Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 4 Multiplication Identities Extra Questions and Answers Notes to clear their doubts.
Kerala SCERT Class 9 Maths Chapter 4 Solutions Multiplication Identities
Multiplication Identities Class 9 Kerala Syllabus Questions and Answers
Kerala State Syllabus 9th Standard Maths Chapter 4 Multiplication Identities Solutions Questions and Answers
Class 9 Maths Chapter 4 Kerala Syllabus – Product Of Sums
Intext Questions And Answers
Question 1.
What is the area of a rectangle of sides 26 cm and 15 cm?
Answer:
Normally, we calculate the area of a rectangle as follows:
Area = Length × Breadth
= 26 × 15
We can calculate the same without directly multiplying it
i. e., 26 × 15 = (20 + 6)(10 + 5)
= (20 × 10) + (20 × 5) + (6 × 10) + (6 × 5)
= 200 + 100 + 60 + 30
= 390 sq.cm
By dividing the rectangle accordingly, the calculations will become much simpler.
But, we can apply this method not only to natural numbers, for example:
6\(\frac{1}{2}\) × 8\(\frac{1}{3}\) = (6 + \(\frac{1}{2}\)) × (8 + \(\frac{1}{3}\))
= (6 × 8) + (6 × \(\frac{1}{3}\)) + (\(\frac{1}{2}\) × 8) + (\(\frac{1}{2}\) × \(\frac{1}{3}\))
= 48 + 2 + 4 + \(\frac{1}{6}\)
= 54\(\frac{1}{6}\)
Question 2.
(x + 1)(y + 1) = xy + x + y + 1
Answer:
For example,
31 × 51 = (30 + 1)(50 + 1)
= (30 × 50) + 30 + 50 + 1
= 1500 + 80 + 1
= 1581
Question 3.
(x + \(\frac{1}{2}\))(y + \(\frac{1}{2}\)) = xy + \(\frac{1}{2}\)(x + y) + \(\frac{1}{4}\)
Answer:
For example,
6\(\frac{1}{2}\) × 8\(\frac{1}{2}\) = (6 + \(\frac{1}{2}\)) × (8 + \(\frac{1}{2}\))
= (6 × 8) + \(\frac{1}{2}\)(6 + 8) + \(\frac{1}{4}\)
= 48 + 7 + \(\frac{1}{4}\)
= 55\(\frac{1}{4}\)
Textual Questions And Answers
Question 1.
Mentally find the products below:
(i) 71 × 91
(ii) 42 × 62
(iii) 10\(\frac{1}{2}\) × 6\(\frac{1}{2}\)
(iv) 9.5 × 3.5
(v) 10\(\frac{1}{4}\) × 6\(\frac{1}{4}\)
Answer:
(i) 71 x 91 = (70 + 1)(90 + 1)
= (70 × 90) + 70 + 90 + 1
= 6300 + 160 + 1
= 6461
(ii) 42 × 62 = (40 + 2)(60 + 2)
= (40 × 60) + 2(40 + 60) + 4
= 2400 + 200 + 4
= 2604
(iii) 110\(\frac{1}{2}\) × 6\(\frac{1}{2}\) = (10 + \(\frac{1}{2}\)) × (6 + \(\frac{1}{2}\))
= (10 × 6) + \(\frac{1}{2}\)(10 + 6) + \(\frac{1}{4}\)
= 60 + 8 + \(\frac{1}{4}\)
= 68\(\frac{1}{4}\)
(iv) 9.5 × 3.5 = (9 + 0.5)(3 + 0.5)
= (9 × 3) + 0.5(9 + 3) + 0.25
= 27 + 6 + 0.25
= 33.25
(v) 10\(\frac{1}{4}\) × 6\(\frac{1}{4}\)
Question 2.
The product of two numbers is 1400 and their sum is 81. What is the product of the numbers next to each?
Answer:
Let one number be x and the other be y
xy = 1400
x + y = 81
We can take consecutive counting numbers next to the given numbers as x + 1 and y + 1 respectively
(x + 1 )(y + 1) = xy + x + y + 1
= 1400 + 81 + 1
= 1482
Question 3.
The product of two odd numbers is 621 and their sum is 50. What is the product of the odd numbers next to each?
Answer:
Let one number be x and the other be y
xy = 621
x + y = 50
Now, take consecutive odd numbers next to the given numbers as x + 2 and y + 2 respectively
(x + 2)(y + 2) = xy + 2(x + y) + 4
= 621 + 100 + 4
= 725
Class 9 Maths Kerala Syllabus Chapter 4 Solutions – Special Operations
Intext Questions And Answers
Question 1.
Prove that the product of any two odd numbers is an odd number.
Answer:
In general, odd numbers can be written as 2n + 1 where n = 0,1,2… So, consider one odd number as 2n + 1 and another as 2m + 1
Now, products of these two odd numbers become
(2 n + 1)(2m + 1) = 4 mn + (2n + 2m) + 1
= 2(2 mn + n + m) + 1
it clear that the product is also an odd number.
Question 2.
Prove that the product of two numbers that have a remainder of 1 when divided by 3 is the same as the number that has a remainder of 1 when divided by 3.
Answer:
The general form of numbers that, when divided by 3, leave a remainder of 1 is 3n + 1 where
n = 0, 1, 2 …
So, consider one number as 3n + 1 and another as 3m + 1
Now, products of these two numbers become
(3n + 1)(3m + 1)
= 9 mn + (3n + 3m) + 1
= 3(3 mn + n + m) + 1
∴ it’s clear that product of numbers leaves a remainder 1 when divided by 3
Question 3.
Prove that for any four consecutive counting numbers, the difference between the product of the outermost numbers and the product of the middle two numbers is two.
Answer:
Let’s take consecutive 4 numbers as x, x + 1, x + 2, x + 3
Product of the two numbers at the ends = x(x + 3) = y² + 3x
Product of middle numbers = (x + 1)(x + 2) = x² + 3x + 2
∴ Difference = x² + 3x + 2 – ( x² + 3x) = 2
Textual Questions And Answers
Question 1.
Calculate each of the following for several numbers and guess a generalization; then prove it using algebra.
(i) The remainder got on division by 3, the product of two numbers one of which leaves remainder 1 on division by 3 and the other, remainder 2.
(ii) The remainder got on division by 4, the product of two numbers one of which leaves remainder 1 on division by 4 and the other, remainder 2.
(iii) The difference of the products of the two numbers at the ends and the two in the middle, of six consecutive natural numbers.
Answer:
(i) Numbers which leave a remainder of 1 when divided by 3 are 1, 4, 7, 10, …
Numbers which leave a reminder of 2 when divided by 3 are 2, 5, 8, 11, …
Now, take the numbers 4, 11
Then, 4 × 11 = 44
Dividing above result by 3, we get remainder = 2
Now, take the numbers 7, 8
Then, 7 × 8 = 56
Again, dividing above result by 3, we get remainder = 2
It is clear that,
The product of a number with a remainder of 1 when divided by 3 and a number with a remainder of 2 when divided by 3 has a remainder of 2.
Using algebra,
Numbers which leave remainder 1 when divided by 3 is taken as 3n + 1 and the remainder of 2 when divided by 3 is taken as 3m + 2
The product of these,
(3n + 1)(3m + 2)
= 9mn + 6n + 3m + 2
= 3(3mn + 2n + m) + 2
∴ Remainder of product divided by 3 = 2
(ii) Numbers that leave a remainder of 1 when divided by 4 are 1, 5, 9, 13, …
Numbers which leave 2 left over when divided by 4 are 2, 6, 10, 14, …
Taking the numbers 5 and 10,
5 × 10 = 50
Divide the product by 4 and the remainder = 2
Taking the numbers 9 and 6,
9 × 6 = 54
Divide the product by 4 and the remainder = 2
The product of a number with a remainder of 1 when divided by 4 and a number with a remainder of 2 when divided by 4 has a remainder of 2.
Using Algebra,
Let the number that leave a remainder of 1 when divided by 4 be 4n + 1 Let the number that leave a remainder of 2 when divided by 4 be 4m + 2
(4n + 1 )(4m + 2)
= 16mn + 8n + 4m + 2
= 4(4mn + 2n + m) + 2
∴ Remainder of product divided by 4 = 2.
(iii) Consider 6 consecutive natural numbers 1, 2, 3, 4, 5, 6.
Product of the two numbers at the ends = 1 × 6 = 6
Product of middle numbers = 3 × 4 = 12
Difference = 12 – 6 = 6.
Thus, the difference of the products of the two numbers at the ends and the two in the middle, of six consecutive natural numbers is 6.
Using Algebra,
Let’s take consecutive 6 numbers as x, x + 1, x + 2, x + 3, x + 4, x + 5
Product of the two numbers at the ends = x(x + 5) = x² + 5x
Product of middle numbers = (x + 2) (x + 3) = x² + 5x + 6
∴ Difference = x² + 5x + 6 – (x² + 5x) = 6
Question 2.
Given below is a method to find the product 36 × 28
(i) Check these for some other products of two-digit numbers.
(ii) Explain why this works using algebra.
(Recall the general algebraic form 10m + n of a two-digit number, seen in Class 7)
Answer:
(ii) All two-digit numbers can be written in the algebraic form 10m + n. In 10m + n here, m represents the tens digit and n represents the units digit.Similarly, in lOp + q here, p represents the tens digit and q represents the emits digit.
Now,
(10m + n)(10p + q) = 100mp + 10(mq + np) + nq
SCERT Class 9 Maths Chapter 4 Solutions – Number Curiosities
Textual Questions And Answers
Question 1.
Write numbers as shown below:
(i) we did in the calendar, mark a square of four numbers and find the difference of the . diagonal products. Do we get the same difference for any such square?
(ii) Explain why this is so, using algebra.
Answer:
Multiply the diagonal pairs, we get
7 × 13 = 91
12 × 8 = 96
Difference = 96 – 91 = 5
Similarly, consider one more square
Multiply the diagonal pairs, we get
9 × 3 = 27
4 × 8 = 32
Difference = 32 – 27 = 5
It’s clear that, the difference between the products of the diagonals in any such square is always the same.
(ii) Algebraically we can write the numbers in the square as follows n n+1 n + 2 n + 3 n + 4
Multiply the diagonal pairs, we get .
n × (n + 6) = n² + 6n
(n + 1) × (n + 5) = n² + 6n + 5
Difference = 5
Question 2.
In the multiplication table we’ve made, draw a square of nine numbers, instead of four, and mark the numbers at the four corners
(i) What is the difference of the diagonal sums?
(ii) Explain using algebra why we get the difference as 4 in all such squares.
(iii) What about a square of sixteen numbers?
Answer:
(i) When adding the diagonal pair, we get
6 + 20 = 26
10 + 12 = 22
Difference = 4
(ii) We can algebraically express the numbers in the square as follows.
| n | n + 2 | n + 4 |
| n + 3 | n + 6 | n + 9 |
| n + 6 | n + 10 | n + 14 |
By adding numbers in the four comers diagonally, we get
n + (n + 14) = 2n + 14
n + 4 + (n + 6) = 2n + 10
Difference = 4
(iii)
By adding the numbers of comer, we get
6 + 30 =36
15 + 12 = 27
Difference = 9
Multiplication Identities Class 9 Kerala Syllabus – Multiplication With A Difference
Textual Questions And Answers
Question 1.
The perimeter of a rectangle is 40 centimetres and its area is 70 square centimetres. Find the area of the rectangle with each side 3 centimetres shorter.
Answer:
Perimeter = 40 centimetres
Area = 70 square centimetres.
Let x be the length and y be the breadth,
2(x + y) = 40 ⇒ x + y = 20
xy = 70
Now, Area of the rectangle with each side 3 centimetres shorter
= (x – 3)(y – -3)
= xy – 3x – 3y + 9 – xy – 3(x + y) + 9
= 70 – 3 × 20 + 9 = 19 square centimetres.
Question 2.
If the sides of a rectangle are decreased by one metre, its area would be 741 square metres; if increased by one metre, it would be 861 square metres.
i) What is the area of the rectangle?
ii) What is its perimeter?
iii) What are the lengths of its sides?
Answer:
Let length be x and y be breath, now we can write,
(x + 1)(y + 1) = 861 ⇒ xy + x + y + l = 861 ….(1) .
(x – 1)(y – 1) = 741 ⇒ xy – (x + y) + 1 = 741 ….(2)
(1) + (2) → 2xy + 2 = 1602 ⇒ xy = 800
(1) – (2) →
2(x + y) = 120 ⇒ x + y = 60
xy = 800 & x + y = 60
(x – y)² = (x + y)² – 4xy
= 60² – (4 x 800) = 400 = 20²
⇒ x – y = 20
x = 40
∴ y = 20
i) Area of rectangle = xy = 800 square metres
ii) Perimeter = 2(x + y) = 120 metre
iii) Length, x = 40 metre Breadth, y = 20 metre
Question 3.
When each of two numbers are increased by one, the product becomes 1271 and when each is decreased by one, the product becomes 1131.
i) What is the product of the numbers?
ii) What is their sum?
iii) What are the numbers?
Answer:
Let the numbers are x andy, then,
(x + 1)(y + 1) = 1271 ⇒ y + x + y + l = 1271 ….(1)
(x – 1)(y – 1) = 1131 ⇒ xy – (x + y) + 1 = 1131 ….(2)
(1) + (2) → 2xy + 2 = 2402
⇒ xy = 1200
(1) – (2) – 2(x + y) = 140
⇒ x + y = 70
∴ (x – y)² = (x + y)² – 4xy
= 70² – (4 × 1200) = 100 = 10²
⇒ x – y = 10
∴ x = 40
y = 30
i) Product of numbers, xy = 1200
ii) Sum of numbers, x + y = 70
iii) Numbers, x = 40 , y = 30
Question 4.
The product of two odd numbers just after each of two odd numbers is 285 and the product of the odd numbers just before each is 165. What are the numbers?
Answer:
Let the numbers be x andy, then,
(x + 2)(y + 2) = 285 ⇒ xy + 2(x + y) + 4 = 285 ….(1)
(x – 2)(y – 2) = 165 ⇒ xy – 2(x + y) + 4 = 165 ….(2)
(1) + (2) → 2xy + 8 = 450
⇒ xy = 221
(1) – (2) → 4(x + y) = 120
⇒ x + y = 30
xy = 221 & x + y = 30
(x – y)² = (x + y)² – 4xy
= 30² – (4 × 221) = 16 = 4²
⇒ x — y = 4
∴ x = 17
y = i3
Numbers are 17, 13.
Question 5.
The product of two numbers is 713 and their difference is 8.
i) What is the product of the larger number increased by one and the smaller number decreased by one?
ii) What is the product of the larger number decreased by one and the smaller number increased by one?
Answer:
Let the larger number be x and smaller number be y, then,
xy = 713
x – y = 8
Product of adding 1 to the larger number and subtracting 1 from the smaller number, we get
(x + 1) (y – 1) = xy – x + y – 1
= xy – (x – y) – 1
= 713 – 8 – 1
= 704
the product of subtracting 1 from the larger number and adding 1 to the smaller number, we get
(x – 1)(y + 1) = xy + x – y – 1
= xy + (x – y) – 1
= 713 + 8 – 1
= 720
Question 6.
The product of two numbers is 5 more than the product of the larger of the numbers increased by one and the smaller decreased by one. How much is the increase in the product if the larger is decreased by one and the smaller increased by one?
Answer:
Let the larger number be x and smaller number be y, then,
Given that, the product of 1 added to the larger number and 1 subtracted from the smaller number is 5 less than the product of the numbers
(x + 1)(y – 1) = xy – 5
⇒- xy – x + y – 1
= xy – 5
⇒ x – y = 4
The product of subtracting 1 from the larger number and adding 1 to the smaller number is
(x – 1)(y + 1) = xy + (x – y) – 1
= xy + 4 – 1
= xy + 3
The product of subtracting 1 from the larger number and adding 1 to the smaller number is 3 more than the product of the original numbers.
Question 7.
The product of the larger of two numbers increased by one and the smaller decreased by one is 540. The product of the larger decreased by one and the smaller increased by one is 560.
i) What is the product of the numbers themselves?
ii) What is their difference?
iii) What are the numbers?
Answer:
i) Let the larger number be x and smaller number be y, then,
The product of adding 1 to the larger number and subtracting 1 from the smaller number, we§ get,
(x + 1)(y – 1) = 540
⇒ xy – x + y – 1 = 540 ……..(1)
The product of subtracting 1 from the larger number and adding 1 to the smaller number, we get
(x – 1)(y + 1) = 560
⇒ xy + (x – y) – 1 = 560 ………….(2)
Adding (1) and (2), we get 2xy – 2 = 1100 ⇒ xy = 551
Product of numbers = 551
ii) (2) – (1) → 2(x – y) = 20
⇒ x – y = 10
Difference of numbers =10
iii) (x + y)² = (x – y)² + 4xy
= 10² + 4 × 551
= 2304
x + y = 48
∴ x = 29
y = 19
Numbers are 29, 19
Question 8.
If the length of a rectangle is increased by 3 metres and the breadth decreased by 2 metres, its area would decrease by 10 square metres. If the length is decreased by 2 metres and the breadth increased by 3 metres, the area would increase by 30 square metres. Calculate the length and breadth of the rectangle.
Answer:
Let length be x and y be breath, now we can write
If the length is increased by 3 meters and the width is decreased by 2 meters, the area will decrease by 10 square meters
(x + 3)(y – 2) = xy – 10
⇒ xy – 2x + 3y – 6 = xy – 10
⇒ 2x – 3y = 4 ….(1)
If the length is reduced by 2 meters and the width is increased by 3 meters, the area will increase by 30 square meters
(x – 2)(y + 3) = xy + 30
⇒ xy + 3x – 2y- 6 = xy + 30
⇒ 3x – 2y = 36 …(2)
From (1) and (2) we get,
x = 20
y = 12
Length =20 metre
Width = 12 metre
Intext Questions And Answers
Question 1.
| For any positive numbers x, y, u, v with x > y and u > v (x – y)(u – v) = xu – xv – yu + yv |
For example:
26 × 17 = (30 – 4)(20 – 3)
= (30 × 20) – (30 × 3) – (4 × 20) + (4 × 3)
= 600 – 90 – 80 + 12
= 442
Let’s try the below questions
(i) 38 × 49
(ii) 47 × 99
(iii) 29 × 46
(iv) 9\(\frac{1}{2}\) × 19\(\frac{1}{2}\)
Answer:
(i) 38 × 49 = (40 – 2)(50 – 1)
= (40 × 50) – (40 × 1) – (2 × 50) + (2 × 1)
= 2000 -40- 100 + 2
= 1862
(ii) 47 × 99 = (50 – 3)(100 – 1)
= (50 × 100) – (50 × 1) – (3 × 100) + (3 × 1)
= 5000- 50 – 300 + 3
= 4653
(iii) 29 × 46 = (30 – 1)(50 – 4)
= (30 × 50) – (30 × 4) – (1 × 50) + (1 × 4)
= 1500 – 120 – 50 + 4
= 1334
(iv) 9\(\frac{1}{2}\) × 19\(\frac{1}{2}\)
= (10 – \(\frac{1}{2}\)) (20 – \(\frac{1}{2}\))
= (10 × 20) – (10 × 1) – (\(\frac{1}{2}\) × 20) + (\(\frac{1}{2}\) × \(\frac{1}{2}\))
= 200 – 5 – 10 + \(\frac{1}{4}\)
= 185\(\frac{1}{4}\)
For any positive numbers x, y, u, v with u > v
(x + y)(u – v) – xu – xv + yu – yv
Example:
14 × 59 = (10 + 4)(60 – 1)
= 600 – 10 + 240 – 4
= 826
Question 2.
Solve the Following Questions
i) 52 × 19
ii) 101 × 48
iii) 97 × 102
iv) 9\(\frac{3}{4}\) × 20\(\frac{1}{2}\)
Answer:
i) 52 × 19 = (50 + 2)(20 – 1)
= (50 × 20) – (50 × 1) + (2 × 20) -(2 × 1)
= 1000-50 + 40 – 2
= 988
ii) 101 × 48 = (100 + 1)(50 – 2)
= (100 × 50) – (100 × 2) + (1 × 50) -(1 × 2)
= 5000 – 200 + 50 – 2
= 4848
iii) 97 × 102 = (100 – 3)(100 + 2)
= (100 × 100) + (100 × 2) – (3 × 100) -(3 × 2)
= 10000 + 200 – 300-6
= 9894
iv) 9\(\frac{3}{4}\) × 20\(\frac{1}{2}\) = (10 – \(\frac{1}{4}\))(20 + \(\frac{1}{2}\))
= (10 × 20) + (10 × \(\frac{1}{2}\)) – (\(\frac{1}{4}\) × 20) – (\(\frac{1}{4}\) × \(\frac{1}{2}\))
= 200 + 5 – 5 – \(\frac{1}{8}\)
= 199\(\frac{7}{8}\)
- (x + 1 )(y – 1) = xy — x + y – 1
- (x – 1)(y + 1) = xy + x – y – 1
Multiplication Identities Class 9 Extra Questions and Answers Kerala Syllabus
Question 1.
50 × 40 = 2000 then find
(a) 51 × 41 =
(b) 52 × 42 =
(c) 49 × 39 =
(d) 48 × 38 =
Answer:
(a) 51 × 41 = 50 × 40 + (50 + 40) + 1²
= 2000 + 90 + 1
= 2091
(b) 52 × 42 = 50 × 40 + 2(50 + 40) + 2²
= 2000 + 180 + 4
= 2184
(c) 49 × 39 = 50 × 40 – (50 + 40) + 1²
= 2000 – 90 + 1
= 1911
(d) 48 × 38 = 50 × 40 – 2(50 + 40) + 2²
= 2000 – 180 + 4
= 1824
Question 2.
The product of two natural numbers is 300, and their sum is 35, then
(a) What is the product of the two natural numbers that come just after each of these two natural numbers?
(b) What is the product of the two natural numbers that come just before each of these two natural numbers?
Answer:
Let the numbers are x and y.
According to the question,
xy = 300 and x + y = 35,
(a) (x + 1)(y + 1) = xy + (x + y) + 1;
= 300 + 35 + 1
= 336
(b) (x – 1)(y – 1) = xy – (x + y) + 1
= 300 – 35 + 1
= 266
Question 3.
The product of two odd numbers is 899, and the sum is 60. What is the product of the two odd numbers that come just after each of these two odd numbers?
Answer:
Let the odd numbers be x and y, then,
xy – 899 and x + y = 60
(x + 2 )(y + 2) = xy + 2(x + y) + z²
= 899 + 2(60) + 4
= 1023
Question 4.
What is the remainder when the product of two numbers is divided by 5, given that one number leaves a remainder of 1 and the other leaves a remainder of 2 when each is divided by 5?
Answer:
Let 5m + 1 be a number that leaves a remainder of 1 when divided by 5 and 5n + 2 be a number
that leaves a remainder of 2 when divided by 5.
Then the product become = (5m + 1)(5n + 2)
⇒ (5m + 1)(5n + 2) = 25mn + 10m + 5n + 2
= 5(5mn + 2m + n) + 2
It is clear that, remainder is 2 when the product is divided by 5
Question 5.
Calculate:
(a) 8\(\frac{1}{2}\) × 4\(\frac{1}{2}\)
(b) 10\(\frac{1}{4}\) × 6\(\frac{1}{4}\)
Answer:
(a) 8\(\frac{1}{2}\) × 4\(\frac{1}{2}\)
= 8 × 4 + \(\frac{1}{2}\)(8 + 4) + (\(\frac{1}{2}\))²
= 32 + 6 + \(\frac{1}{4}\)
= 38\(\frac{1}{4}\)
(b) 10\(\frac{1}{4}\) × 6\(\frac{1}{4}\)
= 10 × 6 + \(\frac{1}{4}\)(10 + 6) + (\(\frac{1}{4}\))²
= 60 + 4 + \(\frac{1}{16}\)
= 64\(\frac{1}{16}\)
Question 6.
The product of two natural numbers just after each of two natural numbers is 2021, and the product of the natural numbers just before each is 2001. What are the product of numbers?
Answer:
Let the numbers be x and y
According to the question
(x + 1)(y + 1) = 2201
(x – 1)(y – 1) = 2001
2xy = 4200
xy = \(\frac{4200}{2}\) = 2100
Question 7.
If the sides of a rectangle are decreased by one metre, its area would be 240 square metres; if increased by one metre, it would be 306 square metres.
i) What is the area of the rectangle?
ii) What is its perimeter?
iii) What are the lengths of its sides?
Answer:
Let length be x and y be breath, now we can write
(x – 1)(y – 1) = 306 ⇒ xy – (x + y) + 1 = 240 ….(1)
(x + 1)(y + 1) = 240 ⇒ xy + x + y + 1 = 306 ….(2)
(1) + (2) → 2xy + 2 = 546
⇒ xy = \(\frac{546 – 2}{2}\) = 272
(2) – (1) → 2(x + y) = 66 .
⇒ x + y = \(\frac{66}{2}\) = 33 …(3)
xy = 272 and x + y = 33
(x – y)² = (x + y)² – 4xy
= 33² – (4 × 272) = 1
⇒ x – y = 1 …(4)
(3) + (4) → 2x = 34
⇒ x = 17
(3) – (4) → 2y = 32
⇒ y = 16
(i) Area of rectangle = xy = 272 square metres
(ii) Perimeter = 2(x + y) = 66 metre
(iii) Length, x = 17 metre
(iv) Breadth, y = 16 metre