \frac{1}{2}\); 8 = 2 × 2²
\(\left(2 \frac{1}{2}\right)^2+\left(3 \frac{1}{2}\right)^2=18 \frac{1}{2}\); 18 = 2 × 3²
(i) Write the next two lines following this pattern.
(ii) Explain the general principle of these using algebra.
Answer:

Question 2.
Some natural numbers can be written as the difference of two perfect squares in two different ways. For example
24 = 7² – 5² = 5² – 1²
32 = 9² – 7² = 6² – 2²
40 = 11² – 9² = 7² – 3²
Write the next few multiples of 8 as the difference of two perfect squares in two different ways. Explain algebraically how we can do this for all multiples of 8 starting with 24.
Answer:
The next few lines are:
48 = 13² – 11² = 8² – 4²
56 = 15² – 13² = 9² – 5²
64 = 17² – 15² = 10² – 6²
Algebraic explanation is: 8x = (2x + 1)² – (2x – 1)² where x is 3, 4, 5,……….

Question 3.
See how some numbers are written as the difference of two perfect squares in two different ways:
15 = 8² – 7² = 4² – 1²
21 = 11² – 10² = 5² – 2²
35 = 18² – 17² = 6² – 1²
(i) What are the different ways in which the numbers 15, 21, and 35 can be written as the product of two factors?
(ii) Find two more numbers of this type, which can be written as the difference of two perfect squares in two different ways.
(iii) What general result can we form from this?
Answer:
(i) 15 = (8 – 7) × (8 + 7) = 1 × 15
15 = (4 – 1) × (4 + 1) = 3 × 5
21 = (11 – 10) × (11 + 10) = 1 × 21
21 = (5 – 2) × (5 + 2) = 3 × 7
35 = (18 – 17) × (18 + 17) = 1 × 35
35 = (6 – 1) × (6 + 1) = 5 × 7

(ii) 45 = 9² – 6² = 7² – 2²
65 = 33² – 32² = 9² – 4²

(iii) a² – b² = (a – b)(a + b)

Question 4.
Compute the following products by writing them as the difference of two squares.
(i) 78 × 22
(ii) 301 × 299
(iii) \(2 \frac{1}{3} \times 1 \frac{2}{3}\)
(iv) 10.7 × 9.3
Answer:

Question 5.
Find the largest of each of the following pairs of products, without actual multiplication:
(i) 75 × 25; 76 × 24
(ii) 76 × 24; 74 × 26
(iii) 10.6 × 9.4; 10.4 × 9.6
Answer:

Differences of Squares (Page No. 58, 59, 60)

Question 1.
Do the computations below in your head:
(i) 68² – 32²
(ii) \(\left(3 \frac{1}{2}\right)^2-\left(2 \frac{1}{2}\right)^2\)
(iii) 3.6² – 1.4²
Answer:
(i) 68² – 32² = (68 – 32)(68 + 32)
= 36 × 100
= 3600

(ii) \(\left(3 \frac{1}{2}\right)^2-\left(2 \frac{1}{2}\right)^2=\left(\frac{7}{2}\right)^2-\left(\frac{5}{2}\right)^2=\left(\frac{7-5}{2}\right)\left(\frac{7+5}{2}\right)\)
= 1 × 6
= 6

(iii) 3.6² – 1.4² = (3. 6 – 1.4)(3.6 + 1.4) = 11

Question 2.
Note the pattern in the computations below:
3² – 2² = 5 = 3 + 2
4² – 3² = 7 = 4 + 3
5² – 4² = 9 = 5 + 4
(i) Write the next two computations following these.
(ii) Write the general principle in these as an algebraic equation.
(iii) Write this general principle in ordinary language.
Answer:
(i) 6² – 5² = 11 = 6 + 5
7² – 6² = 13 = 7 + 6

(ii) n² – (n – 1)² = 2n – 1 = n + (n – 1)

(iii) The difference between the square of an integer and the square of the preceding integer is equal to the sum of those two integers.

Question 3.
In a calendar sheet, take a square containing nine numbers and mark the numbers on the left and right, top, and bottom of the middle number:

Calculate the products of the left and right numbers and the top and bottom numbers, and also their difference
8 × 10 = 80
2 × 16 = 32
80 – 32 = 48
Do this for other squares of nine numbers. Explain using algebra why the difference is 48 in all cases.
(Hint: It is convenient to take the middle number as x)
Answer:
Let the middle number be x.
The number to the left is x – 1.
The number to the right is x + 1.
The number on top is x – 7.
The number on the bottom is x + 7.
Since, in a calendar, the numbers in a row differ by 1, and the numbers in a column differ by 7.
Product of left and right numbers be (x – 1)(x + 1) = x² – 1
Product of top and bottom numbers be (x – 7)(x + 7) = x² – 49
Difference = (Product of left and right) – (Product of top and bottom)
= (x² – 1) – (x² – 49)
= x² – 1 – x² + 49
= 48
Therefore, the difference between the product of the left and right numbers and the product of the top and bottom numbers is always 48.

Question 4.
As in the previous problem, mark a square of nine numbers in a calendar. Mark the four numbers in the corners:

Calculate the products of the diagonal pairs of numbers and find their difference:
15 × 3 = 45
1 × 17 = 17
45 – 17 = 28
Do this for other squares of nine numbers. Explain using algebra why the difference is 28 in all cases.
Answer:
Let the middle number be x.
The number to the top left is x – 8.
The number to the top right is x – 6.
The number on the bottom left is x + 6.
The number on the bottom right is x + 8.
Product of first diagonal part = (x – 8)(x + 8) = x² – 64
Product of second diagonal part = (x – 6)(x + 6) = x² – 36
Difference = (x² – 36) – (x² – 64)
= x² – 36 – x² + 64
= 28
Therefore, the difference between the products of the diagonal pairs of numbers is 28.

Question 5.
A square has a perimeter of 20 centimetres. A rectangle has one side 2 centimetres longer and one side 2 centimetres shorter than a side of this square.
(i) What is the perimeter of the rectangle?
(ii) What are the areas of the square and the rectangle?
Answer:
Perimeter of the square = 20 cm
One side = \(\frac {20}{2}\) = 5 cm
(i) One side of the rectangle = 5 + 2 = 7 cm
∴ Other side = 5 – 2 = 3 cm
Perimeter of the rectangle = 2 × (7 + 3)
= 2 × 10
= 20 cm

(ii) Area of the square = side²
= 5²
= 25 cm²
Area of the rectangle = 7 × 3 = 21 cm²

Question 6.
One side of a rectangle is longer than the side of a square, and the other side is equally shorter than the side of the square.
(i) What can we say about the perimeters of the square and the rectangle?
(ii) Which has the larger area, the square or the rectangle?
Answer:
Let the side of the square be x cm
One side of the rectangle = x + a
Other side = x – a (where a > 0)
(i) The perimeter of the square = 4x
Perimeter of the rectangle = 2[(x + a) + (x – a)]
= 2(2x)
= 4x
∴ The perimeter of the rectangle is equal to the perimeter of the square.

(ii) Area of the square = x²
Area of the rectangle = (x + a)(x – a) = x² – a²
Here x² – a² < x²
∴ The square has the larger area.

Class 8 Maths Chapter 3 Kerala Syllabus Square Identities Questions and Answers

Class 8 Maths Square Identities Questions and Answers

Question 1.
Calculate the squares below in your head:
(i) 101²
(ii) 9.2²
Answer:
(i) 101² = 100² + 2 × 100 × 1 + 1²
= 10000 + 200 + 1
= 10201

(ii) 9.2² = (9 + 0.2)²
= 9² + 2 × 9 × 0.2 + 0.2²
= 81 + 3.6 + 0.04
= 84.64

Question 2.
23² = 529
(i) What is 24²?
(ii) What is 26²?
Answer:
23² = 529
(i) 24² = (23 + 1)²
= 23² + 2 × 23 × 1 + 1²
= 529 + 46 + 1
= 576

(ii) 26² = (23 + 3)²
= 23² + 2 × 23 × 3 + 3²
= 529 + 138 + 9
= 676

Question 3.
\(\left(1 \frac{1}{2}\right)^2+\left(3 \frac{1}{2}\right)^2+\left(5 \frac{1}{2}\right)^2\) = _____________
Answer:

Question 4.
Find the value of \(\left(1 \frac{1}{2}\right)^2+\left(2 \frac{1}{2}\right)^2+\left(3 \frac{1}{2}\right)^2+\left(4 \frac{1}{2}\right)^2\)
Answer:

Question 5.
\(\left(19 \frac{1}{2}\right)^2+\left(20 \frac{1}{2}\right)^2\) = _____________
Answer:

Question 6.
101² – 100²
Answer:
101² – 100² = 101 + 100 = 201

Question 7.
99999² – 99998²
Answer:
99999² – 99998² = 99999 + 99998 = 199,997

Question 8.
\(\left(99 \frac{1}{2}\right)^2+\left(100 \frac{1}{2}\right)^2\) = _____________
Answer:

Question 9.
From the following, which of the numbers will be the sum of the digits in the ones and tens place of a square of a number that ends in 5?
(a) 2
(b) 5
(c) 7
(d) 3
Answer:
(c) 7

Question 10.
If the square of a number ending in 5 is of the form 56xy. Then y – x = _____________
Answer:
y – x = 5 – 2 = 3

Question 11.
If the square of a number ending in 5 is of the form 110pq. Then p × q = _____________
Answer:
p × q = 2 × 5 = 10

Question 12.
10² + 2² = 100 + 4 = 104. By using this, write 208 as the sum of the two perfect squares.
Answer:
208 = 2 × 104
= 2(10² + 2²)
= (10 + 2)² + (10 – 2)²
= 12² + 8²

Question 13.
3² + 2² = 9 + 4 = 13.
By using this, write 26 as the sum of the two perfect squares.
Answer:
26 = 2 × 13
= 2(3² + 2²)
= (3 + 2)² + (3 – 2)²
= 5² + 1²

Question 14.
11² + 10² = 121 + 100 = 221.
By using this, write 442 as the sum of the two perfect squares.
Answer:
442 = 2 × 221
= 2(11² + 10²)
= (11 + 10)² + (11 – 10)²
= 21² + 1²

Class 8 Maths Chapter 3 Notes Kerala Syllabus Square Identities

→ The square of the sum of two numbers is equal to the sum of the squares of these numbers and twice their product.

→ (x + y)² = x² + y² + 2xy for all numbers x and y.

→ To get the square of one more than a number, we need only add the square of the number, twice the number, and one.

→ (x + 1)² = x² + 2x + 1 for all numbers x.

→ The square of the difference of two numbers is twice their product subtracted from the sum of their squares.

→ (x – y)² = x² + y² – 2xy for all numbers x and y.

→ The product of any two numbers is the difference of the squares of half their sum and half their difference.

→ The difference of the squares of two numbers is the product of the sum and difference of the numbers.

→ x² – y² = (x + y)(x – y) for all numbers x and y.

→ Sum of digits of a perfect square is either 1 or 4 or 7, or 9.

→ The difference of the squares of two consecutive natural numbers is their sum.

In the first chapter, we discussed the problems related to finding the squares of a number. But in this chapter, ‘Square Identities’, we are going to discuss the algebraic formulas used to simplify expressions that involve the squares of numbers. These identities follow a particular pattern, making it easier to expand without doing lengthy multiplication each time. In this chapter, we are going to discuss squares of sums, squares of differences, and differences of squares.

Squares of Sums
We have seen a method to calculate the squares of many numbers in the section Squares.

How do we find the square of 36?
Answer:

36² = (30 + 6)²
= 30² + (2 × 30 × 6) + 6²
= 900 + 360 + 36
= 1296
Here we wrote 36 as the sum of 30 + 6 to calculate its square.
So the square of 36 can be written as (30 + 6)² = 30² + (2 × 30 × 6) + 6²

In general, we can say that the square of the sum of two numbers is equal to the sum of the squares of these numbers and twice their product.
In algebraic form, we can write it as (x + y)² = x² + y² + 2xy for all numbers x and y.
We can see it geometrically as a result of areas:

Worksheet – 1

Question 1.
1. 42²
2. 56²
3. 73²
4. 84²
5. 93²
Answer:
1. 42² = 40² + 2 × 40 × 2 + 2²
= 1600 + 160 + 4
= 1764

2. 56² = 50² + 2 × 50 × 6 + 6²
= 2500 + 600 + 36
= 3136

3. 73² = 70² + 2 × 70 × 3 + 3²
= 4900 + 420 + 9
= 5329

4. 84² = 80² + 2 × 80 × 4 + 4²
= 6400 + 640 + 16
= 7056

5. 93² = 90² + 2 × 90 × 3 + 3²
= 8100 + 540 + 9
= 8649

To get the square of one more than a number, we need only add the square of the number, twice the number, and one.
In general, (x + 1)² = x² + 2x + 1 for all numbers x

For example, find 31²?
Answer:
31² = 30² + 2 × 30 + 1²
= 900 + 60 + 1
= 961

Find 21²?
Answer:
21² = 20² + 2 × 20 + 1²
= 400 + 40 + 1
= 441

Find 51²?
Answer:
51² = 50² + 2 × 50 + 1²
= 2500 + 100 + 1
= 2601

Find 61²?
Answer:
61² = 60² + 2 × 60 + 1²
= 3600 + 120 + 1
= 3721

Find 71²?
Answer:
71² = 70² + 2 × 70 + 1²
= 4900 + 140 + 1
= 5041

Find 101²?
Answer:
101² = 100² + 2 × 100 + 1²
= 10000 + 200 + 1
= 10201

Suppose take y as \(\frac {1}{2}\), we get
(x + \(\frac {1}{2}\))² = x² + x + \(\frac {1}{4}\) for all numbers x
It can also be written in the form, (x + \(\frac {1}{2}\))² = x(x + 1) + \(\frac {1}{4}\)

For example, Find \(1 \frac{1^2}{2}\)?
Answer:
\(\mathbf{1} \frac{1}{2}^{\mathbf{2}}=1^2+1+\frac{1}{4}=2 \frac{1}{4}\) = 2.25

Find \(9 \frac{1^2}{2}\)?
Answer:
\(9 \frac{1^2}{2}=9 \times 10+\frac{1}{4}=90 \frac{1}{4}\) = 90.25

Find \(10 \frac{1^2}{2}\)?
Answer:
\(10 \frac{1}{2}^2=10 \times 11+\frac{1}{4}=110 \frac{1}{4}\) = 110.25

Find \(19 \frac{1^2}{2}\)?
Answer:
\(19 \frac{\frac{1}{2}^2}{2}=19 \times 20+\frac{1}{4}=380 \frac{1}{4}\) = 380.25

Find \(6 \frac{1^2}{2}\)?
Answer:
\(6 \frac{1}{2}^2=6^2+6+\frac{1}{4}=42 \frac{1}{4}\) = 42.25

Find \(12 \frac{1^2}{2}\)?
Answer:
\(12 \frac{1^2}{2}\) = 12² + 12 + \(\frac {1}{4}\)
= 144 + 12 + \(\frac {1}{4}\)
= 156\(\frac {1}{4}\)
= 42.25

For any non-zero number x, we get
\(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\)

For example, Find \(3 \frac{1}{3}^2\)?
Answer:
\(3 \frac{1}{3}^2=\left(3+\frac{1}{3}\right)^2=3^2+\frac{1}{3^2}+2=9+\frac{1}{9}+2=11 \frac{1}{9}\)

Find \(4 \frac{1}{4}^2\)?
Answer:
\(4 \frac{1}{4}^2=\left(4+\frac{1}{4}\right)^2=4^2+\frac{1}{4^2}+2=16+\frac{1}{16}+2=18 \frac{1}{16}\)

Find \(10 \frac{1}{10}^2\)?
Answer:
\(10 \frac{1}{10}^2=\left(10+\frac{1}{10}\right)^2=10^2+\frac{1}{10^2}+2=100+\frac{1}{100}+2=102 \cdot \frac{1}{100}\)

Worksheet – 2

Question 1.
1. \(\left(5 \frac{1}{5}\right)^2\)
2. \(\left(30 \frac{1}{30}\right)^2\)
3. \(\left(100 \frac{1}{100}\right)^2\)
4. \(\left(3 \frac{1}{2}\right)^2+\left(3 \frac{1}{3}\right)^2\)
5. \(\left(10 \frac{1}{2}\right)^2+\left(10 \frac{1}{3}\right)^2\)
Answer:

Square of a number ending in 5:
The square of any number with the last digit (the digit in the one’s place) 5 has 25 as the last two digits.
The last digit of any number is the remainder on division by 10.
So a number ending in 5, divided by 10, leaves a remainder of 5.
The general form of such a number is 10n + 5.
The square of a number in this form is:
(10n + 5)² = (10n)² + (2 × 10n × 5) + 5²
= 100n² + 100n + 25
= 100(n² + n) + 25
= 100n(n + 1) + 25
The last two digits of 100n(n + 1) are zeroes.
∴ The last two digits of 100n(n + 1) + 25 is 2 and 5

For example: Find 35²?
Answer:
35² = 100 × 3 × 4 + 25 = 1225

Find 55²?
Answer:
55² = 100 × 5 × 6 + 25 = 3025

Remainders on dividing natural numbers by 3:
If we consider any number, it will be a multiple of 3 or the number that leaves a remainder of 1 or 2 on division by 3.
That means, the general form of a number which is not a multiple of 3 is 3n + 1, where n = 0, 1, 2, 3,… or 3n + 2, where n = 0, 1, 2, 3,……….

The square of any non-multiple of 3 leaves a remainder of 1 on division by 3.
That means, (3n + 1)² = 9n² + 2 × 3n × 1 + 1
= 9n² + 6n + 1
= 3(3n² + 2n) + 1
= multiple of 3 + 1
Here, the remainder is 1.
(3n + 2)² = 9n² + 12n + 4
= 3(3n² + 4n) + 4
= 3(3n² + 4n) + 3 + 1
= 3(3n² + 4n + 1) + 1
= multiple of 3 + 1
Here also the remainder is 1.
Any perfect square leaves a remainder of 0 or 1 on division by 3.
Any number that leaves a remainder of 2 on division by 3 is not a perfect square.

Worksheet – 3

Question 1.
Calculate
(i) 45²
(ii) 205²
(iii) 555²
Answer:
(i) 45² = (100 × 4 × 5) + 25
= 2000 + 25
= 2025

(ii) 205² = (100 × 20 × 21) + 25
= 42000 + 25
= 42025

(iii) 555² = (100 × 55 × 56) + 25
= 308000 + 25
= 308025

Squares of Differences
The square of the difference of two numbers is twice their product subtracted from the sum of their squares.
If x and y are any two numbers.
Then, (x – y)² = x² + y² – 2xy for all numbers x and y.

For example: Compute 58².
Answer:
58² = (60 – 2)²
= 60² – 2 × 60 × 2 + 2²
= 3600 – 240 + 4
= 3364

If the equations of squares of sums and differences are added together, we get (x + y)² + (x – y)² = 2(x² + y²)
Therefore, 2(x² + y²) = (x + y)² + (x – y)²
The differences of this two identities is, 4xy = (x + y)² – (x – y)²
That means xy = \(\left(\frac{x+y}{2}\right)^2-\left(\frac{x-y}{2}\right)^2\)

The product of any two numbers is the difference of the squares of half their sum and half their difference.
If all the multiples of 4 starting with 8 can be written as the difference of two perfect squares.

For example: Compute 125 × 75.
Answer:
125 × 75 = \(\left(\frac{125+75}{2}\right)^2-\left(\frac{125-75}{2}\right)^2\)
= 100² – 25²
= 9375

Worksheet – 3

Compute the following.
(i) 112 × 12
(ii) 144 × 14
(iii) 17²
(iv) 39²
Answer:
(i) 112 × 12 = \(\left(\frac{112+12}{2}\right)^2-\left(\frac{112-12}{2}\right)^2\)
= 62² – 50²
= 1344

(ii) 144 × 14 = \(\left(\frac{114+14}{2}\right)^2-\left(\frac{114-14}{2}\right)^2\)
= 79² – 65²
= 2016

(iii) 17² = (20 – 3)²
= 20² – 2 × 20 × 3 + 3²
= 400 – 120 + 9
= 289

(iv) 39² = (40 – 1)²
= 40² – 2 × 40 × 1 + 1²
= 1600 – 80 + 1
= 1521

Differences of Squares
The difference of the squares of two numbers is the product of the sum and difference of the numbers.
x² – y² = (x + y)(x – y) for all numbers x and y.
If we take y as 1, then we get the equation as x² – 1 = (x + 1)(x – 1)

For three consecutive natural numbers, the product of the first and the last is one less than the square of the middle number.

For example: Compute 68² – 32²?
Answer:
68² – 32² = (68 + 32)(68 – 32)
= 100 × 36
= 3600

Worksheet – 4

Question 1.
Calculate
(i) 42² – 17²
(ii) \(\left(10 \frac{1}{2}\right)^2-\left(9 \frac{1}{2}\right)^2\)
Answer:
(i) 42² – 17² = (42 + 17)(42 – 17)
= 59 × 25
= 1475

(ii) \(\left(10 \frac{1}{2}\right)^2-\left(9 \frac{1}{2}\right)^2=\left(10 \frac{1}{2}+9 \frac{1}{2}\right)\left(10 \frac{1}{2}-9 \frac{1}{2}\right)\)
= 20 × 1
= 20