Kerala Syllabus Class 8 Mathematics - Unit 3 Square Identities - Questions and Answers | Teaching Manual
Questions and Answers for Class 8 Mathematics - Unit 3 Square Identities - Study Notes | Text Books Solution STD 8 - Maths: Unit 3 Square Identities - Questions and Answers | ഗണിതം - വർഗസമവാക്യങ്ങൾ - ചോദ്യോത്തരങ്ങൾ
എട്ടാം ക്ലാസ്സ് Mathematics - Unit 3 Square Identities എന്ന പാഠം ആസ്പദമാക്കി തയ്യാറാക്കിയ ചോദ്യോത്തരങ്ങള്. ഈ അധ്യായത്തിന്റെ Teachers Handbook, Teaching Manual എന്നിവ ഡൗൺലോഡ് ചെയ്യാനുള്ള ലിങ്ക് ചോദ്യോത്തരങ്ങളുടെ അവസാനം നൽകിയിട്ടുണ്ട്. പുതിയ അപ്ഡേറ്റുകൾക്കായി ഞങ്ങളുടെ Telegram Channel ൽ ജോയിൻ ചെയ്യുക.
Kerala Syllabus STD 8 Maths: Unit 3 Square Identities - Textbook Solutions
♦ Textbook Activities (Textbook Page No: 40)
Squares of sums
1) Calculate the squares below in head:
Answer:
(i) 15², (ii) 25², (iii) 33², (iv) (5½)², (v) (10½)², (vi) (25½)², (vii) (5⅕)²
Answer:
(i) 15²
15² = (14 + 1)²
= 14² + 2 × 14 × 1 + 1
= 196 + 28 + 1 = 225
(ii) 25²
25² = (24 + 1)²
= 24² + 2 × 24 × 1 + 1
= 576 + 48 + 1 = 625
(iii) 33²
33² = (32 + 1)²
= 32² + 2 × 32 × 1 + 1
= 1024 + 64 + 1 = 1089
(iv) (5½)²
(5½)² = (5 + ½)²
= 5² + 5 + ¼
= 25 + 5 + ¼ = 30 ¼
(v) (10½)²
(10½)² = (10 + ½)²
= 10² + 10 + ¼
= 100 + 10 + ¼ = 110 ¼
(vi) (25½)²
(25½)² = (25 + ½)²
= 25² + 25 + ¼
= 625 + 25 + ¼ = 650 ¼
(vii) (5⅕)²
(5⅕)² = (5 + ⅕)²
= 5² + 5²/1 + 2
= 25 + ¹⁄₂₅ + 2
= 27¹⁄₂₅
(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 head, using this:
(i) What is 22²?
(ii) What is 52²?
(iii) 25² = 625; What is 27²?
Answer:
Take y as 2 in the identity (x + y)² = x² + y² + 2xy
(x + 2)² = x² + 2² + 2 × x × 2
= x² + 4 + 4x = x² + 4x + 4
(i) 22² = (20 + 2)² = 20² + 4 × 20 + 4
= 400 + 80 + 4 = 484
(ii) 52² = (50 + 2)² = 50² + 4 × 50 + 4
= 2500 + 200 + 4 = 2704
(i) 25² = 625
27² = (25 + 2)² = 25² + 4 × 25 + 4
= 625 + 100 + 4 = 729
(3) 43² = 1849
(i) What is 44²?
(ii) What is 46²?
Answer:
(i) 43² = 1849
44² = (43 + 1)² = 43² + 2 × 43 + 1
= 1849 + 86 + 1 = 1936
(ii) 46² = (43 + 3)² = 43² + 6 × 43 + 9
= 1849 + 258 + 9 = 2116
♦ Textbook Activities (Textbook Page No: 45)
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 remainder 1 on division by 4
(iii) The square of any odd number leaves remainder 1 on division by 8
Answer:
1, 3, 5,… are odd numbers. Their squares, 1, 9, 25,... are also odd numbers.
(i) The general form of all odd numbers is (2n + 1), (n = 0, 1, 2, 3, ....).
Their squares are
(2n + 1)² = (2n)² + (2n × 1 × 2) + 1²
= 4n² + 4n + 1
= 2 (2n² + 2n) + 1
When an odd number is divided by 2, the remainder is 1. Similarly, when the square of an odd number is divided by 2, the remainder is also 1. Therefore, the square of any odd number is always an odd number.
(ii) The general form of all odd numbers is (2n+1), (n = 0, 1, 2, 3, ....) and their squares are
(2n+1)² = 4 (n² + n) + 1.
When the square 4n² + 4n+ 1 is divided by 4, the remainder is 1.
(iii) The square of any odd number can be written in the form 4n² + 4n + 1.
(2n + 1²) = 4(n² + n) + 1
= 4 × 2 (n² + n / 2) + 1 = 8 (n² + n / 2) + 1
When an odd number is divided by 8, the remainder is 1.
2. Calculate the remainders on dividing the squares of some natural numbers by 4. Explain why the statements below are true:
(i) The squares of any natural number divided by 4 leaves remainder 0 or 1.
(ii) Any natural number which leaves remainder 2 or 3 on division by 4 is not a perfect square.
Answer:
1² = 1 = 4 × 0 + 1
2² = 4 = 4 × 1 + 0
3² = 9 = 4 × 2 + 1
4² = 14 = 4 × 4 + 0
5² = 25 = 4 × 6 + 1
(i) The general form of any natural number is either 2n or 2n+1. If we take the square of these as
(2n)² = 4n²
(2n+1)² = 4n² + 4n + 1 = 4(n² + n) + 1
That is, when the square of any natural number is divided by 4, the remainder will be 0 or 1.
(ii) The general form of numbers that gives a remainder of 2 when divided by 4 is 4n + 2.
The general form of numbers that gives a remainder of 3 when divided by 3 is 4n+ 3. We have proven that when a number is even, its square is 4n, and when it's odd, the square is 4n +1. Therefore, there are no other possible forms of squares of natural numbers when divided by 4. That is, they will not be of the form 4n + 2 or 4n+3. Therefore, numbers that leaves a remain- der of 2 or 3 when divided by 4 are not perfect squares.
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).
3 = 2² − 1²
5 = 3² − 2²
7 = 4² − 3²
9 = 5² − 4²
11 = 6² − 5²
13 = 7² − 6²
The general form of all odd numbers is 2n +1.
2n + 1 = (n + 1)² − n² (n = 0, 1, 2, 3, .....)
(n + 1)² − n² = n² + 2n + 1 − n² = 2n + 1
So all odd numbers greater than 1 can be written as the difference of two squares.
4. Give reasons for the fact that the square of any number ending in 1, also end in 1. What about numbers ending in 5?
And numbers ending in 6?
Numbers that ending in 1 are 1, 11, 21, 31, .....
1² =1, 11² = 121, 21² = 441, 31² = 961,.......
The general form of numbers that end in 1 is 10n + 1.
Let's calculate the square
(10n+1)² = (10n)² + 2 × 10n × 1 + 1²
= 100n² + 20n + 1
= 10 (10n² + 2n) + 1,
Now again, we get a number that is 1 added to a multiple of 10. This shows that numbers ending in 1 have squares that also end in 1. Numbers that end in 5 are
5, 15, 25, 35, ..........
5² = 25, 15² = 225, 25² = 625, 35² = 1225
The general form of numbers that end in 5 is 10n + 5.
Let's calculate the square
(10n+5)² = 100n² + 100n + 25= 100 (n² + n) + 25
A number that 25 added to the multiple of 100 will end in 5.
Numbers that ending in 6 are
6, 16, 26, 36, .....
6² = 36, 16² = 256, 26² = 676
The general form of numbers that end in 6 is 10n + 6.
Let's calculate the square (10n+6) = (10n)² + 2 × 10n × 6 + 6²
= 100n² + 120n + 36
= 100n² + 100n + 20n + 20 + 16
= 100 (n² + n) +20 (n + 1) + 16
Numbers that 16 added to the multiple of 100 and 20 end in 6.
♦ Textbook Activities (Textbook Page No: 50)
Squares of differences
The square of the difference of two numbers is twice their product subtracted from the sum of their squares
1) Calculate the squares below in head:
(i) 29², (ii) 38², (iii) 999², (iv) (9½)², (vi) (9.7)²
Answer:
(i) 29²
29² = (30 − 1)²
= 30² + 1² − (2 × 30 × 1)
= 900 + 1 − 60
= 841
(i) 38²
38² = (40 − 2)²
= 40² + 2² − (2 × 40 × 2)
= 1600 + 4 − 160
= 1444
(i) 999²
999² = (1000 − 1)²
= 1000² + 1² − (2 × 1000 × 1)
= 1000000 + 1 − 2000
= 998001
(iv) (9½)²
(9½)² = (10 − ½)²
= 10² + (½)² − (2 × 10 × ½)
= 100 + ¼ − 10
= 90¼
(vi) (9.7)²
(9.7)² = (10 − 0.3)²
= 10² + (0.3)² − (2 × 10 × 0.3)
= 100 + 0.09 − 6
= 94.09
(2) See these computations:
3² − (2 × 3) = 3 = 2² − 1
4² − (2 × 4) = 8 = 3² − 1
3² − (2 × 5) = 15 = 4² − 1
Explain the general principle of these using algebra.
Answer:
Let's assume x is a natural number such that x² − (2 × x) = (x − 1)² − 1
The left side of this equation = x² − 2x
The right side of this equation = (x − 1)² − 1
= x² − 2x + 1 − 1
= x² − 2x
From this
We get x² − 2x = (x − 1)² − 1
If x = 7
7² − (2 × 7) = 35 = 6² − 1
If x = 8
8² − (2 × 8) = 48 = 7² − 1
The product of any two numbers is the difference of the squares of half their sum and half their difference
♦ Textbook Activities (Textbook Page No: 53, 54)
(1) See these computations:
(½)² + (1½)² = 2½ 2 = 2 × 1²
(1½)² + (2½)² = 8½ 8 = 2 × 2²
(2½)² + (3½)² = 18½ 18 = 2 × 3²
(i) Write the next two lines following this pattern.
(ii) Explain the general principle of these using algebra.
Answer:
(i) (3½)² + (4½)² = 32½ 32 = 2 × 4²
(4½)² + (5½)² = 50½ 50 = 2 × 5²
(ii) Let (x − ½) be the first number and (x + ½) be the next number in each.
(x − ½)² = x² + (½)² − (2 × x × ½)
= x² + ¼ − x
(x + ½)² = x² + (½)² + (2 × x × ½)
= x² + ¼ + x
(x − ½)² + (x + ½)² = x² + ¼ − x + x² + ¼ + x
= 2x² + ½
Let's check by giving values of x like 1, 2, 3... The algebraic form of the terms on the right-hand side of this pattern is 2x². Let's verify by substituting x with values like 1, 2, 3, .....
(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:
48 = 13² − 11² = 8² − 4²
56 = 15² − 13² = 9² − 5²
64 = 17² − 15² = 10² − 6²
(x + y)² = x² + 2xy + y²
(x − y)² = x² − 2xy + y²
(x + y)² − (x − y)² = 4xy
4xy = (x + y)² − (x − y)²
Consider the equation 4xy = (x + y)² − (x − y)²
We can write 24 as 4xy
24 = 4 × 6 × 1 (Here x = 6, y = 1)
24 = 4 × 3 × 2 (Here x = 3, y = 2)
If we use the equation 4xy = (x + y)² − (x − y)²
We get
24 = (6 + 1)² − (6 − 1)² = 7² − 5²
24 = (3 + 2)² − (3 − 2)² = 5² − 1²
Expressing 32 in the form 4xy
32 = 4 × 8 × 1 (here x = 8, y = 1)
32 = 4 × 4 × 2 (here x = 4, y = 2)
From this we get
32 = (8 + 1)² − (8 − 1)² = 9² − 7²
24 = (4 + 2)² − (4 − 2)² = 6² − 2²
Expressing 40 in the form 4xy
40 = 4 × 10 × 1 (here x = 10, y = 1)
32 = 4 × 5 × 2 (here x = 5, y = 2)
From this we get
40 = (10 + 1)² − (10 − 1)² = 11² − 9²
40 = (5 + 2)² − (5 − 2)² = 7² − 3²
Express the numbers 48, 56, and 64 in the same form.
(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 = 1 × 15 = 3 × 5, 21 = 1 × 21 = 3 × 7, 35 = 1 × 35 = 5 × 7
(ii) • 32 = 16 × 2 = ((16 + 2) / 2)² − ((16 − 2) / 2)²
32 = 9² − 7²
• 32 = 8 × 4 = ((8 + 4) / 2)² − ((8 − 4) / 2)²
32 = 6² − 2²
• 64 = 16 × 4 = ((16 + 4) / 2)² − ((16 − 4) / 2)² = 10² − 6²
• 64 = 32 × 2 = ((32 + 2) / 2)² − ((32 − 2) / 2)² = 17² − 15²
(iii) If the sum of two distinct factors x and y of a natural number is odd, the number itself can't be expressed as a difference of two perfect squares like ((x + y)/2)² − ((x − y) / 2)²
Example: 20 = 10 × 2 = 5 × 4
We can write 20 = ((10 + 2)/2)² − ((10 − 2) / 2)² = 6² − 4²
If we take 20 = 20 = 5 × 4
20 = ((5 + 4)/2)² − ((5 − 4) / 2)²
= (4½)² − (½)² not a natural number.
So, 20 can't be written as the difference of two perfect squares using the factors 5 and 4.
If the sum of the two distinct factors of a natural number is odd, the number itself can't be expressed as a difference of two perfect squares.
(4) Compute the following products by writing them as the difference of two squares.
(i) 78 × 22, (ii) 301 × 299, (iii) 2⅓ × 1⅔, (iv) 10.7 × 9.3
Answer:
(i) 78 × 22
78 × 22 = ((78 + 22)/2)² − ((78 − 22)/2)²
= 50² − 28²
= 2500 − 784
= 1716
28² = (30 − 2)²
= 30² + 2² − (2 × 30 × 2)
= 900 + 4 − 120
= 784
(ii) 301 × 299
301 × 299 = ((301 + 299)/2)² − ((301 − 299)/2)²
= 300² − 1²
= 90000 − 1
= 89999
(ii) 2⅓ × 1⅔
2⅓ × 1⅔ = ⁷⁄₃ × ⁵⁄₃ ((⁷⁄₃ + ⁵⁄₃)/2)² − ((⁷⁄₃ − ⁵⁄₃)/2)² = 2² − (⅓)²
= 4 − ⅑
= ³⁵⁄₉ = 3⁸⁄₉
(iv) 10.7 × 9.3
10.7 × 9.3 = ((10.7 + 9.3)/2)² − ((10.7 − 9.3)/2)²
= 10² − (0.7)²
= 100 − 0.49
= 99.51
(5) Find the largest of each of the following pair 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:
(i) 75 × 25 76 × 24
75 × 25 = ((75 + 25)/2)² − ((75 − 25)/2)²
= 50² − 25²
76 × 24 = ((76 + 24)/2)² − ((76 − 24)/2)²
= 50² − 26²
Here, the answer becomes larger when 25² is subtracted from 50². The larger product comes from 75 × 25.
(ii) 76 × 24 74 × 26
76 × 24 = ((76 + 24)/2)² − ((76 − 24)/2)²
= 50² − 26²
74 × 26 = ((74 + 26)/2)² − ((74 − 26)/2)²
= 50² − 24²
Here, the answer becomes larger when 24² is subtracted from 50². The larger product comes from 74 × 26.
(iii) 10.6 × 9.4 10.4 × 9.6
10.6 × 9.4 = ((10.6 + 9.4)/2)² − ((10.6 − 9.4)/2)²
= 10² − (0.6)²
10.4 × 9.6 = ((10.4 + 9.6)/2)² − ((10.4 − 9.6)/2)²
= 10² − (0.4)²
Here, the answer becomes larger when (0.4)² is subtracted from 10². The larger product comes from 10.4 × 9.6
♦ Textbook Activities (Textbook Page No: 58, 59, 60)
Differences of squares
The difference of the squares of two numbers is the product of the sum and difference of the numbers
(1) Do the computations below in head:
(i) 68² − 32², (ii) (3½)² − (2½)², (iii) 3.6² − 1.4²
Answer:
(i) 68² − 32²
68² − 32² = (68 + 32) × (68 − 32)
= 100 × 36 = 3600
(ii) (3½)² − (2½)²
(3½)² − (2½)² = (3½ + 2½) × (3½ − 2½)
= 6 × 1 = 6
(iii) 3.6² − 1.4²
3.6² − 1.4² = (3.6 + 1.4) × (3.6 − 1.4)
= 5 × 2.2 = 11
(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) If x is a natural number, the natural number just preceding it is (x − 1).
x² − (x − 1)² = x² − (x² − 2x + 1)
= x² − x² + 2x − 1 = 2x − 1
x + (x − 1) = 2x − 1
From this, we get x² − (x − 1)² = 2x − 1 = x + (x − 1).
(iii) The difference between the squares of a number and the number just preceding it is equal to the sum of those two numbers.
(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:
numbers, and also their difference
8 x 10 = 80
80 − 32 = 48
2 x 16 = 32
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:
The product of the numbers on the left and right of the middle number 14 = 13 x 15 = 195
The product of the numbers on the top and bottom of the middle number 14 = 7 x 21 = 48
Algebraic method
= (x − 7) (x + 7)
= x² − 7² = x² − 49
The difference between these = x² − 1 − (x² − 49)
= x² − 1 − x² + 49 = 48
The difference between these is always 48.
(4) As in the previous problem, mark a square of nine numbers in a calendar. Mark the four numbers in the corners:
15 × 3 = 45
45 − 17 = 28
1 × 17 = 17
Do this for other squares of nine numbers. Explain using algebra why the
difference is 28 in all cases.
Answer:
Here, the diagonal pairs of numbers are 9, 25 and 11, 23.
9 x 25 = 225
11 x 23 = 253
The difference between these = 253 − 225 = 28
Algebraic method
(x − 6) (x + 6) = x² − 6² = x² − 36
The difference between these
= x² − 36 − (x² − 64)
= x² − 36 − x² + 64 = 28
The difference between these is always 28.
(5) A square has perimeter 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:
(i) The perimeter of the square = 20 cm
20
One side of the square = ²⁰⁄₄ = 5 cm.
One side of the rectangle is 2 cm longer than one side of the square = 5 + 2 = 7 cm.
One side of the rectangle is 2 cm shorter than one side of the square = 5 − 2 = 3 cm
The perimeter of the rectangle = 2 (7 + 3) = 2 x 10 = 20 cm.
(ii) The area of the square = 5 x 5 = 25 sq. cm
The area of the rectangle = 7 x 3 = 21 sq.cm
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:
(i) Let the side of the square be x cm
The longer side of the rectangle = (x + y) cm
The shorter side of the rectangle = (x − y) cm
The perimeter of the square= 4x cm
The perimeter of the rectangle = 2(x + y + x − y) = 2(2x) = 4x cm
The perimeters of the square and the rectangle are equal.
(ii) The area of the square = x² sq.cm
The area of the rectangle = (x + y) × (x − y) = x² - y²
Here, the area of the square will always be more than the area of the rectangle. Therefore, the square has more area.
♦ We have seen that all odd numbers greater than 1 and certain even numbers can be written as the difference of two perfect squares. What kind of natural numbers cannot be written as the difference of two perfect squares?
1, 2, 4, 6, 10, 14,....
Examine the factors of these numbers except 1.
If the factors are x and y, then ((x + y) / 2) and ((x − y) / 2) are not natural numbers.
In these, the sum and difference of the factors are odd numbers.
Now let's look at the conclusion: Let x and y be two different factors of a natural number.
If their sum is an odd number, then the natural number cannot be written as the difference of two
perfect squares of the form ((x + y) / 2)² − ((x − y) / 2)²
Consider the natural number 20. Taking the factors
x = 10 and y = 2, ((x + y) / 2) = 6 and ((x − y) / 2) = 4.
So, 20 = 6² − 4².
But if we take the factors of 20 as x = 5 and y = 4
Then ((x + y) / 2) = ((5 + 4) / 2) = 4½, not a natural number.
((x − y) / 2) = ((5 − 4) / 2) = ½, not a natural number.
So 20 cannot be written as the difference of two perfect squares using the factors 5 and 4.
If the sum of the factors of a number is an odd number, then the number cannot be written as the difference of two perfect squares.
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