Unit 1 Number Systems

​ Chapter 1: Real Numbers

Euclids Division Lemma

What is a dividend? Let us understand it with the help of a simple example. Can you divide 14 by 6?

​

After division, we get 2 as the quotient and 2 as the remainder.

Thus, we can also write 14 as 6 × 2 + 2.

A dividend can thus be written as:

Dividend = Divisor × Quotient + Remainder

Can you think of any other number which, when multiplied with 6, gives 14 as the dividend and 2 as the remainder?

Let us try it out with some other sets of dividends and divisors.

(1) Divide 100 by 20: 100 = 20 × 5 + 0

(2) Divide 117 by 15: 117 = 15 × 7 + 12

(3) Divide 67 by 17: 67 = 17 × 3 + 16

Thus, if we have a dividend and a divisor, then there will be a unique pair of a quotient and a remainder that will fit into the above equation.

This brings us to Euclid’s division lemma.

If a and b are positive integers, then there exist two unique integers, q and r,
such that a = bq + r
This lemma is very useful for finding the H.C.F. of large numbers where breaking them into factors is difficult. This method is known as Euclid’s Division Algorithm.

Let us look at some more examples.

Example 1: Find the H.C.F. of 4032 and 262 using Euclid’s division algorithm.

Solution:

Step 1:

First, apply Euclid’s division lemma on 4032 and 262.

4032 = 262 × 15 + 102

Step 2:

As the remainder is non-zero, we apply Euclid’s division lemma on 262 and 102.

262 = 102 × 2 + 58

Step 3:

Apply Euclid’s division lemma on 102 and 58.

102 = 58 × 1 + 44

Step 4:

Apply Euclid’s division lemma on 58 and 44.

58 = 44 × 1 + 14

Step 5:

Apply Euclid’s division lemma on 44 and 14.

44 = 14 × 3 + 2

Step 6:

Apply Euclid’s division lemma on 14 and 2.

14 = 2 × 7 + 0

In the problem given above, to obtain 0 as the remainder, the divisor has to be taken as 2. Hence, 2 is the H.C.F. of 4032 and 262.

Note that Euclid’s division algorithm can be applied to polynomials also.

Example 2: A rectangular garden of dimensions 190 m × 60 m is to be divided in square blocks to plant different flowers in each block. Into how many blocks can this garden be divided so that no land is wasted?

Solution:

If we do not want to waste any land, we need to find the largest number that completely divides both 190 and 60 and gives the remainder 0, i.e., the H.C.F. of (190, 60).

To find the H.C.F., let us apply Euclid’s algorithm.

190 = 60 × 3 + 10

60 = 10 × 6 + 0

Therefore, the H.C.F. of 190 and 60 is 10.

Therefore, there will be​= 19 square blocks along the length of the garden and ​= 6 blocks along its breadth.

Hence, the total number of blocks in the garden will be 19 × 6 = 114.

Example 3: Find the H.C.F. of 336 and 90 using Euclid’s division algorithm.

Solution:

As 336 > 90, we apply the division lemma to 336 and 90.

336 = 90 × 3 + 66

Applying Euclid’s division lemma to 90 and 66: 90 = 66 × 1 + 24

Applying Euclid’s division lemma to 66 and 24:

66 = 24 × 2 + 18

Applying Euclid’s division lemma to 24 and 18:

24 = 18 × 1 + 6

Applying Euclid’s division lemma to 18 and 6:

18 = 6 × 3 + 0

As the remainder is zero, we need not apply Euclid’s division lemma anymore. The divisor (6) is the required H.C.F.

Example 4: Find the H.C.F. of 45, 81, and 117 using Euclid’s division algorithm.

Solution:

Let us begin by choosing any two out of the three given numbers, say 45 and 81. As 81 > 45, we apply Euclid’s division lemma to 81 and 45.

81 = 45 × 1 + 36

Applying Euclid’s division lemma to 45 and 36:

45 = 36 × 1 + 9

Applying Euclid’s division lemma to 36 and 9:

36 = 9 × 4 + 0

As the remainder is zero, the H.C.F. of 45 and 81 is 9.

Now, we again need to apply Euclid’s division algorithm on the H.C.F. of the two numbers and the remaining number.

Since the H.C.F. of 45 and 81 is 9 and the third number is 117, we apply Euclid’s division lemma to 117 and 9.

117 = 9 × 13 + 0

As the remainder is zero, the H.C.F. of 9 and 117 is 9.

Here, the second H.C.F. (the H.C.F. of the H.C.F. of the first two numbers and the third number) is the H.C.F. of the three numbers.

Thus, we can say that the H.C.F. of the three numbers is 9.

Example 5: In an inter-school essay writing competition, the numbers of participants from schools A, B, and C are 20, 16, and 28 respectively. If the participants in each room are from the same school, then find the minimum number of rooms required such that each room has the same number of participants.

Solution:

If we need to find the minimum number of rooms, then we need to keep the maximum number of participants in each room i.e., we need to find the largest number that completely divides 20, 16, and 28.

Thus, we start by choosing any two out of the given three numbers, say 20 and 16. Applying Euclid’s division lemma to 20 and 16:

20 = 16 × 1 + 4

Applying Euclid’s division lemma to 16 and 4:

16 = 4 × 4 + 0

Hence, 4 is the H.C.F. of 20 and 16.

Applying Euclid’s division algorithm to 28 and 4:

28 = 4 × 7 + 0

Hence, 4 is the H.C.F. of 28 and 4.

∴H.C.F. of 20, 16, 28 = 4

Therefore, each room would have 4 participants.

The number of rooms in which the participants from schools A, B, and C can be

accommodated is ​ = 5,​ = 4, and ​= 7 respectively.

Therefore, a total of 5 + 4 + 7 =16 rooms are required.

The Use Of Euclids Division Lemma To Prove Mathematical Relationships

​Whenever we divide 32 by 5, we will get 6 as the quotient and 2 as the remainder. Therefore, we can also write 32 as 6 x 5 + 2.

We can thus write this rule about the dividend as

 

This expression is unique, i.e., whenever we divide an integer (dividend) by another integer (divisor), we always get a fixed quotient and a fixed remainder.

The above statement can be proven by taking an example.

Whenever we divide 54 by 11, we will get 4 as the quotient and 10 as the remainder. We will never get a quotient other than 4 and a remainder other than 10 when we divide 54 by 11.

This brings us to Euclid’s division lemma.

​

This lemma has several applications, one of which is to prove mathematical relationships among numbers.

Let us discuss this concept with the help of a few examples.

Example 1: Prove that every positive integer is of the form 3p, 3p + 1, or 3p + 2, where p is any integer.

Solution:

Let a be any positive integer and let b = 3.

Applying Euclid’s algorithm to a and 3:

a = 3p + r; for some integer p and 0 ≤ r < 3

Therefore, a can be 3p, 3p + 1, or 3p + 2.

As a is a positive integer, we can say that any positive integer is of the form 3p, 3p + 1, or 3p + 2.

Example 2: Prove that every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1, where m is any integer.

Solution:

Let a be any positive integer and let b = 2.

According to Euclid’s division lemma, there exist two unique integers m and r such that

a = bm + r = 2m + r, where 0 ≤ r < 2. Thus, r = 0 or 1

If r = 0, i.e., if a = 2m, then the expression is divisible by 2. Thus, it is an even number.

If r = 1, i.e., if a = 2m + 1, then the expression is not divisible by 2. Thus, it is an odd number.

Thus, every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1.

Example 3: Prove that the expression y(y + 1) always represents an even number, where y is any positive integer.

Solution:

Let y be an integer. Thus, it may be either odd or even.

When y is an odd number, i.e., when y is of the form 2p + 1, where p is an integer:

y(y + 1) = (2p + 1)(2p + 1 + 1) = (2p + 1)(2p + 2) = 2(2p + 1)(p + 1)

The above expression is a multiple of 2. Thus, the expression y(y + 1) represents an even number.

When y is an even number, i.e., when y is of the form 2q, where q is an integer:

y(y + 1) = (2q)(2q + 1) = 2q(2q + 1)

The above expression is a multiple of 2. Thus, the expression y(y + 1) represents an even number.

Thus, for a positive integer y, the expression y(y + 1) always represents an even number.

Example 4:

Show that

(a)    The sum, the difference, and the product of two even numbers is always even.

(b)    The sum and the difference of two odd numbers is always even, whereas their product is always odd.

Solution:

(a)  Let there be two even numbers, x and y, such that x = 2m and y = 2n, where m and n are two positive integers.

Now, x + y = 2m + 2n = 2(m + n) = 2p, where p = m + n is an integer. Thus, x + y is always even.

Similarly, x − y = 2m − 2n = 2(m − n) = 2q, where q = m − n is an integer. Thus, x − y is always even.

Likewise, xy = 2m × 2n = 4mn = 2(2mn) = 2t, where t = 2mn is an integer. Thus, xy is always even.

Hence, the sum, the difference, and the product of two even numbers is always even.

(b) Let there be two odd numbers x and y such that x = 2m + 1 and y = 2n + 1, where m and n are two positive integers.

Now, x + y = (2m + 1) + (2n + 1) = 2(m + n + 1) = 2p, where p = m + n + 1 is an integer. Thus, x + y is even.

Similarly, x − y = (2m + 1) − (2n + 1) = 2(m − n) = 2q, where q = m − n is an integer.

Thus, x − y is even.

Likewise, xy = (2m + 1)(2n + 1) = 2(2mn + m + n) + 1 = 2t + 1, where t = 2mn + m + n is an integer.

Thus, xy is always odd.

Hence, the sum and the difference of two odd numbers is always even, whereas their product is always odd.

Prime Factorisation of Numbers Using Fundamental Theorem of Arithmetic

We know that all composite numbers can be represented as the product of two or more prime numbers. Let us understand this concept by taking the example of 36 and factorising it in different ways.

We can see that whichever way we factorise the number 36, it will be broken down as the product of the same prime numbers, which is unique. The only difference is that the ordering of the prime numbers will be different for different ways of factorising the number. In fact, this is true for all numbers. We can check this by taking the example of a larger number, say 21560, which can be uniquely broken down into its prime factors as

2³ × 5 × 7² × 11

Hence, we can say that any composite number can be written in the form of the product of prime numbers, which is unique, except the order in which they occur. By this, we mean that 2 × 3 × 7 × 11 is the same as 7 × 11 × 2 × 3.

This is the fundamental theorem of arithmetic. It can be formally stated as:

Thus, this theorem can be used to write the prime factorisation of any number. Let us try to build on this concept with the help of some examples.

Example 1:

Write the prime factorization of 31250. What are its prime factors? Solution:

31250 = 2 × 15625

= 2 × 5 × 3125

= 2 × 5 × 5 × 625

= 2 × 5 × 5 × 5 × 125

= 2 × 5 × 5 × 5 × 5 × 25

= 2 × 5 × 5 × 5 × 5 × 5 × 5

= 2 × 5⁶

Hence, 2 × 5⁶ is the prime factorisation of 31250. Its prime factors are 2 and 5.

Example 2: If it is given that 13125 = 2^a ×3^b× 5^c× 7^d, then find the value of a +2b +7c + 11d.

Solution:

2^a × 3^b × 5^c × 7^d = 13125

= 3 × 4375

= 3 × 5 × 875

= 3 × 5 × 5 ×175

= 3 × 5 × 5 × 5 × 35

= 3 × 5 × 5 × 5 × 5 × 7

= 3¹ × 54 × 7¹

∴ 2^a × 3^b × 5^c × 7^d = 2⁰ × 3¹ × 5⁴ × 7¹

Comparing exponents of the bases (integers): a = 0, b = 1, c = 4, and d =1

Hence, a +2b +7c +11d = 0 + 2 ×1 + 7 × 4 + 11 × 1

= 0 + 2 + 28 + 11

= 41

Example 3: Show that the expressions given below are composite numbers.

(a) 3 × 5 × 7 × 23 + 2 × 7 × 11 × 13

(b) 29 × 35 + 14

(c) 3⁴ + 6³

Solution:

(a) 3 × 5 × 7 × 23 + 2 × 7 × 11 × 13 = 7( 3 × 5 × 23 + 2 × 11 × 13)

= 7(345 + 286)

= 7 × 631

Since both 7 and 631 are prime numbers, we have expressed the given expression as the product of two prime numbers. We know that according to the fundamental theorem of arithmetic, every composite number can be uniquely written as the product of its prime factors. Thus, the given expression represents a composite number.

(b) 29 × 35 + 14 = 29 × 5 × 7 + 2 × 7

= 7(29 × 5 + 2)

= 7 × 147

= 7 × 3 × 7 × 7

= 3 × 7³

Since both 3 and 7 are prime numbers, we have expressed the given expression as the product of its prime factors. We know that according to the fundamental theorem of arithmetic, every composite number can be uniquely written as the product of its prime factors. Thus, the given expression represents a composite number.

(c) 3⁴ + 6³ = 3⁴ + (2 × 3) ³

= 3⁴ + 2³ × 3³

= 3³(3 + 2³)

= 3³ × 11

Since both 3 and 11 are prime numbers, we have expressed the given expression as the product of its prime factors. It is known that according to the fundamental theorem of arithmetic, every composite number can be uniquely written as the product of its prime factors. Thus, the given expression represents a composite number.

Application Of The Fundamental Theorem Of Arithmetic To Find The HCF And LCM Of Numbers

All composite numbers can be written as the product of two or more prime numbers. For example, 20 can be written as 2² × 5; 54 can be written as 2 × 3³, and so on.

Note that if we do not consider the way in which the prime factors are written, then we can prime factorise every number in only one way. This applies to other numbers as well.

This leads to the fundamental theorem of arithmetic, which states that:

Even though we did not notice it before, whenever we prime factorise a number, we use the fundamental theorem of arithmetic to do so.

For example, the prime factorisation of 980 is represented as

980 = 2 × 490

= 2 × 2 × 245

= 2 × 2 × 5 × 49

= 2 × 2 × 5 × 7 × 7

= 2² × 5¹ × 7²

Hence, 2² × 5¹ × 7² is the prime factorisation of 980; and 2, 5, and 7 are its prime factors.

By applying the fundamental theorem of arithmetic to the prime factorized numbers, we can also find their HCF and LCM.

This is known as the prime factorisation method, which states that:

Let us understand this method with the help of some examples.

Example 1: Find the LCM and the HCF of 432 and 676 using the prime factorization method.

Solution:

We can write these numbers as

432 = 24 × 3³

676 = 2² × 13²

To calculate the HCF

We observe that the only common prime factor is 2 and the smallest power of this prime factor is also 2.

Thus, HCF (432, 676) = 2² = 4

To calculate the LCM

We observe that the prime factors of 432 and 676 are 2, 3, and 13. The greatest powers of these factors are 4, 3, and 2 respectively.

LCM is the product of the greatest power of each prime factor.

Thus, LCM (432, 676) = 2⁴ × 3³ × 13² = 73008

Example 2: Find the HCF and the LCM of 28, 42, and 64 using the prime factorization method.

Solution:

We can write these numbers as

28 = 2² × 7¹

42 = 2 × 3¹ × 7¹

64 = 2⁶

HCF is the product of the smallest power of each common prime factor. Here, the only common prime factor is 2 and its power is 1.

Thus, HCF (28, 42, 64) = 2¹ = 2

LCM is the product of the greatest power of each prime factor. Thus, LCM (28, 42, 64) = 2⁶ × 3¹ × 7¹ = 1344

Example 3: Find the HCF and the LCM of 1080 and 900 using the prime factorization and show that HCF × LCM = Product of two numbers.

Solution:

1080 = 2³ × 3³ × 5

900 = 2² × 3² × 5²

Hence, HCF (1080, 900) = 2² × 3² × 5 = 180

LCM (1080, 900) = 2³ × 3³ × 5² = 5400

HCF × LCM = 180 × 5400 = 972000

Product of numbers = 1080 × 900 = 972000

Hence, HCF × LCM = Product of two numbers

Example 4: The HCF of 273 and another number is 7, while their LCM is 3003. Find the other number.

Solution:

Let the first number (a) be 273 and the second number be b.

It is given that HCF (a, b) = 7 and LCM (a, b) = 3003.

We know that HCF LCM = Product of two numbers.

⇒ HCF (a, b) × LCM (a, b) = a × b

⇒ 7 × 3003 = 273 × b

⇒ b = 77

Hence, the other number is 77.

Example 5: Anurag takes 6 minutes to complete one round of jogging around the circular track of a park, while Twinkle takes 8 minutes to do the same. If both of them start jogging at the same time from the same point, then how much time will they take before they meet at the point from which they started?

Solution:

Since Anurag and Twinkle take 6 minutes and 8 minutes respectively to complete one round of the circular track, the time after which they will meet at the starting point will be the lowest multiple of 6 and 8, i.e., their LCM.

6 = 2 × 3

8 = 2³

∴ LCM (6, 8) = 2³ × 3 = 24

Thus, they will meet at the starting point after 24 minutes.

Example 6: There are 120 students in a class. When the students were arranged according to their roll numbers, it was observed that every second student got distinction in Mathematics, every third student got distinction in Science, and every fifth student got distinction in English. How many students got distinction in all three subjects?

Solution:

As every second, third, and fifth student got distinction in Math, Science, and English respectively, the roll numbers of the students who got distinction in all three subjects will be equal to the multiples of the LCM of 2, 3, and 5.

LCM (2, 3, 5) = 2 × 3 × 5 = 30

Thus, every 30^th student got distinction in all three subjects.

Thus, a total of students got distinction in all three subjects.

Properties Of Prime Numbers

Consider the number 8ⁿ, where n is a natural number.

Is there any value of n for which 8ⁿ ends with zero?

It is difficult to answer this question directly. However, we can answer this question by making use of the fundamental theorem of arithmetic. It states that

This means that if we are given a composite number, then that number can be written as a product of prime numbers in only one way (except for the order of prime numbers).

For example: the composite number 255 can be written as the product of primes as follows. 255 = 3 × 5 × 17

Also, 255 can be written as 3 × 17 × 5 or 5 × 3 × 17 or 5 × 17 × 3 or 17 × 3 × 5 or 17 × 5 × 3.

Thus, we can see that 255 can be expressed as a product of unique prime numbers 3, 5, and 17 but the order of representation may differ.

Now, by making use of the above theorem, we can answer the question which we were discussing in the beginning. Let us see how.

Suppose the number 8n ends with zero for some value of n. Since the number ends with zero, it should be divisible by 10. Now, 10 = 2 × 5

Thus, this number should be divisible by 2 and 5 also.

Therefore, the prime factorization of 8n should contain both the prime numbers 2 and 5. We have, 8ⁿ = (2³)ⁿ = 2³ⁿ

⇒ The only prime in the factorization of 8ⁿ is 2.

Thus, by fundamental theorem of arithmetic, there is no other prime in the factorization of 8ⁿ.

Hence, there is no natural number n for which 8n ends with the digit zero. In this way, we can make use of the above theorem.

Let us now look at some more examples to understand this concept better.

Example 1: Prove that the number 9ⁿ, where n is a natural number, cannot end with a zero.

Solution:

Suppose the number 9ⁿ ends with a zero for some value of n.

Since the number ends with zero, it should be divisible by 10.

Now, 10 = 2 × 5

Thus, this number should be divisible by 2 and 5 also.

Therefore, the prime factorization of 9ⁿ should contain both the prime numbers 2 and 5.

We have, 9ⁿ = (32)ⁿ = 3²ⁿ

⇒ The only prime in the factorization of 9ⁿ is 3.

Thus, by fundamental theorem of arithmetic, there is no other prime in the factorization of 9ⁿ.

Hence, there is no natural number n for which 9ⁿ ends with the digit zero.

Example 2: Check whether the numbers 49ⁿ, where n is a natural number, can end with a zero.

Solution:

Suppose the number 49n ends with a zero for some value of n.

Since the number ends with zero, it should be divisible by 10.

Now, 10 = 2 × 5

Thus, this number should be divisible by 2 and 5 also.

Therefore, the prime factorization of 49n should contain both the prime numbers 2 and 5. We have, 49ⁿ = (7²)ⁿ = 7²ⁿ

⇒ The only prime in the factorization of 49ⁿ is 7.

Thus, by fundamental theorem of arithmetic, there is no other prime in the factorization of 49ⁿ.

Thus, there is no natural number n for which 49ⁿ ends with the digit zero.

Irrational Numbers

We know that a number which cannot be written in the form of , where p and q are integers and q ≠ 0, is known as an irrational number.

For example: all numbers of the form , where p is a prime number such as etc., are irrational numbers.

How can we prove that these are irrational numbers?

We can prove this by making use of a theorem which can be stated as follows.

“If p divides a2, then p divides a (where p is a prime number and a is a positive integer)”.

So go through the given video to understand the application of the above stated property.

Similarly, we can prove that square roots of other prime numbers like , etc. are irrational numbers.

Besides these irrational numbers, there are some other irrational numbers like etc.

We can also prove why these numbers are irrational. Before this, let us first see what happens to irrational numbers, when we apply certain mathematical operations on them.

We will now prove that is irrational.

We know that is irrational (as proved before).

Now, the multiplication of a rational and an irrational number gives an irrational number. Therefore, is an irrational number.

Let us now try to understand the concept further through some more examples.

Example 1: Prove that  is irrational. Solution:

Let us assume that is not irrational, i.e. is a rational number.

Then we can write , where a and b are integers andb ≠ 0.

Let a and b have a common factor other than 1.

After dividing by the common factor, we obtain

, where c and d are co-prime numbers.

As c, d and 2 are integers, and are rational numbers.

Thus, is rational.

is rational as the difference of two rational numbers is again a rational number.

This is a contradiction as is irrational.

Therefore, our assumption that is rational is wrong.

Hence, is irrational.

Example 2: Prove that is irrational.

Solution:

Let us assume  is rational. Then, we can write

,

where a and b are co-prime and b ≠ 0.

⇒

Now, as a and b are integers, is rational or is a rational number.

This means that is rational.

This is a contradiction as is irrational.

Therefore, our assumption that  is rational is wrong.

Hence, is an irrational number.

Decimal Expansions of Rational Numbers

The Need for Converting Rational Numbers into Decimals

A carpenter wishes to make a point on the edge of a wooden plank at 95 mm from any end. He has a centimeter tape, but how can he use that to mark the required point?

Simple! He should convert 95 mm into its corresponding centimeter value, i.e., 9.5 cm and then measure and mark the required length on the wooden plank.

This is just one of the many situations in life when we face the need to convert numbers into decimals. In this lesson, we will learn to convert rational numbers into decimals, observe the types of decimal numbers, and solve a few examples based on this concept.

Know More

Two rational numbers and are equal if and only if ad = bc.

Take, for example, the rational numbers and . Let us see if they are equal or not. Here, a = 2, b = 4, c = 3 and d = 6

Now, we have:

ad = 2 × 6 = 12

bc = 4 × 3 = 12

Since ad = bc, we obtain = .

We know that the form represents the division of integer p by the integer q. By solving

this division, we can find the decimal equivalent of the rational number . Now, let us convert the numbers , and into decimals using the long division method.

While the remainder is zero in the division of 5 by 8, it is not so in case of the other two divisions. Thus, we can get two different cases in the decimal expansions of rational numbers.

Observing the Decimal Expansions of Rational Numbers

We can get the following two cases in the decimal expansions of rational numbers.

Case I: When the remainder is zero

In this case, the remainder becomes zero and the quotient or decimal expansion terminates after a finite number of digits after the decimal point. For example, in the decimal

expansion of , we get the remainder as zero and the quotient as 0.625.

Case II: When the remainder is never zero

In this case, the remainder never becomes zero and the corresponding decimal expansion

is non-terminating. For example, in the decimal expansions of and , we see that the remainder never becomes zero and their corresponding quotients are non-terminating decimals.

When we divide 4 by 3 and 2 by 7, we get 1.3333… and 0.285714285714… as the respective quotients. In these decimal numbers, the digit ‘3’ and the group of digits

‘285714’ get repeated. Therefore, we can write and

Here, the symbol  indicates the digit or group of digits

that gets repeated.

Solved Examples

Example 1: Write the decimal expansion of and find if it is terminating or non- terminating and repeating.

Solution:

Here is the long division method to find the decimal expansion of .

Hence, the decimal expansion of is 49.48. Since the remainder is obtained as zero, the decimal numberis terminating.

Example 2: Write the decimal expansion of and find if it is terminating or non- terminating and repeating.

Solution:

Here is the long division method to find the decimal expansion of .

Hence, the decimal expansion of is 87.33 Since the remainder 9 is obtained again

and again, the decimal numberis non-terminating and repeating. The decimal number can also be written as .

Medium

Example 1: Find the decimal expansion of each of the following rational numbers and write the nature of the same.

1.

2.

3.

4.

Solution:

We have = 0.64356435... =

The group of digits ‘6435’ repeats after the decimal point. Hence, the decimal expansion of the given rational number is non-terminating and repeating.

We have = 2.3075

Hence, the given rational number has a terminating decimal expansion.

We have = 0.3737... =

The pair of digits ‘37’ repeats after the decimal point. Hence, the decimal expansion of the given rational number is non-terminating and repeating.

We have = 0.67

Hence, the given rational number has a terminating decimal expansion.

Terminating and Non-terminating Repeating Decimal Expansions of Rational Numbers

We can find the decimal expansion of rational numbers using long division method.

However, it is possible to check whether the decimal expansion is terminating or non- terminating without actually carrying out long division also.

Let us start by taking a few rational numbers in the decimal form.

(a)

(b)

0.275

On prime factorizing the numerator and the denominator, we obtain

Can you see a pattern in the two examples?

We notice that the given examples are rational numbers with terminating decimal expansions. When they are written in the form, where p and q are co-prime

(the HCF of p and q is 1), the denominator, when written in the form of prime factors, has 2 or 5 or both.

The above observation brings us to the given theorem.

If x is a rational number with terminating decimal expansion, then it can be expressedin the form, where p and q are co-prime (the HCF of p and q is 1) and the prime factorisation of q is of the form 2n5m, where n and m are non-negative integers.

Contrary to this, if the prime factorisation of q is not of the form 2n5m, where n and m are non-negative integers, then the decimal expansion is a

non-terminating one.

Let us see a few examples that will help verify this theorem.

(a)

(b)

(c) 

(d) 

Note that in examples (b) and (d), each of the denominators is composed only of the prime factors 2 and 5, because of which, the decimal expansion is terminating. However, in examples (a) and (c), each of the denominators has at least one prime factor other than 2 and 5 in their prime factorisation, because of which, the decimal expansion is non- terminating and repetitive.

To summarize the above results, we can say that:

Let us solve a few examples to understand this concept better.

Example 1: Without carrying out the actual division, find if the following rational numbers have a terminating or a non-terminating decimal expansion.

(a)

(b)  

Solution:

(a)

As the denominator can be written in the form 2ⁿ5^m, where n = 6 and m = 2 are non-negative integers, the given rational number has a terminating decimal expansion.

(b)

As denominator cannot be written in the form 2ⁿ5^m, where n and m are non-negative integers, the given rational number has a non-terminating decimal expansion.

Example 2: Without carrying out the actual division, find if the expression  has a terminating or a non-terminating decimal expansion.

Solution:=

As the denominator can be written in the form 2ⁿ5^m, where n = 7and m = 0 are non-negative integers, the given rational number has a terminating decimal expansion.

Hence, 5.5859375 is the decimal expansion of the given rational number.

Graphic Interpretation Of The Number Of Zeros Of A Polynomial

A polynomial in variable x is denoted by p(x).

The highest power of the variable in the polynomial is known as its degree. For example, the highest power of variable x in the polynomial x² − 3x + 2 is 2. Hence, it’s degree is 2.

Let us better understand this concept with a few more examples.

Example1: Find the number of zeroes of the polynomial whose graph is given below.

Solution:

The given graph intersects the x- axis at 4 distinct points. Hence, the number of zeroes of the given polynomial is 4.

Example 2: Find the number of zeroes of the polynomial whose graph is given below.

Solution:

The given graph does not intersect the x-axis.

Hence, the given polynomial has no zeroes or in other words, the number of zeroes of the given polynomial is zero.

Example 3: Find the number of zeroes of the polynomials whose graphs are given below.

Solution:

(a)    The graph intersects the x-axis at only 1 point.

Hence, the number of zeroes of this polynomial is 1.

(b)    The graph intersects the x-axis at 4 distinct points.
 
Hence, the number of zeroes of this polynomial is 4.

(c)    The graph intersects the x-axis at only 1 point.

Hence, the number of zeroes of this polynomial is 1.

(d)    The graph of the polynomial does not intersect the x-axis.

Hence, this polynomial has no zeroes.

(e)    The graph intersects the x-axis at 2 distinct points.

Hence, the number of zeroes of this polynomial is 2.

(f)    The graph intersects the x-axis at 2 distinct points.

Hence, the number of zeroes of this polynomial is 2.

(g)    The graph of the polynomial does not intersect the x-axis.

Hence, this polynomial has no zeroes.

(h)    The graph intersects the x-axis at only 1 point.

Hence, the number of zeroes of this polynomial is 1.

(i)    The graph intersects the x-axis at only 1 point.

Hence, the number of zeroes of this polynomial is 1.

(j)    The graph intersects the x-axis at 4 distinct points.

Hence, the number of zeroes of this polynomial is 4.

(k)    The graph intersects the x-axis at 5 distinct points.

Hence, the number of zeroes of this polynomial is 5.

(l)    The graph intersects the x-axis at 3 distinct points.

Hence, the number of zeroes of this polynomial is 3.

Geometrical Interpretation Of Zeros Of Polynomials

Let us consider a polynomial p(x) = 2x – 5. Its graph can be drawn as follows:

If we notice the above graph carefully, then we can see that the graph of the polynomial p(x) = 2x – 5 passes through the x-axis at a point .

Do you know what is the value of the polynomial p(x) at the x-coordinate of this point? Let us find out how.

Thus, is a zero of the polynomial.

Thus, we can say that

Using the above statement, we can find out the zeroes of the polynomials. The method of finding zeroes can be easily understood with the help of the following example.

Let us find the zeroes of the polynomial p(x) = x² – 4x + 3. The graph of this polynomial can be drawn as follows:

From the graph, we can see that the graph of the polynomial p(x) = x² – 4x + 3 cuts the x- axis at two points respectively (1, 0) and (3, 0). Now, the x-coordinates of these points are 1 and 3 respectively. Thus, 1 and 3 are the zeroes of the given polynomial p(x) = x² – 4x + 3.

In this way we can interpret the zeroes of the polynomial geometrically.

Let us discuss one more example based on geometrical interpretation of zeroes of a polynomial.

Example1: Find the zeroes of the polynomials whose graphs are shown below.

Solution:

(i)    The graph of the polynomial intersects the x-axis at the point (2, 0).

The x coordinate of the point (2, 0) is 2.

Thus, 2 is a zero of the polynomial whose graph has been shown.

(ii)    The graph of the polynomial intersects the x-axis at the point (0, 0).

The x co-ordinate of the point (0, 0) is 0. Thus, 0 is a zero of the polynomial.

(iii)    The graph does not intersect the x-axis, so there is no zero of the polynomial whose graph has been shown.

Pair of Linear Equations in Two Variables

Inconsistent, Consistent, And Dependent Pairs Of Linear Equations

Linear equations in two variables are equations where we have two variables. We require two equations to find the solution of linear equations in two variables. Go through the following video to understand the basic concept of linear equations in two variables.

Remember: To solve a linear equation in one variable, only one equation is required. To solve linear equations in two variables, two linear equations in the same two variables are required. Likewise, we can say that to solve linear equations with n number of

variables, n numbers of linear equations in the same n number of variables are required. The general form of a pair of linear equations in two variables is written as

, where a1, b1 are real numbers and (i.e., a1, b1 ≠ 0)

, where a2, b2 are real numbers and (i.e., a2, b2 ≠ 0)

We know that the solution of a linear equation must satisfy the equation. Conversely, we can say that the value of the variable in the equation, which satisfies the equation, is the solution of the equation.

In case of a pair of linear equations, the values of x and y, which satisfy both the equations, are the solutions of the equation.

Geometrically, a linear equation represents a straight line. Every solution of an equation is a point on the line represented by the equation.

Likewise, a pair of linear equations represents two straight lines. When their graphs are drawn, there are three possibilities. Let us go through the video to understand the various possibilities.

Example 1: Find whether the given pairs of linear equations are consistent or inconsistent?

(a) 5x + 2y = 2 20x + 8y = 1

(b) 4x + y = 8

7x – 2y = 1

(c) 5x + 6y = 9

10x + 12y = 18

Solution:

(b) 4 x + y = 8

7x – 2y = 1

Here, a₁ = 4 b₁ = 1 c₁ = –8

a₂ = 7 b₂ = –2 c₂ = –1

Therefore, according to the above rule, this system of equations has a unique solution. It is represented by intersecting lines.

(c) 5x + 6y = 9

10x + 12y = 18

Here, a₁ = 5 b₁ = 6 c₁ = –9

a₂ = 10 b₂ = 12 c₂ = –18

Therefore, according to the above rule, this system of equations has infinite solutions. It is represented by coincident lines.

Example 2: There are two lines drawn in each of the following graphs. Each line represents a linear equation in two variables. State whether the pair of linear equations represented in each graph is consistent, inconsistent, or dependent.

(a)

(b)

(c)

Solution:

(a) In the given graph, the lines representing the two equations are parallel to each other and never intersect. Thus, the lines in this graph represent an inconsistent pair of linear equations in two variables as they do not have any solution.

(b) In the given graph, the lines representing the two equations intersect at a point. Thus, the lines in this graph represent a consistent pair of linear equations in two variables as they have a unique solution.

(c) In the given graph, the lines are coincident i.e., the same line represents the two equations. Thus, the lines in this graph represent a dependent pair of linear equations in two variables as they have infinite number of solutions.

Expressing Given Situations Mathematically

We come across many situations in real life when it is easy to find the solution, if we express them mathematically.

Let us see such a situation.

The coach of the school cricket team buys 5 bats and 20 leather balls for Rs 3500. After some time, some more boys joined the team for practice, so he buys another 4 bats and 15 balls for Rs 2750. Suppose that the price of bat and ball does not change in the time period.

Can we express this situation mathematically to find out the individual prices of a ball and a bat?

Let the price of a bat be Rs x and that of a ball be Rs y.

It is given that 5 bats and 20 balls cost Rs 3500.

∴ Cost of 5 bats = 5x

And cost of 20 balls = 20y

⇒ Cost of 5 bats and 20 balls = 5x + 20y

⇒ 5x + 20y = 3500

Similarly, it is also given that 4 bats and 15 balls cost Rs 2750.

⇒ 4x + 15y = 2750

Therefore, the algebraic representation of the given situation is

5x + 20y = 3500 … (1)

4x + 15y = 2750 … (2)

After solving these equations, we can find out the individual prices of the ball and the bat. Let us see some more examples.

Example 1: Aman and Yash each have certain number of chocolates. Aman says to Yash, if you give me 10 of your chocolates, I will have twice the number of chocolates left with you. Yash replies, if you give me 10 of your chocolates, I will have the same number of chocolates as left with you. Write this situation mathematically?

Solution:

Suppose Aman has x number of chocolates and Yash has y number of chocolates.

According to the first condition, Yash gives 10 chocolates to Aman so that Aman has twice the number of chocolates than what Yash has.

⇒ x + 10 = 2(y – 10)

According to the second condition, Aman gives 10 chocolates to Yash such that both have equal number of chocolates.

⇒ y +10 = x – 10

Thus, the algebraic representation of the given situation is

x + 10 = 2(y – 10) … (1)

y +10 = x – 10 … (2)

Example 2: Cadets in the military academy are made to stand in rows. If one cadet is extra in each row, then there would be 2 rows less. If one cadet is less in each row, then there would be 3 more rows. Express the given situation mathematically?

Solution:

Let the number of cadets in each row be x and the number of rows be y.

Total number of cadets = number of rows  number of cadets in each row

It is given to us that when one cadet is extra in each row, there are 2 rows less.

∴ xy = (y – 2) (x + 1) 

 xy = xy – 2x + y – 2

2x – y = – 2 … (1)

It is also given to us that if one cadet is less in each row, then there are 3 more rows.

∴ xy = (y + 3) (x – 1)

xy = xy + 3x – y – 3

3x – y = 3 … (2)

Thus, the algebraic representation of the given situation is

2x – y = – 2 … (1)

3x – y = 3 … (2)

Graphical Solution of a Pair of Linear Equations In Two Variables

Suppose you go to a stationary shop to buy some registers and pens. If you buy two registers and two pens, then you have to pay Rs 30. However, if you buy two registers and four pens, then you have to pay Rs 40.

Now, how can we represent this information algebraically, i.e., in the form of equations? Let us denote the cost of each register by Rs x and that of each pen by Rs y.

Now, it is given to us that two registers and two pens cost Rs 30.

∴ 2x + 2y = 30

⇒ x + y − 15 = 0 … (1)

We also know that two registers and four pens cost Rs 40.

∴ 2x + 4y = 40

⇒ x + 2y − 20 = 0 … (2)

Thus, we now have two linear equations in two variables from the given situation.

x + y − 15 = 0 … (1)

x + 2y − 20 = 0 … (2)

These two equations are known as a pair of linear equations in two variables. If we try to individually solve each equation, we will be unable to do so. We can arrive at an answer only if we try and solve both these equations together.

The general form of a pair of linear equations in two variables is written as

, where a1, b1, c1 are real numbers, and (i.e., a1, b1 ≠ 0)

, where a2, b2, c2 are real numbers, and (i.e., a2, b2 ≠ 0)

A pair of linear equations represents two straight lines. When their graphs are drawn, there are three possibilities.

(i)    The two lines may intersect at one point.

(ii)    The two lines may be parallel.

(iii)    They may represent the same line, i.e. they may be coincident.

The solution of a pair of linear equations is the point of intersection of their graphs.

Similarly, you can also graphically prove two lines to be inconsistent. Try it out yourself.

Limitations of graphical method:

• If the large values of variables are involved in the solution of system of linear equations then it is not convenient to use graphical method.
• If the fractional values of variables are involved in the solution of system of linear equations then it is not possible to obtain accuracy through graphical method.
• If the angle between two lines is small then the exact point of intersection cannot be obtained through graphical method.

Now, let us solve some examples to understand this concept better.

Example 1: Anshuman and Vikram have 10 marbles with them. Twice the number of marbles with Anshuman is equal to three times the number of marbles with Vikram. Solve this question graphically.

Solution:

Let us suppose that Anshuman has x marbles and Vikram has y marbles.

From the given information, we can form two equations as

x + y = 10

⇒ x + y − 10 = 0 … (i)

2x = 3y

⇒ 2x − 3y = 0 … (ii)

A table can be drawn for the corresponding values of x and y for equation (i) as

Similarly, a table can be drawn for the corresponding values of x and y for equation (ii) as

The two equations can be plotted on a graph as

As seen in the graph, the two lines intersect at the point (6, 4).

This implies that the solution for the pair of linear equations is x = 6 and y = 4. Thus, Anshuman has 6 marbles while Vikram has 4 marbles.

Example 2: Find graphically whether the following pairs of linear equations are consistent or inconsistent? If the equations are consistent, then find their solutions as well.

(a) 2x + 3y = 2

4x + 3y = 4

(b) x + y = 5

2x + 2y = 10

Solution:

(a) 2x + 3y = 2

4x + 3y = 4

A table can be drawn for the corresponding values of x and y for the equation 2x + 3y = 2 as

Similarly, a table can be drawn for the corresponding values of x and y for the equation 4x + 3y = 4 as

These points can now be plotted and joined to obtain the graphs of the equations as

As seen in the graph, the two lines intersect at the point (1, 0). Thus, the given pair of linear equations is consistent with its solution as x = 1 and y = 0.

(b)x + y = 5

2x + 2y = 10

A table can be drawn for the corresponding values of x and y for the equation x + y = 5 as

Similarly, a table can be drawn for the corresponding values of x and y for the equation 2x + 2y = 10 as

These points can now be plotted and joined to obtain the graphs of the lines as

As seen in the graph, the two equations are represented by the same line. This means that the given pair of linear equations is dependent. Any point lying on this line is a solution to the pair of equations.

Substitution Method Of Solving Pairs Of Linear Equations

In a class, the number of boys is 7 more than twice the number of girls.

Can you find the number of boys and girls in the class? Let the number of boys be x and the number of girls be y. Now, according to the given condition,

x – 2y = 7

We cannot find the unique values of x and y by solving this equation because there are multiple values of x and y for which this equation holds true.

We can write the above equation as x = 2y + 7. Thus, by taking different values of y, we will obtain different values of x.

However, if we are given one more condition, then the values of x and y can be evaluated.

To solve a linear equation in two variables, two linear equations in the same two variables are required.

Let us consider one more condition. Suppose the total number of students in the class is 52.

Now, can you find the number of boys and girls in the class?

According to this condition, we can write x + y = 52

Therefore, we obtain a pair of linear equations in two variables.

x – 2y = 7

x + y = 52

Therefore, number of boys = 37 and number of girls = 15

In this method, we have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why this method is known as the substitution method.

Now, let us solve some examples to understand this concept better.

Example 1: The sum of the digits of a two-digit number is 8. If 18 is added to the number, then the digits interchange their places. Find the number.

Example 2: Solve the following pair of linear equations.

7x – 5y + 12 = 0

3x + 8y – 5 = 0

Solution:

Given, 7x – 5y + 12 = 0 … (i)

3x + 8y – 5 = 0 … (ii)

From equation (i), we obtain

7x – 5y + 12 = 0

On substituting the value of y in equation (ii), we obtain

⇒ 71x + 71 = 0

⇒ 71x = – 71

⇒ x = – 1

Putting the value of x in equation (iii), we obtain

Therefore, x = – 1 and y = 1

Example 3: Arnab has few coins of Rs 5 and Rs 2 which together amount to Rs 167. If he has a total of 46 coins then find the number of both types of coins with him.

Solution:

Let the number of coins of Rs 2 be x and that of Rs 5 be y.

According to the question, we have

2x + 5y = 167 ...(i)

x + y = 46 ...(ii)

From equation (ii), we obtain

x = 46 – y ...(iii)

On substituting the value of x from equation (iii) in equation (i), we obtain

2(46 – y) + 5y = 167

⇒ 92 – 2y + 5y = 167

⇒ 92 + 3y = 167

⇒ 3y = 75

⇒ y = 25

On substituting the value of y in equation (ii), we obtain

x + 25 = 46

⇒ x = 21

Hence, Arnab has 21 coins of Rs 2 and 25 coins of Rs 5.

Example 4: A boat can travel 28 km downstream and 20 km upstream in 7 hours. Also, this boat can travel 21 km downstream and 18 km upstream in 6 hours. What is the speed of the boat in still water and the speed of the stream?

Solution:

Let x km/hr and y km/hr be the speed of the boat in still water and speed of the stream respectively.

Therefore,

Speed of the boat downstream = (x + y) km/hr

Speed of the boat upstream = (x – y) km/hr

∴ Time taken by the boat to travel 28 km downstream = And, time taken by the boat to travel 20 km upstream = According to the question, we have

Now,

Time taken by the boat to travel 21 km downstream = And, time taken by the boat to travel 18 km upstream = According to the question, we have

On substituting in equation (i) and (ii), we obtain

28a + 20b = 7 ...(iii)

21a + 18b = 6 ...(iv)

From equation (iv), we obtain

7a + 6b = 2

⇒ 7a = 2 – 6b

On substituting the value of a from (v) in (iii), we obtain

On putting this value of b in equation (v), we obtain

On resubstituting the values of a and b, we obtain

From (vii), we obtain x = y + 4. On substituting this value of x in (vi), we obtain

y + 4 + y = 14

⇒ 2y = 10

⇒ y = 5

On substituting y = 5 in (vii), we get

x – 5 = 4

⇒ x = 9

Thus, the speed of the boat in still water is 9 km/hr and the speed of stream is 5 km/hr.

Elimination Method To Solve A Pair Of Linear Equations

There are many real life situations which can be represented in the form of linear equations. Let us begin with such a situation.

Suppose you go to the market with your friend to buy clothes. You buy 3 shirts and 6 pairs of jeans from a shop and the total cost of your purchase comes out to be Rs 7000. Your friend, on the other hand buys 2 shirts and 3 pairs of jeans and the total cost of his purchase comes out to be Rs 3855. Now, we can represent this information mathematically as pair of linear equations in two variables. We can then solve the equations and get the

cost of each shirt and each pair of jeans. Let’s look at the following video to learn how to mathematically represent the information and then solve them using the elimination method.

Example 2: Solve the following equations using elimination method.

3x + 4y = 0

4x – 3y = 50

Solution:

Given,

3x + 4y = 0 … (i)

4x – 3y = 50 … (ii)

On multiplying equation (i) by 3 and equation (ii) by 4, we obtain

9x + 12y = 0 … (iii)

16x – 12y = 200 … (iv)

On adding equation (iii) and (iv), we obtain

25x = 200

Putting the value of x in equation (i), we obtain

3 × 8 + 4y = 0

⇒ 4y = − 24

Thus, x = 8 and y = − 6.

Example 3: Solve the following equations using elimination method.

7x + 5y = 43

5x + 7y = 41

Solution:

Given,

7x + 5y = 43 … (i)

5x + 7y = 41 … (ii)

It can be observed that coefficients of x and y in equation (i) are interchanged in equation (ii).

Let us add both the equations as follows:

⇒ 12(x + y) = 84

⇒ x + y = 7 ...(iii)

Now, let us subtract equation (ii) from equation (i) as follows:

⇒ 2(x – y) = 2

⇒ x – y = 1 ...(iv)

On adding equations (iii) and (iv), we obtain

⇒ x = 4

On putting the value of x in equation (i), we obtain

7 × 4 + 5y = 43

⇒ 28 + 5y = 43

⇒ 5y = 15

⇒ y = 3

Thus, x = 4 and y = 3.

Example 5: Five units of a solution are obtained by mixing two liquids A and B. If the liquids A and B are used in the ratio as 2 : 3 then the total cost comes out to be Rs 50.

Also, if the liquids A and B are used in the ratio as 3 : 2 then the total cost increases by 10%. Find the cost of per unit of each of the liquids A and B.

Solution:

Let x and y be costs of per unit of liquids A and B respectively. According to the question, we have

2x + 3y = 50 ...(i)

And

3x + 2y = 50 + 10% of 50

3x + 2y = 50 + 5

3x + 2y = 55 ...(ii)

On multiplying equation (i) by 3 and equation (ii) by 2, we obtain

6x + 9y = 150 ...(iii)

6x + 4y = 110 ...(iv)

On subtracting equation (iv) from equation (iii), we obtain

5y = 40

⇒ y = 8

On substituting y = 8 in equation (ii), we obtain

3x + 2(8) = 55

⇒ 3x + 16 = 55

⇒ 3x = 39

⇒ x = 13

Hence, the cost of per unit of liquid A is Rs 13 an that of liquid B is Rs 8.

Example 6: Some friends ate 5 pizzas and 3 burgers for which they paid Rs 445. One of them said "If the numbers of pizzas and burgers would have been exchanged, we would pay Rs 50 less." What is the cost of a pizza and that of a burger?

Solution:

Let the cost of a pizza be Rs x and that of a burger be Rs y.

∴ Cost of 5 pizzas = Rs 5x And, cost of 3 burgers = Rs 3y

According to the first condition, we have

5x + 3y = 445 ...(i)

Similarly, for the second condition, we have

3x + 5y = 445 – 50

⇒ 3x + 5y = 395 ...(ii)

On multiplying equation (i) by 3 and equation (ii) by 5, we obtain

15x + 9y = 1335 ...(iii)

15x + 25y = 1975 ...(iv)

On subtracting equation (iii) from equation (iv), we obtain

16y = 640

⇒ y = 40

On substituting y = 40 in equation (ii), we obtain

3x + 5(40) = 395

⇒ 3x + 200 = 395

⇒ 3x = 195

⇒ x = 65

Thus, cost of a pizza is Rs 65 and that of a burger is Rs 40.

Cross-Multiplication Method Of Solving Pairs Of Linear Equations

We can represent many situations in real life as linear equation in two variables. Let us consider such a situation.

Suppose Samay is older than Sumit by 30 years. After 5 years, Samay will be thrice as old as Sumit.

Can we find the present ages of Samay and Sumit?

Yes, we can find the present ages. However, before that, we have to represent this situation in the form of linear equations in two variables. Let us see how we can represent the above given situations as linear equations.

Let the present age of Samay be x years and the present age of Sumit be y years.

It is given that Samay is older than Sumit by 30 years. Thus, according to this condition, we have

x – y = 30

After 5 years, the age of Samay will be(x + 5) years and the age of Sumit will be (y + 5) years. However, it is given that after 5 years, Samay will be thrice as old as Sumit. Thus, we have

x + 5 = 3 (y + 5)

⇒ x + 5 = 3y + 15

⇒ x – 3y = 10

Now, we obtain the pair of linear equations as follows:

x – y = 30 … (1)

x – 3y = 10 … (2)

We can find the present ages of Samay and Sumit by solving the above equations for variables x and y.

We know three methods to solve a pair of linear equations.

(i)    Substitution method

(ii)    Elimination method

(iii)    Graphical method

Now, we will follow another method to solve linear equations in two variables. This method is known as Cross-multiplication method. Firstly, let us discuss this method and after that we will solve the above equations for x and y. So, go through the following video to understand the cross-multiplication method and how it is applied to solve the above given equations.

Therefore, the present ages of Samay and Sumit are 40 years and 10 years respectively.

In this way, we can solve a pair of linear equations in two variables by cross-multiplication method.

We could solve these linear equations in two variables by other methods also. But why should we use cross-multiplication method to solve linear equations?

The cross-multiplication method is easier as compared to the other methods if the coefficients of variables in the equations are large numbers or numbers that look complex. To understand this concept better, let us see the following examples.

Example 2: Solve the following pair of linear equations in two variables by cross- multiplication method.

ax + by = a² + b² and

x + y = 2a Solution:

The above two equations can be written as

ax + by – (a² + b²) = 0 … (1)

x + y – 2a = 0 … (2)

From the above two equations, we obtain

a₁ = a, b₁ = b, c₁ = – (a² + b²) and

a₂ = 1, b₂ = 1, c₂ = – 2a

Equations Reducible To A Pair Of Linear Equations In Two Variables

Consider the following equations.

x = y; x = 21y + 15; m + 20n = –13; etc.

We can see that the above equations are linear equations in two variables. However, there are many equations, which do not appear to be linear equations at first glance such as

Now, can we solve these equations?

Example1: Solve the following pair of linear equations.

Solution:

From equation (1), we can write,

From equation (2), we can write,

Now, equation (3) becomes 5a + b = 7 … (5)

and equation (4) becomes 2a – 3b = – 4 … (6)

On multiplying (5) by 3, we obtain

15a + 3b = 21 … (7)

On adding equations (6) and (7), we obtain

17a = 17

⇒ a = 1

On putting the value of a in equation (5), we obtain

5 × 1 + b = 7

⇒ 5 + b = 7

⇒ b = 7 – 5 = 2

Since = a = 1

⇒ x = 1

and = b = 2

⇒ y =

Thus, x = 1 and y =

Example2: Solve the following pair of linear equations.

Solution:

The given equations can be written as

Thus, the equations become

2a + 5b = 7 … (3)

And, 3a + 10b = 9 … (4)

Now, on multiplying equation (3) by 2, we obtain

4a + 10b = 14 … (5)

On subtracting equation (4) from equation (5), we obtain

a = 14 – 9

a = 5

Putting the value of a in equation (3), we obtain

10 + 5b = 7

5b = 7 – 10

b =

On adding these two equations, we obtain

2x =

Thus,

Introduction to Quadratic Equations

Consider the equations

In equation (1), variable occurs only in second degree. The quadratic equations involving a variable only in second degree is a Pure quadratic equation.

Thus, an equation that can be expressed in the form , where a and c are real numbers and a ≠ 0 is a pure quadratic equation.

In equation (2) , variable occurs both in second and first degree. The quadratic equation involving a variable in both second and first degree is an Adfected quadratic equation.

Thus, an equation that can be expressed in the form , where a, b and c are real numbers and a ≠ 0 is an Adfected quadratic equation.

Now, let us learn about the roots of a quadratic equation.

Roots of a given quadratic equation are the values of the variables which satisfy that quadratic equation.

For example, consider the quadratic equation x² – 8x + 15 = 0. Here, L.H.S. = x2 – 8x + 15 and R.H.S. = 0

Let us substitute the different values of variable x in this equation and observe the results. When x = 1 then x² – 8x + 15 = 12 – 8(1) + 15 = 1 – 8 + 15 = 8 ∴ L.H.S. ≠ R.H.S.

When x = –1 then x² – 8x + 15 = (–1)² – 8(–1) + 15 = 1 + 8 + 15 = 24 ∴ L.H.S. ≠ R.H.S.

When x = 3 then x² – 8x + 15 = 3² – 8(3) + 15 = 9 – 24 + 15 = 0 ∴ L.H.S. = R.H.S.

When x = 4 then x² – 8x + 15 = 4² – 8(4) + 15 = 16 – 32 + 15 = –1 ∴ L.H.S. ≠ R.H.S.

When x = 5 then x² – 8x + 15 = 5² – 8(5) + 15 = 25 – 40 + 15 = 0 ∴ L.H.S. = R.H.S.

When x = –5 then x² – 8x + 15 = (–5)² – 8(–5) + 15 = 25 + 40 + 15 = 80 ∴ L.H.S. ≠ R.H.S.

It can be observed that for x = 3 and x = 5, we have L.H.S. = R.H.S. which means that the equation x² – 8x + 15 = 0 is satisfied. Thus, 3 and 5 are roots of this quadratic equation.

In other words, we can say that x = 3 and x = 5 are the solutions of quadratic equation x² – 8x + 15 = 0.

Also, for x = 1, x = –1, x = 4 and x = –5, we have L.H.S. ≠ R.H.S. which means that the equation x² – 8x + 15 = 0 is not satisfied. Thus, 1, –1, 4 and –5 are not the roots or solution of this quadratic equation.

Let us go through few examples to understand the concept better.

Example 1: Is the equation a quadratic equation?

Solution:

The given equation is

We can also write it as

Hence, we can say that the given equation is a quadratic equation.

Example 2: Among the values given below find the roots of the quadratic equation 2x2 + 5x – 3 = 0.

x = 1, , x = 2, x = –3,

Solution:

We have,

Equation: 2x² + 5x – 3 = 0

Values: x = 1, , x = 2, x = –3,

Let us substitute the given values of x in the equation and observe the results. When x = 1 then 2x² + 5x – 3 = 2(1)² + 5(1) – 3 = 2 + 5 – 3 = 4 ∴ L.H.S. ≠ R.H.S.

When then

∴ L.H.S. = R.H.S.

When x = 2 then 2x² + 5x – 3 = 2(2)² + 5(2) – 3 = 8 + 10 – 3 = 15 ∴ L.H.S. ≠ R.H.S.

When x = –3 then 2x² + 5x – 3 = 2(–3)² + 5(–3) – 3 = 18 – 15 – 3 = 0 ∴ L.H.S. = R.H.S.

When then

∴ L.H.S. ≠ R.H.S.

Only, and x = –3 satisfy the given quadratic equation.

Thus, and –3 are the roots of the given quadratic equation.

Example 3: If one root of the quadratic equation 3x² + 4x – p = 0 is 2323 then find the value of p.

Solution:

Since 2323 is a root of the equation 3x² + 4x – p = 0, it must satisfy the equation.

Example 4: State whether m is a root of the equation 2x² – (2m + 3)x + 3m = 0 or not.

Solution:

If m is a root of the equation 2x² – (2m + 3)x + 3m = 0 then it would satisfy the equation. Let us check it by substituting x = m in the given equation

2m² – (2m + 3)m + 3m

= 2m² – 2m² – 3m + 3m

= 0

∴ L.H.S. = R.H.S.

Hence, m is a root of the equation 2x² – (2m + 3)x + 3m = 0.

Expressing Given Situations as Quadratic Equations in One Variable

Quadratic equations arise in several situations around us and in different fields of mathematics. But how is a quadratic equation formulated?

Let us learn to express a situation mathematically.

Suppose there are two squares as shown in the following figure.

The sum of the areas of these two squares is 482 m² and the difference of their perimeters is 32 m. Can we form a quadratic equation for the given mathematical situation?

Example1: An express train takes 1 hour less than a passenger train to travel a distance of 132 km. If the average speed of the express train is 11 km/h more than that of a passenger train, then form a quadratic equation to find the average speed of the express train?

Solution:

Let the average speed of the express train be x km/h.

Since it is given that the speed of the express train is 11 km/h more than that of a passenger train,

Therefore, the speed of the passenger train will be x −11 km/h. Also we know that

Time taken by the express train to cover 132 km =

Time taken by the passenger train to cover 132 km =

And the express train takes 1 hour less than the passenger train. Therefore,

This is the required quadratic equation.

Example 2: A chess board contains 64 equal squares and the area of each square is 6.25 cm². A border 2 cm wide is made around the board. Formulate a quadratic equation to find the length of each side of the board?

Solution:

Let the length of each side of the chess board be x cm.

Therefore, the length of the chess board without border will be x − 4 cm as shown in the following figure.

Area of the chess board without border = (x − 4)²

It is given that the area of each square is 6.25 cm².

Area of 64 squares on the chess board = 6.25 × 64

Therefore,

(x − 4)2 = 64 6.25

This is the required quadratic equation.

Solution of Quadratic Equations by the Method of Factorization

Quadratic equations come from many real life situations. Let us consider one such situation.

The area of a rectangular garden is 500 m² and its length is 5 m more than its breadth. Can we find the length and breadth of the garden?

The garden can be drawn as

Let the breadth of the garden be x m.

Then, according to the given information, its length must be (x + 5) m.

Now, area of the garden = length × breadth = (x + 5) × x

The area of the garden is given as 500 m².

∴ (x + 5) × x = 500

We can rewrite this equation as

x² + 5x = 500

⇒ x² + 5x – 500 = 0

This is a quadratic equation in variable x.

The length and the breadth of the garden can be calculated by solving the given quadratic equation.

How do we solve a quadratic equation? There are three methods of solving a quadratic equation, one of which is the method of factorization.

We have now expressed the equation x² + 5x – 500 = 0 as the product of two factors, (x + 25) and (x – 20). We can find the solutions of this equation by equating each factor to 0.

∴ x + 25 = 0 and x – 20 = 0

⇒ x = –25 and x = 20

However, the breadth of a park cannot be negative.

∴ x = 20

Thus, breadth of the park = 20 m and length of the park = (20 + 5) m = 25 m.

Let us now look at some more examples to get a grasp on this concept.

Example 1: Find the roots of the equation .

Solution:

First, we rearrange the given equation according to the standard format.

x² – x = 12

⇒ x² – x – 12 = 0

We need a pair of numbers whose product is –12 and whose sum is –1.

These numbers are 3 and –4.

We can now rewrite the equation as

x² – x – 12 = 0

⇒ x² + 3x – 4x – 12 = 0

⇒ x(x + 3) – 4(x + 3) = 0

⇒ (x – 4)(x + 3) = 0

The solutions of the given equation can now be determined by equating the two factors to 0.

∴ (x – 4) = 0 and (x + 3) = 0

⇒ x = 4 and x = –3

Thus, the two roots of the quadratic equation are 4 and –3.

Example 2: Find the roots of the equation .

Solution:

First, we rearrange the given equation according to the standard format.

6x² + 5x – 4 = 0

Here, we need to find two such numbers whose product is 6 × (–4) = –24 and whose sum is 5.

The required numbers are 8 and –3.

Now, [8 × (–3) = –24 and 8 + (–3) = 5]

6x² + 5x – 4 = 0

⇒ 6x² + 8x –3x – 4 = 0

⇒ 2x(3x + 4) – 1(3x + 4) = 0

⇒ (2x – 1)(3x + 4) = 0

⇒ 2x – 1 = 0 and 3x + 4 = 0

Thus, the roots of the given equation are .

Example 3: Find two positive consecutive odd integers whose product is 99.

Solution:

Let the first integer be x.

The next odd integer will thus be x + 2.

∴ x(x + 2) = 99

x² + 2x – 99 = 0

x² + 11x – 9x – 99 = 0

x(x + 11) – 9(x + 11) = 0

(x – 9)(x + 11) = 0

x – 9 = 0 and x + 11 = 0

x = 9 and x = –11

However, we know that the two integers are positive.

∴ x = 9

Thus, the two integers are 9 and (9 + 2) = 11.

Example 4: The width of a rectangle is 16 feet less than 3 times its length. If the area of the rectangle is 35 ft², then what are the dimensions of the rectangle?

Solution:

Here, the width of the rectangle has been expressed in terms of its length.

Hence, let us take the length of the rectangle to be x.

Thus, width of the rectangle = 3x – 16

The area of the rectangle is 35 ft².

∴ 35 = x(3x –16)

35 = 3x² – 16x

3x² – 16x – 35 = 0

3x² – 21x + 5x – 35 = 0

3x(x – 7) + 5(x – 7) = 0

(3x + 5)(x – 7) = 0

3x + 5 = 0 and x – 7 = 0

However, the length of a rectangle cannot be negative.

∴ x = 7

Thus, length of the rectangle is x = 7 ft and the width of the rectangle is 3x – 16 = (3 × 7 – 16) ft = 5 ft

Example 5: Difference between the ages of a father and his son is 22 years. If the product of their ages is 555 then what are their ages?

Solution:

Let the age of the son be x years.

Then age of the father will be 22 + x years.

Since the product of their ages is 555, we have

x(22 + x) = 555

⇒22x + x² = 555

⇒x² + 22x – 555 = 0

⇒x² + 37x – 15x – 555 = 0

⇒x(x + 37) – 15(x + 37) = 0

⇒(x + 37) (x – 15) = 0

⇒(x + 37) = 0 or (x – 15) = 0

⇒x = –37 or x = 15

Age cannot be negative, so we consider x = 15.

∴ Age of son = 15 years

∴ Age of father = (22 + 15) years = 37 years

Example 6: A bus takes 3 hours more than a train to travel a distance of 770 km. The speed of the train is more than that of bus by 15 km/hr. What are the average speeds of the bus and train?

Solution:

Let the average speed of the bus be x km/hr.

Then the average speed of the train will be (x + 15) km/hr.

Time taken by bus to cover 770 km =

 Time taken by train to cover 770 km =

According to the question, we have

⇒ x(x + 70) – 55(x + 70) = 0

⇒ (x + 70) (x – 55) = 0

⇒ (x + 70) = 0 or (x – 55) = 0

⇒ x = –70 or x = 55

Speed cannot be negative, so we will consider x = 55.

∴ Average speed of bus = 55 km/hr

∴ Average speed of train = (55 + 15) km/hr = 70 km/hr

The Concept Of Arithmetic Progression

Any given series of numbers may exhibit some special properties. A sequence of numbers in arithmetic progression is one of these special series.

Let us look at the following example.

Example 1: Check whether the following sequences form an A.P. or not.

(i)

(ii)

(iii)

(iv) A list of prime numbers greater than 2

(v)  A list of the squares of natural numbers 

(i) The given sequence is

Difference between the second and the first term

Difference between the third and the second term

Difference between the fourth and the third term

Since the difference between any two consecutive terms is a constant, the sequence is an arithmetic progression.
(ii)    The given sequence is 

Difference between the second and the first term

Difference between the third and the second term

Difference between the fourth and the third term

Since the difference between any two consecutive terms is a constant, the sequence is an arithmetic progression.
(iii)    The given sequence is 

This sequence can be rewritten as

Difference between the second and the first term

Difference between the third and the second term

Difference between the fourth and the third term

Since the difference between the consecutive terms is not a constant, the sequence is not an arithmetic progression.

(iv) The list of the prime numbers greater than 2 is: 3, 5, 7, 11 …

Difference between the second and the first term = 5 – 3 = 2

Difference between the third and the second term = 7 – 5 = 2

Difference between the fourth and the third term = 11 – 7 = 4

Since the difference between the consecutive terms is not a constant, the sequence is not an arithmetic progression.

(v)  The list of the squares of natural numbers is: 1², 2², 3², 4² …

= 1, 4, 9, 16 …

Difference between the second and the first term = 4 – 1 = 3

Difference between the third and the second term = 9 – 4 = 5

Difference between the fourth and the third term = 16 – 9 = 7

Since the difference between the consecutive terms is not a constant, the sequence is not an arithmetic progression.

The Terminologies Related To Arithmetic Progressions

An arithmetic progression (A.P.) is a sequence in which the difference between any two consecutive terms is constant.

Now there are many common terms associated with an arithmetic progression, which one has to understand to solve any A.P. related problem.

Let us solve some more examples to understand this concept better.

Example 1: Find the common difference of the A.P.

Also, state whether it is finite or infinite and write down its first term.

Solution:

The given A.P. is

a₁ =

a₂ =

a₃ =

a₄ =

∴Common difference of the A.P. =

First term of the A.P. =

The given A.P. is infinite because its last term cannot be calculated.

Example 2: State whether the following statements are true or false.

(i)    The sequence  forms an A.P. with the first term as and the common difference as .

(ii)  The common difference of the A.P. is .

(iii) The  n^th term of an A.P. is given by 3n – 4. Its common difference is 4.

Solution:

(i)  The given  sequence is

Difference between the first and the second term

Difference between the second and the third term

Difference between the third and the fourth term

Since the difference between the consecutive terms of the sequence is constant, the given sequence is an A.P. with the first term as and the common difference as .

Thus, the statement is false.

(ii)   The given A.P. is

Common difference of the A.P. =

Thus, the statement is true.

(iii) It is given that the nth term of the A.P. is given by 3n – 4.

Thus, first term of the A.P. = a = 3 × 1 – 4 = 3 – 4 = –1

Second term of the A.P. = a₂ = 3 × 2 – 4 = 6 – 4 = 2

Third term of the A.P. = a₃ = 3 × 3 – 4 = 9 – 4 = 5

Fourth term of the A.P. = a₄ = 3 × 4 – 4 = 12 – 4 = 8

Thus, common difference of the A.P. = 8 – 5 = 5 – 2 = 2 – (–1) = 3

Thus, the statement is false.

Missing Terms In Arithmetic Progressions

An arithmetic progression (A.P.) is a sequence in which the difference between any two consecutive terms is constant.

Now if we are given an A.P. 3, 6, 9, ?, 15 and asked to find the missing term, then how will we go about it?

Let us now solve some problems based on this concept.

Example 1: Find the missing terms in the following arithmetic progressions.

(i) 

(ii) 

Solution:

(i)  In the given A.P.,

Thus, missing term, a₃

(ii) In the given A.P.,

Thus, missing term, a₂ =

Example 2: Find the first two terms of an A.P. in which the third term is 10 and the common difference is 7.

Solution:

Here, d = 7 and a₃ = 10

Hence, the value of a₂ can be found by subtracting 7 from a₃.

∴ a₂ = 10 − 7 = 3

Similarly, the value of a1 can be found by subtracting 7 from a₂.

∴ a₁ = 3 − 7 = −4

Thus, the first two terms of the A.P. are –4 and 3.

Difference Between Similarity and Congruence

Similar figures and congruent figures may appear to be closely related concepts, but there is an important difference between them.

Congruency of line segments:

“Two line segments are congruent to each other if their lengths are equal”.

Consider the following line segments.

Here, the line segments AB and PQ will be congruent to each other, if they are of equal length.

Conversely, we can say that, “Two line segments are of equal length if they are congruent to each other”.

i.e. if , then AB = PQ.

Congruency of angles:

“Two angles are said to be congruent to each other if they have the same measure”.

The angles shown in the following figures are congruent to each other as both the angles are of the same measure 45°.

Thus, we can write ∠BAC ≅ ∠QPR.

Its converse is also true.

“If two angles are congruent to each other, then their measures are also equal”.

There is one special thing about congruent figures that their corresponding parts are always equal.

For example, if two triangles are congruent then their corresponding sides will be equal. Also, their corresponding angles will be equal.

Look at the following triangles.

Here, ∆ABC ∆DEF under the correspondence ∆ABC ↔ ∆DEF. This correspondence rule represents that in given triangles, AB ↔ DE (AB corresponds to DE), BC ↔ EF, CA ↔ FD, ∠A

— ∠D, ∠B ↔ ∠E, ∠C ↔ ∠F. These are corresponding parts of congruent triangles (CPCT), ∆ABC and∆DEF.

Since ∆ABC and ∆DEF are congruent, their corresponding parts are equal. Therefore, AB = DE, BC = EF, CA = FD

And, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F

Similarly, we can apply the method of CPCT on other congruent triangles also. Let us now try and apply what we have just learnt in some examples.

Example 1: Find which of the pairs of line segments are congruent.

(i)

(ii)

Solution:

(i)    Lengths of the two line segments are not same. Therefore, they are not congruent.

(ii)    Each of the line segments is of length 3.1 cm, i.e. they are equal. Therefore, they are congruent.
 

Example 2: If and = 9 cm, then find the length of . Solution:

Since , i.e. line segment AB is congruent to line segment PQ,

therefore, and are of equal length.

∴ = 9 cm

Example 3: If ∠ABC ∠PQR and ∠PQR = 75o, then find the measure of ∠ABC. Solution:

If two angles are congruent, then their measures are equal.

Since ∠ABC ≅ ∠PQR,

∴ ∠ABC = ∠PQR T

herefore, ∠ABC = 75^o

Example 4: Which of the following pairs of angles are congruent?

(i)

(ii)

Solution:

(i)    The measure of both the angles is the same. Therefore, they are congruent.

(ii)    The measures of the two angles are different. Therefore, they are not congruent.

Example 5: Identify the pairs of similar and congruent figures from the following. 

(i)

(ii)

(iii)    

(iv)    

(v)    

(vi)    

(vii)    

(viii)    

Solution:

Figures (i) and (iii) are similar because their corresponding angles are equal and their corresponding sides are in the same ratio. However, these figures are not congruent as they are of different sizes.

Figures (ii) and (viii) are congruent as they are of the same shape and size (circles with radius 1 cm each).

Example 6:

Are the following figures similar or congruent?

Solution:

The two given figures show two one-rupee coins. As both the figures represent the same coin in two different sizes, they are similar to each other. However, the pictures are not congruent because of their different sizes.

Example 7: In the following figure, ΔPQR and ΔSTU are congruent.

If PQ = 8 cm, QR = 6 cm then find the perimeter of ΔSTU.

Solution:

In ΔPQR, we have

PQ = 8 cm, QR = 6 cm and ∠Q = 90°

Applying Pythagoras theorem in ΔPQR, we obtain

RP² = PQ² + QR²

⇒ RP² = 8² + 6²

⇒ RP² = 64 + 36

⇒ RP² = 100

⇒ RP = 10 cm

Since ΔPQR and ΔSTU are congruent, their corresponding parts will be equal. Therefore,

PQ = 8 cm = ST (CPCT)

QR = 6 cm = TU and (CPCT) RP = 10 cm = US (CPCT)

∴ Perimeter of ΔSTU = ST + TU + US = 8 cm + 6 cm + 10 cm = 24 cm

Basic Proportionality Theorem and Its Converse

Consider the following figure.

In the above figure, DE is parallel to AF and DF is parallel to AC. Can we say that point E divides BF in the same ratio in which point F divides BC?

For this, we have to prove that .

To prove it, we should have the knowledge of basic proportionality theorem (Thales theorem).

Now, let us solve the problem discussed in the beginning with the help of BPT. In ABF, we know that AF is parallel to DE.

Thus, using BPT,

… (1)

Similarly in ABC, DF is parallel to AC. Thus, using BPT,

… (2)

From equations (1) and (2), we obtain

Thus, we can say that point E divides BF in the same ratio in which point F divides BC. The converse of BPT is also true, which can be stated as follows.

“If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side”.

Corollary of BPT:

If a line is drawn parallel to a side of a triangle, then the sides of the new triangle formed are proportional to the sides of the given triangle.

I.e, In the given figure, if DE || AB, then

Applications of Basic Proportionality Theorem:

There are two very important properties based on BPT which are as follows:

1. Property of intercepts made by three parallel lines on a transversal.
2. Property of angle bisector of a triangle.

Let us discuss these properties in detail along with their proofs.

Property 1: Intercept Theorem

The lengths of the intercepts made by three parallel lines on one transversal are in the same ratio as the lengths of the corresponding intercepts made by the same lines on any other transversal.

Let us prove this property.

Given: line l || line m || line n

Transversal x intersects these lines at points P, Q and R while transversal y intersects these lines at points S, T and U.

To prove:

Construction: Draw a line segment PU intersecting line m at point M.

Proof:

In ΔPRU, we have QM || RU

Similarly, in ΔUSP, we have TM || SP

Hence proved.

Property 2: Angle Bisector Theorem

In a triangle, the angle bisector divides the side opposite to the angle in the ratio same as the ratio of remaining sides.

Let us prove this property.

Given: In ΔPQR, ray RT bisects ∠PRQ.

To prove:

Construction: Draw a ray from Q parallel to ray RT such that it intersects extended PR at S.

Proof:

We have,

RT || QS and PS is transversal

∴ ∠PRT = ∠RSQ ...(1) (Corresponding angles)

Considering other transversal RQ, we obtain

∠TRQ = ∠RQS ...(2) (Alternate angles) But ∠PRT = ∠TRQ (RT bisects ∠PRQ)

∴ ∠RSQ = ∠RQS [Using (1) and (2)]

Thus, in ΔRQS,

RS = RQ ...(3) (Side opposite to equal angles are equal) Now, in ΔPQS, we have

RT || QS

Hence proved.

Property 3: Converse of Angle Bisector Theorem

If a straight line through one vertex of a triangle divides the opposite side in the ratio of the other two sides, then the line bisects the angle at the vertex.

Let us prove this property.

Given: In ΔPQR, line PT divides the opposite side BC internally such that.

Hence proved.

These properties are very useful sometimes.

Let us now solve some examples based on BPT, its converse and properties related to BPT.

Example 1: In triangle ABC, D and E are points on the sides AB and AC, such that AB =11.2 cm, AD = 2.8 cm, AC = 14.4 cm, and AE = 3.6 cm. Show that DE is parallel to BC.

Solution:

It is given that,

AB = 11.2 cm, AD = 2.8 cm, AC =14.4 cm, and AE = 3.6 cm

Therefore, BD = AB – AD = 11.2 – 2.8 = 8.4 cm

And,

EC = AC –AE =14.4 – 3.6 = 10.8 cm

Now,

Thus, DE divides sides AB and AC of triangle ABC in the same ratio. Therefore, by the converse of BPT, we obtain that DE is parallel to BC.

Example 2: In the figure shown below, find the length of PM, if it is given that LM || QR. The corresponding measures are shown in the figure.

Solution:

Here, LM || QR

Then, using basic proportionality theorem, we obtain

Let PM = x cm

Then, MR = 12 – x cm And, PL = 2 cm

LQ = 10 cm – 2 cm = 8 cm

On putting these values in equation (i), we obtain

2(12 – x) = 8x

24 – 2x = 8x

24 = 8x + 2x

24 = 10x

x = 2.4 cm

Thus, PM = 2.4 cm

Example 3: If ABCD is a trapezium with AD || BC, then prove that , where O is the point of intersection of diagonals AC and BD.

Solution:

A trapezium has been shown in the following figure.

A line LM is drawn parallel to AD and BC and passing through O.

Here, LO || BC.

Using BPT in ΔABC,

Similarly, using BPT in ΔABD as LO || AD, we obtain

From equations (i) and (ii), we obtain

Hence, proved

Example 4: In ΔPQR, LM || QR and L is the mid-point of side PQ. Show that PM = MR.

Solution:

Here, LM || QR

Using basic proportionality theorem (BPT),

Now, L is the mid-point of PQ.

∴ PL = LQ

Using this in equation (i), we obtain

∴ PM = MR

Hence, proved

Example 5: In trapezium ABCD, AB || EF || DC. Find the length of BF and FC.

Solution:

In trapezium ABCD, AB || EF || DC.

Here, AD and BC are transversals to parallel segments AB, EF and DC.

Intercepts made by AD are AE and ED while intercepts made by BC are BF and FC. Using property of intercepts made by three parallel lines on a transversal, we obtain

Thus, BF = 6 cm and FC = 4 cm.

Example 6: In ΔABC, BD bisects ∠ABC. Find the length of AD.

Solution:

In ΔABC, BD bisects ∠ABC

Thus, by using the property of angle bisector of a triangle, we obtain

⇒ AD = 1.5

Hence, the length of AD is 1.5 cm.

Example 7.

In the given figure, DE || AB. If perimeter of Δ ABC : perimeter of ΔCDE = 4:5 and DE = 1.2 cm, then find the length of AB.

Answer:

It is given that, perimeter of ΔABC : perimeter of ΔCDE = 4:5 and DE = 1.2. InΔ ABC, DE || AB.

By applying the corollary of basic proportionality theorem, we get

Distance Formula

Let us consider the following graph.

In the above figure, O is the origin, A is a point at a distance of 3 units on x-axis, and B is a point at a distance of 4 units on y-axis.

Can we find the distance between the two points A and B?

In coordinate geometry, the distance formula has a lot of applications. Let us look at one of its applications.

What can you say about the three points (–6, 10), (–1, 1), and (3, –8)? Are they collinear?

Let us see.

If the sum of the distances of any point from the other two is equal to the distance between the other two points, then we can say that the given points are collinear.

For example, let (–6, 10), (–1, 1), and (3, –8) be denoted by A, B, and C respectively.

Now, using distance formula, , we obtain

Here, AB + BC

≠ AC

Therefore, the points A, B, and C are not collinear.

Let us solve some more examples based on the distance formula.

Example 1: Find the distance between the point (7, –1) and origin.

Solution:

Let the points (7, –1) and origin (0, 0) be denoted by A and O respectively. Then, using distance formula, we obtain

Thus, the distance between the points (7, –1) and (0, 0) is units.

Example 2: Find the value of a, if the distance between (a, –1) and (8, 3) is .

Solution:

Let points (a, –1) and (8, 3) be denoted by A and B respectively. Then, using distance formula, we obtain

The distance is given as . Therefore, we can write

On squaring both sides, we obtain

16 × 5 = 64 + a² – 16a + 16

⇒ 80 – 64 – 16 = a² – 16a

⇒ 80 – 80 = a² – 16a

⇒ a² – 16a = 0

⇒ a (a – 16) = 0

∴ a = 0 and 16

Thus, the values of a are 0 and 16.

Example 3: Find the point on y-axis that is equidistant from (–5, 2) and (9, –2).

Solution:

The x-coordinate of all points on the y-axis is zero. Therefore, let (0, b) be the point which is equidistant from (–5, 2) and (9, –2).

∴ Distance between (0, b) and (–5, 2) = Distance between (0, b) and (9, –2) Using distance formula, we obtain

On squaring both sides, we obtain

29 + b² – 4b = 85 + b² + 4b

⇒ 8b = 29 – 85

⇒ 8b = –56

⇒ b = –7

Thus, the point (0, –7) on the y-axisis equidistant from (–5, 2) and (9, –2).

Example 4: Check whether the points (–3, –2), (–2, 3), and (3, 4) are the vertices of an isosceles triangle or not.

Solution:

Let the vertices (–3, –2), (–2, 3), and (3, 4) be denoted as A, B, and C respectively as shown in the figure.

Using distance formula, we obtain

BC

AC

Now, AB = BC = units

Thus, (–3, –2), (–2, 3), and (3, 4) are the vertices of an isosceles triangle.

Trigonometric Ratios

Suppose a boy is standing in front of a lamp post at a certain distance. The height of the boy is 170 cm and the length of his shadow is 150 cm.

You can see from the above figure that the boy and his shadow form a right-angled triangle as shown in the figure below.

The ratio of the height of the boy to his shadow is 170:150 i.e., 17:15.

Is this ratio related to either of the angles of ΔABC?

We can also conclude the following:

cos A , tan A

Also, note that

tan A = and cot A =

Let us now solve some more examples based on trigonometric ratios.

Example 1: In a triangle ABC, right-angled at B, side AB = 40 cm and BC = 9 cm. Find the value of sin A, cos A, and tan A.

Solution:

It is given that AB = 40 cm and BC = 9 cm

Using Pythagoras theorem in ΔABC, we obtain

(AC)² = (AB)² + (BC)²

(AC)² = (40)² + (9)²

(AC)² = 1600 + 81

(AC)² = 1681

(AC)² = (41)²

AC = 41 cm

Now, sin A

cos A

tan A

Example 2: From the given figure, find the values of cosec C and cot C, if AC = BC + 1.

Solution:

Now, it is given that AB = 5 cm and

AC = BC + 1 … (1)

By Pythagoras theorem, we obtain

(AB)² + (BC)² = (AC)²

⇒ (AC)² − (BC)² = (AB)²

⇒ (BC + 1)² – (BC)² = (5)² [Using (1)]

⇒ (BC)² + 1 + 2BC – (BC)² = 25

⇒ 2BC = 25 – 1

⇒ 2BC = 24

⇒ BC = 12 cm

∴ AC = 12 + 1 = 13 cm

Thus, cosec C =

=

And, cot C =

=

Example 3: In a right-angled triangle ABC, which is right-angled at B, tan A . Find the value of cos A and sec A.

Solution:

It is given that tan A

We know that tan A

⇒

Let BC = 12k and AB = 5k

Using Pythagoras theorem in ΔABC, we obtain

(AC)² = (AB)² + (BC)²

= (5k)² + (12k)²

= 25k² + 144k²

(AC)² = 169k²

AC = 13k

Now, cos A

Sec A

Use Trigonometric Ratios In Solving Problems

If we know the value of a trigonometric ratio, then we can find the values of other trigonometric ratios and the value of any expression involving these trigonometric ratios.

Example1: If cot A , then find the value of .

Solution:

It is given that cot A .

We know that

tan A

tan A = 3

Then,

Example 2: Find the value of (sin² θ + cos² θ), if sec θ .

Solution:

We have sec θ … (i)

We know that sec θ

Let ABC be a triangle in which ∠C = θ, therefore we have

sec θ

From equations (i) and (ii), we have

Let us take AC = 25k and BC = 24k.

Using Pythagoras theorem, we have,

(AC)² = (AB)² + (BC)²

(25k⁾² = (AB)² + (24k)²

625k² = (AB)² + 576k²

AB² = (625 − 576)k²

AB² = 49k²

AB = 7k

∴ sin θ

cos θ

Now, sin² θ + cos² θ

= 1

∴ sin² θ + cos² θ = 1

Some Applications of Trigonometry

Solving Real World Problems Involving Heights and Distances

Tushar is looking at the top of a tree. The height of Tushar is 1.6 m. The distance between the foot of the tree and Tushar is 12 m. The line joining his eye and the top of the tree makes an angle of 30° with the horizontal line. Tushar wants to know the height of the tree.

Can we help him?

In this way, we can find the unknown distance or height using the concept of trigonometric ratios. Let us discuss some examples based on problems related to heights and distances.

Example 1: A boy is looking at a ball on the ground from the top of a building which is    high. If the distance between his eye and the ball is 40 m, then what will be the angle of depression and the distance between the foot of the building and the ball?

Solution:

The diagram of this situation can be drawn as follows

Let the angle of depression be θ and the distance between the foot of the building and the ball be x m.

It is given that,

AC = line of sight = 40 m

AB = height of the building = 

The angle of depression = ∠ACB = θ
 
Now, in right-angled triangle ABC, we have

Thus, the angle of depression is 60°.

Again, in    , we have

Thus, the distance between the foot of the building and the ball is 20 m.

Example 2: A man is standing at a distance of 20 m at a point C in front of a lamp post AB. The angle of elevation at point C is 60º. Find the distance covered by the man if he walks away from the lamp post so that the angle of elevation becomes 30°.

Solution:

The figure of this situation can be drawn as follows.

Let the man walk x m and reach at point D. Let the height of the lamp post AB be h. Now, in right-angled triangle ABC, we have 

Thus, the height of the lamp post is   . In    , we have

Thus, the required distance is 40 m.

Example 3: Two buildings are at a distance of    from each other. A boy standing at the top of the smaller building is looking at the top and bottom of the other building. The angle of elevation and depression are 60° and 30° respectively. What will be the heights of the two buildings?

Solution:

The given situation can be represented geometrically by the following figure.

Let the height of smaller building be x m and larger building be y m.

Here, the angle of elevation, i.e., ∠CAB = 60°

and the angle of depression, i.e., ∠BAD = 30°.
 
∠BAD and ∠ADE are alternate interior angles.

∴ ∠ADE = ∠BAD

= 30°

From the figure, we have

BD = AE = x
∴ BC = CD − BD

= y − x

Now, in right-angled    , we have

Now, in right angled   , we have

Using equation (1), we have

y − 15 = 45

⇒ y =45 + 15 = 60 m
 
Thus, the heights of the two buildings are 15 m and 60 m respectively.
 
 

Concept Of Tangent At Any Point Of The Circle

A circle has important elements such as chords, secants, and tangents.

A tangent at any point of a circle is perpendicular to the radius through the point of contact.

In the above figure O is the centre of circle, line l is the tangent and P is point of contact.

∴ l ∴ OP

Proof:

It is given that O is the centre of the circle, l is the tangent to this circle and P is the point of contact.

Let us assume l is not perpendicular to the radius of the circle.

In this case, let us draw perpendicular OA to tangent l. Thus, point A is distinct from point P.

Let B be any point on tangent such that BAP is a line and BA = AP.

Now, in ΔOAB and ΔOAP, we have

OA = OA (Common side)

∠OAB = ∠OAP (OA ⊥ tangent l)

BA = AP (By construction)

∴ ΔOAB ΔOAP

∴ OB = OP (By CPCT)

Since OB = OP, point B also lies on the circle.

Also, point B is different from point P.

Thus, tangent l touches the circle at two distinct points. This contradicts the definition of tangent.

Hence, our assumption is wrong.

Therefore, tangent l ⊥OP.

Hence proved.

The converse of this theorem is also true which states that:

The line perpendicular to the radius of a circle at its outer end is tangent to the circle.

Proof:

Let O be the centre of the circle, OP be the radius and l be the line perpendicular to OP such as it passes through point P on the circle.

Also, let A be any point on line l distinct from P.

From the figure, it can be observed that ΔOAP is a right angled triangle.

∴ OA is hypotenuse for ΔOAP.

∴ OA > OP (Radius)

∴ OA is not radius.

∴ Point A does not lie on the circle.

∴ No point of line l other than P lies on the circle.

∴ P is the only point common to the circle and line l.

∴ Line l is tangent to the circle at point P. Hence proved.

Let us now solve some examples related to the tangents of the circle.

Example 1: Draw a circle with centre O, and two lines such that one is a tangent and other is a secant.

Solution:

The figure can be drawn as follows.

Here, is the secant, which intersects the circle at C and D and is a tangent whose point of contact with the circle is P.

Example 2: Which of the following statements are correct?

(i)  There can be only one tangent at a point on the circle.

(ii)  Diameter is also a secant of the circle.

Solution:

(i)    Correct

(ii)    Incorrect

Example 3: A line XY is a tangent of the circle with centre O and radius 6 cm. The point of contact is P, and Q is any point on the tangent XY. If OQ = 10 cm, then what is the length of PQ?

Solution:

The figure can be drawn as follows.

We know that the radius through the point of contact is perpendicular to the tangent.

∴ OP⊥XY

Using Pythagoras theorem in right-angled triangle OPQ, we obtain (OQ)² = (OP)² + (PQ)²

(PQ)² = (OQ⁾² – (OP)²

(PQ)² = (10)² – (6)²

(PQ)² = 100 – 36

(PQ)² = 64

PQ = 8 cm

Thus, the length of PQ is 8 cm.

Example 4: Two tangents XY and PQ are drawn at the ends of a diameter AB of the circle

with centre O. Show that XY PQ.

Solution:

The figure can be drawn as follows.

XY is a tangent at A and OA is the radius.

We know that the radius through the point of contact is perpendicular to the tangent.

∴ XY⊥OA

⇒ XY⊥AB … (1)

Similarly, PQ⊥AB … (2)

Now, the lines perpendicular to the same line are parallel.

∴ From equations (1) and (2), we obtain

XY || PQ

Hence, proved

Example 5: A circle is inscribed in a triangle such as it touches all the three sides of the triangle. Prove that the area of the triangle is half the product of its perimeter and the radius of circle.

Solution:

Let O be the centre of the circle inscribed in ΔABC.

It can be seen that AB, BC and CA are tangents which touches the circle at P, Q and R respectively.

Also, OP, OQ and OR are the radii of the circle.

We know that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, AB ⊥ OP, BC ⊥ OQ and CA ⊥ OR.

Let s be the perimeter of ΔABC.

∴ s = AB + BC + CA

Now,

Hence proved.

Example 6: Observe the given figure.

Find the angle between radii.

Solution:

In the given figure, O is the centre of the circle and OP and OR are the radii. Also, QP and SR are the tangents to the circle at points P and R respectively which intersect each other at point A.

It is given that,

∠PAS = 110°

We need to find the angle between radii OP and OR i.e., ∠POR.

From the figure, we have

∠PAS + ∠PAR = 180°

⇒ 110° + ∠PAR = 180°

⇒ ∠PAR = 180° – 110°

⇒ ∠PAR = 70°

We know that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

Therefore, QP ⊥ OP and SR ⊥ OR.

∴ ∠APO = 90° and ∠ARO = 90°

Now, in quadrilateral APOR, we have

∠POR +∠APO + ∠ARO + ∠PAR = 360° (By angle sum property of quadrilaterals)

⇒ ∠POR + 90° + 90° + 70° = 360°

⇒ ∠POR + 250° = 360°

⇒ ∠POR = 110°

Thus, the angle between the radii is 110°.

Division of a Line Segment into a Given Ratio

Vijay’s class teacher told him to mark two points A and B on a line segment PQ of length 10

cm such that and . He does not know how to construct it. Can we help him? Now let us try to mark a point B in the same way.

It is given that .

Let us take PB = x and BQ = 2x. It is also given that PQ = 10 cm.

But, PQ = PB + BQ

10 cm = x + 2x 3x = 10 cm

x = cm

Now, it is difficult to mark a point B on the line segment PQ, since the value of x is in decimal form and even non-terminating.

We can observe that the marking of point A was only possible due to two factors.

1.We were able to measure the length of AQ and PA.

2. The length of the line PQ was divisible by the sum of the numerator and denominator in the ratio 3:2 i.e., the sum of numerator and denominator is 3 + 2 = 5. The length of PQ i.e., 10 cm was divisible by 5.

But in the second example, length of PQ is not divisible by the sum of the denominator and numerator of the given ratio. In order to carry out such constructions, we follow two methods. Let us learn each of these methods with the help of examples.

Let us suppose that we have to divide a line segment of 8.8 cm in the ratio 4:7.

Let us discuss one more example to understand the concept better.

Example: Draw a line segment of length 6.2 cm and divide it in the ratio 5:3.

Solution:

To divide a line segment of length 6.2 cm in the ratio 5:3, we follow the below given steps.

1. We draw a line segment PQ of length 6.2 cm.

2. Now, we draw a ray PX making an acute angle with PQ and draw a ray QY parallel to PX by making ∠PQY equal to ∠QPX.

3.  We locate 5 points P₁, P₂, P₃, P₄, and P5 on PX and 3 points Q₁, Q₂, and Q₃ on QY such that PP₁ = P₁P₂ = P₂P₃ = P₃P₄ = P₄P₅ = QQ₁ = Q₁Q₂ = Q₂Q₃.

4. Now, we join P5Q3 which intersects PQ at R.

Now, R is the point on PQ which divides the line segment PQ in the ratio 5:3.

Construction Of A Triangle Similar To The Given Triangle

Consider the following triangle ABC.

In the above figure, the measures of the sides AB, BC, and CA are given as 3.8 cm, 7.9 cm, and 6.5 cm respectively.

Can we construct a triangle similar to ΔABC and whose sides are of the sides of ΔABC?

Example 1: Construct a triangle similar to the following triangle with sides of the sides of the given triangle.

Solution:

Firstly, we draw a ray QX making an acute angle with QR on the side opposite to vertex P and mark 5 points (corresponding to the greater of 2 or 5) X1, X2, X3, X4, and X5 on QX such that QX1 = X1X2 = X2X3 = X3X4 = X4X5.

Now, we join X2R (corresponding to the denominator of the given ratio) and draw a line passing through X5 (corresponding to the numerator of the given ratio) and parallel to X2R. Let this line intersect the extended line QR at point R′. Now, we draw a line passing through R′ and parallel to RP, which intersects the extended line PQ at point P′.

Thus, P′QR′ is the required triangle which is similar to the given triangle and with sides of the sides of the given triangle.

Areas Related to Circles

Length of Arc and Area of Sector of Circle

Some important formulae to remember are:

In case of quadrant, θ = 90°. Therefore,

In case of semicircle, θ = 180°. Therefore,

Relation between the area of sector and the length of an arc:
Look at the following figure.

Here, O is the centre of the circle of radius r and APB is an arc of length l.

Clearly, measure of arc APB = Central angle = θ

Let the area of sector OAPB be A.

Now, we have

 From length of arc and area of corresponding sector, we can also conclude that.

These formulae are very helpful to calculate various attributes related to circle. Let us solve some examples to learn the concept better.
Example 1: Find the area of the sectors OACB and OADB. Also find the length of the arc ACB. (Take π =    )

Solution:

Given, radius of the circle, r = 6.3 cm 

Angle of sector OACB, θ = 40°

θ360×πr²

 = 13.86 cm²

And, area of sector OADB = πr² − area of the sector OACB

= 124.74 − 13.86 cm²

= 110.88 cm²

Length of arc ACB =  

= 4.4 cm

Example 2: Find the area swept by the hour hand of length 10 cm of a clock when it moves from 2 p.m. to 4 p.m. Also find the length covered by the tip of the hour hand during that time. (Take π =3.14)

Solution: 

When an hour hand moves from 2 p.m. to 4 p.m., it moves for 2 hours.

The hour hand takes 12 hours to complete 1 rotation i.e., to rotate 360°.

In 1 hour, it will rotate = 

In 2 hours, it will rotate = 30° × 2 = 60°

Therefore, when it moves from 2 p.m. to 4 p.m., it sweeps an area equal to the area of a sector of angle 60°.
 
Radius of this sector = length of the hour hand = 10 cm

∴ Area swept 

= 52.33 cm² (approx.)

Length covered by the tip 

= 10.47 cm (approx.)

Example 3: If the area and length of the arc of a sector are   cm² and   cm respectively, then find out the radius of the circle and the angle of sector.

Solution:

Let r be the radius of the circle and θ be the angle of sector.

It is given that the area of the sector is  cm².

Also, the length of the arc is  cm.

⇒   = … (2)
On dividing Equation (1) with Equation (2), we obtain

cm
 
On putting the value of r in equation (1), we obtain

   =  

Thus, the radius of the circle is 16 cm and the angle of the sector is 105°.

Example 4: Find the area of the sector OAXB and length of the arc AXB for the given figure.

Solution:

In the figure, OA = AB = OB (All are of length 7 cm) Hence, ΔOAB is an equilateral triangle.
∴ 

⇒ θ = 60°

Area of the sector =  cm²

Length of the arc =  cm

Example 5: A truck has two wipers which do not overlap and each wiper has a blade of length 28 cm sweeping through an angle of 90°. What is the total area cleaned at each sweep of the blades?
 
Solution:

It is clear that each wiper sweeps a quadrant of a circle of radius 28 cm.

Area cleaned at each sweep =  

Example 6: Radius of a circle is 14 cm. What is the area of the sector corresponding to the arc of length 20 cm?

Solution:

Radius of circle (r) = 14 cm

Length of arc (l) = 20 cm

Example 7: Area of a sector of a circle whose radius is 12 cm is 132 cm2. Find the measure of central angle and length of the corresponding arc.

Solution:

Area of sector (A) = 132 cm²

Radius of circle (r) = 12 cm
Let the measure of central angle be θ and length of arc be l.

Now, we have

Thus, the central angle measures 105°.

Also, we have

 

Curved Surface Area Of Combination Of Solids

We know how to find the curved surface areas of three-dimensional figures such as cone, cylinder, sphere, etc. Now, we will learn to find out the curved surface area of a combination of solids.

Let us consider the following figure.

The given figure shows a temple. In this figure, the base of the temple is a right circular cylinder and the top is a right circular cone. The height of the conical part is half the height of the cylindrical part and the radius of the base of the temple is 2 ft more than the height of the conical part. The temple has to be plastered from outside at the rate of Rs 5 per square foot.

Can we find the approximate cost of plastering the temple from outside?

In order to find the cost, we have to find the curved surface area of the temple. By adding the curved surface areas of cone and cylinder, we can obtain the curved surface area of the temple.

First of all, let us review the formulae to find the curved surface areas of different solids. The formulae for different solids are given in the following table.

In this way, we can find the curved surface area of a combination of solids. Now, let us solve some more examples to understand the concept better.

Example 1: A toy is in the form of a hemisphere mounted by a cone. The diameter of the hemisphere is 21 cm and height of the whole toy is 24.5 cm. If the surface of the

toy is painted at the rate of Rs 1 per 5 cm2, then find the cost required to paint the entire toy.

Solution:

It is given that the diameter of the hemisphere is 21 cm.

∴ Radius of the hemisphere, r = 10.5 cm

Radius of the base of the cone, r = 10.5 cm

From the figure, height of cone, h = 24.5 − 10.5 = 14 cm

Now, l = = cm

∴ Surface area of the toy = C.S.A. of hemisphere + C.S.A. of cone

= 2πr² + πrl

= πr (2r + l)

=

= 22 × 1.5 × 38.5

= 1270.5 cm²

∴ Cost required for painting = 1270.5 × = Rs 254 approximately

Example 2: The given figure shows a post office box. It was painted with red colour and it is observed that the ratio between the cost of painting the hemispherical part and the cylindrical part is 2:7. What is the ratio between the height of the box and the radius of the hemispherical part?

Solution:

As the box is painted with a constant cost, irrespective of the hemispherical part and the cylindrical part, we obtain

, where r is the radius of the hemispherical part and h is height of the

cylindrical part

Thus, the ratio between the height of the box and the radius of the hemispherical part is 9:2.

Example 3: A tent is in the shape of a cylinder surmounted by a conical top. If the ratio of the height of the cylinder, the radius of cylinder, and the height of the cone is 8:4:3 and the cost of canvas required for making the tent is Rs 10560 at the rate of Rs 10 per m2, then find the height of the tent.

Solution:

Let us consider, height of cylinder = h’, radius of cylinder = r, and height of cone = h

Let h’ = 8x, r = 4x, and h = 3x

Then, l = = = 5x

Amount of canvas = m²

However, amount of canvas = C.S.A. of cylinder + C.S.A. of cone

1056 = 2πrh’ + πrl

1056 = πr (2h’ + l)

1056 =

1056 =

264x² = 1056

x² =

x = m

Height of tent = height of cylindrical part + height of conical part

= 8x + 3x

= 11x

= 11× 2

= 22

Thus, the height of the tent is 22 m.

Board question(s) related to this lesson:

Total Surface Areas Of Combination Of Solids

Total surface area of any solid is the sum of its curved surface area and bases. In many situations, we are required to find the total surface area of a combination of solids. Let us consider such a situation.

The largest possible hemisphere is taken out from a cube of side 10 cm. Can we find out the total surface area of the remaining solid?

To find it, consider the following figure.

From the cube, the largest hemisphere is taken out.

That means, diameter of hemisphere = edge of the cube = 10 cm

∴ Radius of the hemisphere =

Total Surface area of the remaining solid = T.S.A. of cube without the circular base + C.S.A. of hemisphere

= 678.5 cm²

Thus, the required area is 678.5 cm².

Let us solve some more examples to understand the concept better.

Example 1: From a cylinder of radius 7 m and height 24 m, a cone of same radius and same height is taken out. Find the surface area of the remaining part. If the inner portion of this remaining part is painted in pink colour at the rate of Rs 23 per 5 m² and the exterior portion is painted in green colour at the rate of Rs 40 per 11 m², then find the total cost required to paint such a solid.

Solution:

It is given that the radius and height of the cylinder and cone are the same. Let radius r = 7 m and height h = 24 m

Now, slant height of the cone, l == m

After taking out the cone from the cylinder,

T.S.A. of remaining outer part = C.S.A. of cylinder + Area of Circle

= 2πrh + πr²

= πr (2h + r)

=

= 22 × 55

= 1210 m²

T.S.A. of remaining inner part = C.S.A. of cone

= πrl

=

= 550 m²

Cost of painting the exterior part =

Cost of painting the interior part =

∴ Cost of painting the remaining solid = 4400 + 2530 = Rs 6930

Thus, the required cost is Rs 6930.

Example 2: A right triangle, whose perpendicular sides are 30 cm and 40 cm, is made to revolve about its hypotenuse. Find the surface area of the double cone so obtained.

Solution:

Let ABD be the right-angled triangle such that AB = 30 cm and AD = 40 cm Using Pythagoras theorem, we obtain

BD² = AB² + AD²

BD² =

BD = cm

Let OB = x and OA = y

Applying Pythagoras theorem in ΔAOB, we obtain

OA² + OB² = AB²

x² + y² = 30²

y² = 900 − x² … (1)

Applying Pythagoras theorem in ΔAOD, we obtain

OA² + OD² = AD²

y² + (50 − x) ² = 40²

900 − x² + 2500 + x² − 100x = 1600 {From equation (1), we have y² = 900 − x²}

100x = 1800

x = 18

On putting this value in equation (1), we obtain

y² = 900 − x² = 900 − 18² = 900 − 324 = 576

y = cm

Here, y is the radius of both the cones.

Now, surface area of the solid = C.S.A. of cone ABC + C.S.A. of cone ADC

= π × 24 × 30 + π × 24 × 40

= 1680 π

= 1680 ×

= 5280 cm²

Thus, the surface area of the double cone so obtained is 5280 cm2.

Example 3: The inner and outer radii of a cylindrical iron container are 10 cm and 11 cm respectively. The thickness of the base is 1.5 cm and the total height of the container is 21.5 cm. Find the cost, if the container has to be thoroughly electroplated with silver metal at the rate of 70 paise per cm².

Solution:

To find out the cost, we have to find out the T.S.A. of the container.

Height of the base, h = 1.5 cm

∴Height of the hollow cylinder, H = 21.5 − 1.5 = 20 cm

External radius, R = 11 cm and internal radius, r = 10 cm

T.S.A. of the container = Outer C.S.A. of hollow cylinder + Inner C.S.A. of hollow cylinder + Area of the ring + C.S.A. of solid cylinder + Area of inner base + Area of outer base

= 2π RH + 2πrH + π (R² − r²) + 2πRh + πr² + πR²

= 2π (R + r) H + π (R² − r²) + π (2Rh + r² + R²)

=

=

= 2640 + 66 +

= cm²

∴ Cost of electroplating = × 70 = 245300 paise = Rs 2453

Thus, the required cost is Rs 2453.

Volume Of Combination Of Solids

We come across many figures in our daily life that are made up of two or more solid figures. Let us consider such an example.

A company produces metallic solid toys that are in the shape of a cylinder with one hemisphere and one cone stuck to their opposite ends. The length of the entire toy is 30 cm; the diameter of the cylinder is 14 cm, while the height of the cone is 10 cm.

Can we find out how much metal should the company order to make 200 toys of this type?

In this way, we can find the volume of any solid figure which is formed by combining two or more basic solids. Let us now look at some more examples.

Example 1: A largest cone is to be taken out from a cube of edge 15 cm. Find the volume of the remaining portion. (Use π = 3.14)

Solution:

The figure can be drawn as follows:

It is given that length of the cube, l = 15 cm

The base of the cone is a circle whose diameter is equal to the length of the edge of the cube.

∴ Radius of cone, r = cm

The height of the cone would be equal to the height of cube.

∴ Height of cone, h = 15 cm

Volume of the remaining portion = Volume of Cube − Volume of Cone

=

=

= 3375 − 883.125

= 2491.875 cm³

Thus, the volume of the remaining portion is 2491.875 cm3.

Example 2: A plastic toy is in the following shape. The diameter of the cylindrical shape is 7 cm, but the bottom of the toy has a hemispherical raised portion. The top of the toy is a cone of same base. If the height of the cylinder is 21 cm and cone is 4

cm, find the amount of air inside the toy. (Use π = )

Solution:

Diameter of the cylinder = 7 cm

Radius of cylinder, r = cm

r = , is also the radius for cone and hemisphere.

Height of the cylinder, H = 21 cm

Height of the cone, h = 4 cm

∴ Volume of the toy = Volume of cylinder + Volume of Cone − Volume of Hemisphere

= 770 cm³

Thus, the volume of air inside the given toy is 770 cm^3.

Example 3: A right triangle, whose perpendicular sides are 30 cm and 40 cm, is made to revolve about its hypotenuse. Find the volume of figure so obtained in terms of π.

Solution:

Let ABD be a right-angled triangle, such that AB = 30 cm and AD = 40 cm.

Using Pythagoras theorem, we have

BD² = AB² + AD²

BD² =

BD = cm

After revolution, we have the following figure.

Let OB = x and OA = y

Using Pythagoras theorem in triangle AOB, we obtain

OA² + OB² = AB²

x² + y² = 30²

y² = 900 − x² … (1)

Using Pythagoras theorem in triangle AOD, we obtain

OA² + OD² = AD²

y² + (50 − x)² = 40²

900 − x² + 2500 + x² − 100x = 1600 {using equation (1)}

100x = 1800

x = 18

On putting the value of x in equation (1), we obtain

y² = 900 − x² = 900 − 182 = 900 − 324 = 576

y = cm

Here, y is the radius of the cones ABC and ADC.

Now, height of the cone ABC, x = 18 cm

Height of the cone ADC, 50 − x = 32 cm

Volume of the double cone = Volume of cone ABC + Volume of cone ADC

=

=

=

= 9600π cm³

Thus, the volume of the figure so obtained is 9600π cm³.

Example 4: An iron container is cylindrical in shape as shown in the figure. The inner and outer diameters are 12 cm and 14 cm respectively. The container has a solid base of width 1.5 cm and the total height of the container is 22.5 cm. If the mass of 1 cm³ of iron is 8 gm, find the weight of the container.

Solution:

The iron container consists of two cylinders.

1. A solid cylinder and
2. A hollow cylinder

External and internal diameters of the hollow cylinder are 14 and 12 cm respectively.

Hence, internal radius, r = 6 cm and external radius R = 7 cm

Height of the solid cylinder, h = 1.5 cm

∴ Height of the hollow cylinder, H = 22.5 − 1.5 = 21 cm

Volume of the iron used in the container = Volume of solid cylinder + Volume of hollow cylinder

=

=

= 231 + 858

= 1089 cm³

Mass of 1 cm³ of iron = 8 gm

Mass of 1089 cm³ of iron = 1089 × 8 = 8712 gm = 8.712 kg

Thus, the weight of the container is 8.712 kg.

Mean Of Grouped Data Using The Direct Method

A cricket player played 7 matches and scored 88, 72, 90, 94, 56, 62, and 76 runs. What is the average score or mean score of the player in these 7 matches?

We know that,

“The mean is the sum of all observations divided by the number of observations”.

We can use the following formula to find the mean of the given data.

Using this formula, the mean of scores obtained by the player

= 76.86

Thus, the average score of the player is 76.86. There are many situations where the number of observations is very large and then it is not possible to use this formula to find the mean of the given data.

Let us consider such a situation.

We have the marks distribution of 100 students of a class in a Mathematics test. The

distribution is given in the form of groups like 0 − 10, 10 − 20 …. To find the mean in such situations, we follow another method.

Example 1: The following table shows the various income brackets of the employees of a company.

Find the average monthly income of each employee.

Solution:

Firstly, we will calculate the class marks of each class interval and then the product of the class marks with the corresponding frequencies. By doing so, we obtain the following table.

The mean can now be easily calculated using the formula.

Mean

Hence, the average income of the employees is Rs 23100.

Example 2: The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Find the value of f1 and f2.

Solution:

Firstly, let us find the class marks of each class interval and then the product of class marks with the corresponding frequencies for each class interval as shown in the following table.

It is given that the sum of the frequencies is 50.

∴ 30 + f₁ + f₂ = 50

f₁ + f₂ = 20

f₁ = 20 − f₂ … (1)

The mean is given as 62.8.

∴ = 62.8

Using equation (1),

30(20 − f₂) + 70 f₂ = 1080

600 + 40 f₂ = 1080

40 f₂ = 1080 − 600 = 480

f₂ =

On putting the value of f₂ in equation (1), we obtain

f₁ = 20 − f₂ = 20 − 12 = 8

Thus, the values of f1 and f2 are 8 and 12 respectively.

Mean Of Grouped Data Using Assumed Mean Method

We know that direct method can be used to find the mean of any data given in grouped form, but the calculation in direct method becomes very tough when the data is given in the form of large numbers, because finding the product of xi and fi becomes difficult and time consuming.

Therefore, we introduce assumed mean method to find the mean of grouped data. This method is also known as shift of origin method. This is an easier method to find the mean as it involves less calculation.

Example 1: The following table shows the life time (in hours) of 20 bulbs manufactured by a reputed company.

Find the average life time of a bulb in hours. Solution:

Firstly, we calculate the class mark of each class-interval and assume one of them as the assumed mean. Here, we take a = 3500. The calculations of the deviations and the product of the deviations with the corresponding frequencies have been represented in the following table.

Using the formula,

= 3500 + 350

= 3850

Thus, the average lifetime of a bulb is 3850 h.

Example 2:

The following table shows the percentage of girls in the top 33 schools of Delhi.

Find the mean percentage of girls by using the assumed mean method.

Solution:

Firstly, we calculate the class mark of each class-interval and assume one of them as the assumed mean. Here, we take a = 50. The calculations of the deviations and the product of the deviations with the corresponding frequencies have been represented in the following table.

Using the formula, , we obtain

Thus, the mean of the given data is 42.12 i.e., the mean percentage of girls in the 33 schools of Delhi is 42.12.

Mean Of Grouped Data Using Step Deviation Method

In the assumed mean method, we assume the mean as a from the class-marks and we use the following formula to find the actual mean.

Where, di is the deviation of a from xi.

When the data is given in the form of large numbers, then sometimes it is difficult and even lengthy to find the mean by using the assumed mean method. In such cases, we follow another method called step deviation method or shift of origin and scale method. Using this method, we can further reduce the calculation.

Example 1: Calculate the mean of the following data using step deviation method.

Solution:

Let us assume a = 50

Let us calculate the deviations of class marks from the assumed mean a. The calculations are shown in the following table.

Highest common factor of all the di’s = h = 20

Now, using the formula , we obtain

Thus, the mean of the given data is 64.6.

 Theoretical and Experimental Probability

Consider an experiment of tossing a coin. Before tossing a coin, we are not sure whether head or tail will come up. To measure this uncertainty, we will find the probability of getting a head and the probability of getting a tail.

A student tosses a coin 1000 times out of which 520 times head comes up and 480 times tail comes up.

The probability of getting a head is the ratio of the number of times head comes up to the total number of times he tosses the coin.

Probability of getting a head

= 0.52

Similarly, probability of getting a tail

= 0.48

These are the probabilities obtained from the result of an experiment when we actually perform the experiment. The probabilities that we found above are called experimental (or empirical) probabilities.

On the other hand, the probability we find through the theoretical approach without actually performing the experiment is called theoretical probability.

The theoretical probability (or classical probability) of an event E, is denoted by P(E) and is defined as

P(E)

Here, we assume that the outcomes of the experiment are equally likely.

When a coin is tossed, there are two possible outcomes. We can either get a head or a tail and these two outcomes are equally likely. The chance of getting a head or a tail is 1.

Thus, probability of getting a head

P(E)

Similarly, probability of getting a tail =

Here, (or 0.5) is the theoretical probability.

Relation between Experimental and Theoretical Probabilities:

There is a fact that the experimental probability may or may not be equal to the theoretical probability.

For example, if we take a coin and toss it by a particular number of times then the theoretical probability of getting a head or a tail will be in each trial, but if we observe the outcomes of all the trials and calculate the experimental probability for head or tail then it will not be exactly equal to theoretical probability.

For approval, we can consider the theoretical and experimental probabilities of getting tail in the above experiment.

We have:

Theoretical probability of getting head = 0.52

Also, in our experiment we obtained:

Experimental probability of getting tail = 0.5

It can be observed that experimental probability is not exactly equal to theoretical probability, but very close to it.

Also, experimental probability of the same event can vary according to the number of trials.

Probability

Consider an experiment where two coins are tossed simultaneously. So, can you find out the probability of getting at least one head?

In this experiment, we are not actually performing the experiment. Therefore, here we can find the probability of getting at least one head using theoretical approach.

The theoretical probability for an event E can be defined as follows.

Using this formula, let us try to find the probability of getting at least one head.

The numbers of possible outcomes by tossing two coins simultaneously are as follows. HH, HT, TH, TT

∴ Total number of outcomes = 4

The favourable outcomes for getting at least one head are HH, HT, TH

∴ Number of favourable outcomes = 3

Probability of getting at least one head =

= 0.75

Let us consider some more examples and calculate the probability using theoretical approach.

Example 1: A dice is thrown and the number on the upper face of the dice is noted.

Find the probability of getting

(a)  3

(b)  7

(c) an odd number

(d) an even number

(e) a number less than or equal to 5

(f)  a number greater than 2.

Solution:

When a dice is thrown, any of the numbers 1, 2, 3, 4, 5, and 6 can be on the upper face of the dice.

∴ Total number of outcomes = 6

(a)  Getting 3

Here, the favourable outcome is only 3.

∴ Number of favourable outcomes = 1

∴ Probability of getting a number 3 on the upper face =

(b)  Getting 7

There is no favourable outcome as we cannot get 7 by throwing a dice.

∴ Number of favourable outcomes = 0

∴ Probability of getting 7 on the upper face of the dice =

(c)  Getting an odd number

The favourable outcomes are 1, 3, and 5.

∴ Number of favourable outcomes = 3

∴ Probability of getting an odd number =

(d) Getting an even number

The favourable outcomes are 2, 4, and 6.

∴ Number of favourable outcomes = 3

∴ Probability of getting an even number =

(e)  Getting a number less than equal to 5

The favourable outcomes are 1, 2, 3, 4, and 5.

∴ Number of favourable outcomes = 5

∴ Probability of getting a number less than equal to 5 =

(f)  Getting a number greater than 2

The favourable outcomes are 3, 4, 5, and 6.

∴ Number of favourable outcomes = 4

∴ Probability of getting a number greater than 2 =

Example 2: In tossing three coins simultaneously, find the probability of getting

Solution:

(a)    at least one head

(b)    at most one tail

(c)    at most two heads

(d)    exactly two heads

(e)    at least one head but at most one tail

(f)    no head at all

(g)    at least one head but at most two heads 

In tossing three coins, the possible outcomes are HHH, HTT, THT, TTH, HHT, HTH, THH, and TTT.

∴ Number of all possible outcomes = 8

(a) Let E be the event of getting at least one head.

The outcomes favourable to E are HHH, HTT, THT, TTH, HHT, HTH, and THH.

∴ Number of outcomes favourable to E = 7

(b) Let E be the event of getting at most one tail.

The outcomes favourable to event E are HHH, HHT, HTH, and THH.

∴ Number of outcomes favourable to E = 4

(c) Let E be the event of getting at most two heads.

The outcomes favourable to E are HTT, THT, TTH, HHT, HTH, THH and TTT.

∴ Number of outcomes favourable to E = 7

(d) Let E be the event of getting exactly two heads.

The outcomes favourable to E are HHT, HTH, and THH.

∴ Number of outcomes favourable to E = 3

(e) Let E be the event of getting at least one head but at most one tail.

The outcomes favourable to E are HHH, HHT, HTH and THH

∴ Number of outcomes favourable to E = 4

(f) Let E be the event of getting no head at all.

The outcome favourable to E is TTT.

∴ Number of outcomes favourable to E = 1

(g) Let E be the event of getting at least one head but at most two heads.

The outcomes favourable to E are HTT, THT, TTH, HHT, HTH, and THH.

∴ Number of outcomes favourable to E = 6

Example 3: In the experiment of throwing two dices, find the probability that the sum of two numbers appearing on the upper face is

(i)  7

(ii)  greater than 11 

Solution:

Let E denote the event of getting a sum 7 and F denote the event of getting a sum greater than 11.

The possible outcomes of this experiment have been shown in the following table.

Number of all possible outcomes = 36

(i) The outcomes favourable to E are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1).

Number of outcomes favourable to event E = 6

(ii) The outcome favourable to F is (6, 6).

Number of outcomes favourable to event F = 1

Example 4: 8 defective bulbs are accidentally mixed with 92 good bulbs. When one bulb is drawn at random, it is not possible to look at the bulb and say whether it is defective or non defective. What is the probability that the bulb drawn is non- defective?

Solution:

Let E be the event of drawing a non defective bulb. Total number of bulbs = 92 + 8

= 100

Number of non defective bulbs = 92

∴

Example 5: From a pack of cards, five of club and spade, jack of heart, ten of diamond and spade, and queen of club is put aside and then the remaining cards are well shuffled. Find the probability of getting

(a) card of red queen

(b) a card of black queen

(c) a ten

(d) a five of club

(e) a jack

(f)  an ace

(g) a face card

(h) a black face card 

Solution:

After taking out 6 cards (five of club and spade, jack of heart, ten of diamond and spade, and queen of club) from 52 cards, we are left with 46 cards.

∴ Number of possible outcomes = 46

(a) Let E be the event of getting a card of red queen.

Number of outcomes favourable to E = 2 (A queen of diamond and a queen of heart)

∴

(b) Let E be the event of getting a card of black queen.

Number of outcomes favourable to E = 1 (Already queen of club is taken out)

∴

(c) Let E be the event of getting a card of ten.

Number of outcomes favourable to E = 2 (Ten of heart and club)

∴

(d) Let E be the event of getting a five of club.

Number of outcomes favourable to E = 0 (Already five of club is taken out)

∴

(e) Let E be the event of getting a jack.

Number of outcomes favourable to E = 3 (Already jack of heart is taken out)

∴

(f) Let E be the event of getting an ace.

Number of outcomes favourable to E = 4

∴

(g) Let E be the event of getting a face card.

Number of outcomes favourable to E = 10 (Already two face cards have been removed)

∴

 (h) Let E be the event of getting a black face card.

Number of outcomes favourable to E = 5 (Already queen of club has been removed)

∴

Example 6: A bag contains 25 balls of red and green colour. If 3 red balls and 2 green balls are added, then the probability of getting green is twice the probability of getting red. Find the number of balls of each colour.

Solution:

Total numbers of balls in the bag = 25

Let the bag contain x number of red balls.

∴ Number of green balls = 25 − x

By adding 3 red balls and 2 green balls,

Total numbers of balls in the bag = 25 + 2 + 3 = 30

Number of red balls = x + 3

∴ Number of green balls = 25 − x + 2= 27 − x

The probability of ball of each colour is

and

But it is given that

= 2 ×

= 2 ×

27 − x = 2x + 6

3x = 21

x = 7 (Number of red balls)

∴ Number of green balls = 25 − x = 25 − 7 = 18

Thus, the bag contains 7 red balls and 18 green balls.

Example 7: 25 cards numbered 6, 7, 8, 9, 10 … 30 are put in a box and mixed thoroughly. If one card is drawn at random, then find the probability that the number on the card is

(a)  odd

(b)  even

(c)  prime

(d)  divisible by 5

(e)  divisible by both 2 and 5 

Solution

We have, total number of cards = 25

∴ Total number of outcomes = 25

(a) Let E be the event of getting an odd number.

The favourable outcomes are 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, and 29.

∴ Number of outcomes favourable to E = 12

∴

(b) Let X be the event of getting an even number.

The favourable outcomes are 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, and 30.

∴ Number of outcomes favourable to X = 13

∴

(c) Let F be the event of getting a prime number.

The favourable outcomes are 7, 11, 13, 17, 19, 23, and 29.

∴ Number of outcomes favourable to F = 7

∴

(d) Let Y be the event of getting a number divisible by 5.

The favourable outcomes are 10, 15, 20, 25, and 30.

∴ Number of outcomes favourable to Y = 5

∴

(e) Let G be the event of getting a number divisible by both 2 and 5 (i.e. getting a number divisible by 10).

The favourable outcomes are 10, 20, and 30.

∴ Number of outcomes favourable to G = 3

∴

Finding Probability Using Complement of a Known Event

Consider the experiment of throwing a dice. Any of the numbers 1, 2, 3, 4, 5, or 6 can come up on the upper face of the dice. We can easily find the probability of getting a number 5 on the upper face of the dice?

Mathematically, probability of any event E can be defined as follows.

Here, S represents the sample space and n(S) represents the number of outcomes in the sample space.

For this experiment, we have

Sample space (S) = {1, 2, 3, 4, 5, 6}. Thus, S is a finite set.

So, we can say that the possible outcomes of this experiment are 1, 2, 3, 4, 5, and 6.

Number of all possible outcomes = 6

Number of favourable outcomes of getting the number 5 = 1

Probability (getting 5)

Similarly, we can find the probability of getting other numbers also.

P (getting 1) , P (getting 2) , P (getting 3) , P (getting 4) and

P (getting 6)

Let us add the probability of each separate observation.

This will give us the sum of the probabilities of all possible outcomes.

P (getting 1) + P (getting 2) + P (getting 3) + P (getting 4) + P (getting 5) + P (getting 6) = + + + + + = 1

“Sum of the probabilities of all elementary events is 1”.

Now, let us find the probability of not getting 5 on the upper face.

The outcomes favourable to this event are 1, 2, 3, 4, and 6.

Number of favourable outcomes = 5

P (not getting 5)

We can also see that P (getting 5) + P (not getting 5)

“Sum of probabilities of occurrence and non occurrence of an event is 1”.

i.e. If E is the event, then P (E) + P (not E) = 1 … (1)

or we can write P(E) = 1 − P (not E)

Here, the events of getting a number 5 and not getting 5 are complements of each other as we cannot find an observation which is common to the two observations.

Thus, event not E is the complement of event E. Complement of event E is denoted by or E'.

Using equation (1), we can write

or

This is a very important property about the probability of complement of an event and it is stated as follows:

If E is an event of finite sample space S, then P ( ) = 1 – P(E) where  is the complement of event E.

Now, let us prove this property algebraically.

Proof:

We have,

E ∪ = S and E ∩ =

⇒ n(E ∪ ) = n(S) and n(E ∩ ) = n()

⇒ n(E ∪ ) = n(S) and n(E ∩ ) = 0 ...(1)

Now,

n(E ∪ ) = n(S)

⇒ n(E) + n( ) – n(E ∩) = n(S)

⇒ n(E) + n( ) – 0 = n(S) [Using (1)]

⇒ n( ) = n(S) – n(E)

On dividing both sides by n(S), we get

⇒ P( ) = 1 – P(E)

Hence proved.

Let us solve some examples based on this concept.

Example 1: One card is drawn from a well shuffled deck. What is the probability that the card will be

(i)  a king?

(ii)  not a king?

Solution:

Let E be the event ‘the card is a king’ and F be the event ‘the card is not a king’.

(i) Since there are 4 kings in a deck.

Number of outcomes favourable to E = 4

Number of possible outcomes = 52

P (E)

(ii) Here, the events E and F are complements of each other.

∴ P(E) + P(F) = 1

P(F) = 1 −

Example 2: If the probability of an event A is 0.12 and B is 0.88 and they belong to the same set of observations, then show that A and B are complementary events.

Solution:

It is given that P (A) = 0.12 and P (B) = 0.88

Now, P(A) + P(B) = 0.12 + 0.88 = 1

The events A and B are complementary events.

Example 3: Savita and Babita are playing badminton. The probability of Savita winning the match is 0.52. What is the probability of Babita winning the match?

Solution:

Let E be the event ‘Savita winning the match’ and F be the event ‘Babita wining the match’. It is given that P (E) = 0.52

Here, E and F are complementary events because if Babita wins the match, Savita will surely lose the match and vice versa.

P (E) + P (F) = 1

0.52 + P (F) = 1

P (F) = 1 − 0.52 = 0.48

Thus, the probability of Babita winning the match is 0.48.

Example 4: In a box, there are 2 red, 5 blue, and 7 black marbles. One marble is drawn from the box at random. What is the probability that the marble drawn will be (i) red (ii) blue (iii) black (iv) not blue?

Solution:

Since the marble is drawn at random, all the marbles are equally likely to be drawn. Total number of marbles = 2 + 5 + 7 = 14

Let A be the event ‘the marble is red’, B be the event ‘the marble is blue’ and C be the event ‘the marble is black.

(i) Number of outcomes favourable to event A = 2

P (A)

(ii) Number of outcomes favourable to event B = 5

∴ P (B)

(iii) Number of outcomes favourable to event C = 7

P (C)

(iv)  We have, P (B)

The event of drawing a marble which is not blue is the complement of event B.

P = 1 − P (B) = 1 −

Thus, the probability of drawing a marble which is not blue is .

Relationship Between Zeroes Of A Polynomial And Its Coefficients

We have learnt to find the zeroes or roots of a polynomial. Also, we know that zeroes or roots of a polynomial are those values of its variables for which the polynomial results to zero.

Now, let us learn about the relation between zeroes and the coefficients of the polynomial.

Relationship between the zero and the coefficient of a linear polynomial:

Let p(x) = ax + b be a linear polynomial such that a ≠ 0.

Zero of this polynomial can be obtained by equating it with 0. i.e, p(x) = 0

⇒ ax + b = 0

⇒ ax = –b

Using this relation, we can find the zero of a linear polynomial.

Relationship between the zeroes and the coefficients of a quadratic polynomial:

Consider a quadratic polynomial p(x) = 3x²– 5x – 12.

Can we find out the sum and the product of the zeroes of this polynomial?

Yes, we can find the sum and the product of zeroes but firstly we have to find out the zeroes of the polynomial.

Here, the zeroes of polynomial p(x) are 3 and .

Now, the sum of zeroes = 3 +

And the product of zeroes = 3 × = – 4

Can we find out the sum and the product of zeroes by any other method?

Yes, there is also a method in which there is no need to find out the zeroes. In that method we use the coefficients of the polynomial to find the sum and the product of zeroes.

Firstly let us see the relation between the sum and product of zeroes and the coefficients of the polynomial.

Let us first consider a quadratic polynomial p(x) = ax² + bx + c, where a, b and c are constants.

If αand β are the zeroes of p(x) = ax² + bx + c, then,

Now, let us find the sum and product of zeroes of the polynomial given in the beginning, using these relations.

The polynomial is p(x) = 3x² – 5x – 12.

On comparing this equation with ax² + bx + c, we have

a = 3, b = –5 and c = –12

∴ Sum of zeroes = – =

And the product of zeroes =

Using these relations we obtained the same values as we found after calculating the zeroes.

Now, let us know the relations between the sum and the product of zeroes and the coefficients of a cubic polynomial.

The general form of a cubic polynomial is p(x) = ax³ + bx² + cx + d where a, b, c and d are constants.

If α, β and γ are the three zeroes of cubic polynomial p(x), then the relations are given by

Let us solve some more problems to have a better understanding of the concept.

Example 1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(a) x² – 9x + 20 (b) 5x² + 21x – 20.

Solution:

(a)    The given quadratic polynomial is p(x) = x² – 9x + 20.

Firstly we will find the zeroes by the method of splitting the middle term and then verify the relationship between the zeroes and the coefficients of the polynomial.

In this method, we have to find two numbers whose product is 20 and sum (or difference) is 9.

Such two numbers are 4 and 5 so, we have

x² – 9x + 20 = x² – (4 + 5)x + 20

= x² – 4x – 5x + 20

= x (x – 4) – 5 (x – 4)

= (x – 4) (x – 5)

Thus, for finding zeroes, we put

x² – 9x + 20 = 0.

(x – 4) (x – 5) = 0

(x – 4) = 0 or (x – 5) = 0

x = 4 or 5

Thus, 4 and 5 are the zeroes of the polynomial p(x) = x² – 9x + 20.

Now sum of zeroes = 4 + 5 = 9

And product of zeroes = 4 × 5 = 20

Now, on comparing p(x) = x² – 9x + 20 with the general quadratic polynomial

ax² + bx + c = 0, we have a = 1, b = –9 and c = 20.

Using the formulae, we have

Sum of zeroes

= 9

Product of zeroes

= 20

Hence, it is verified that Sum of zeroes = and

Product of zeroes = αβ = .

(b) Comparing the given polynomial with ax² + bx + c, we have, a = 5, b = 21

and c = – 20.

Using the method of splitting the middle term we have,

p(x) = 5x² + 21 x – 20 = 5x² + (25 – 4)x – 20

= 5x² + 25x –4 x – 20

= 5x(x + 5) –4 (x + 5)

= (x + 5) (5x – 4)

For finding zeroes, put p(x) = 0 (x + 5) (5x – 4) = 0

x = – 5 or x =

Thus, the zeroes of the polynomial 5x² + 21 x – 20 are – 5 and .

Now, sum of zeroes = – 5 + = =

And product of zeroes = – 5 × = –4 = =

Hence, verified.

Example 2: If 1, 2, and 6 are the zeroes of the cubic polynomial x³ – 9x² + 20x – 12, then verify the relations between the zeroes and the coefficients.

Solution:

The given cubic polynomial is p(x) = x³ – 9x² + 20x – 12.

On comparing the given polynomial with the general form ax³ + bx² + cx + d of cubic polynomial, we have a = 1, b = – 9, c = 20, d = – 12.

Now the zeroes are 1, 2, and 6.

Let α = 1, β = 2 and γ = 6

Now α+ β + γ = 1 + 2 + 6

= 1 × 2 + 2 × 6 + 6 × 1

= 2 + 12 + 6

= 20

and αβ γ = 1 × 2 × 6

= 12

Hence, this verifies the relations between the coefficients and the zeroes of the cubic polynomial.

Formation Of Polynomial Using The Sum And Product Of Zeroes

5 and –3 are the zeroes of a quadratic polynomial p(x). Can we form the quadratic polynomial with the help of these zeroes?

Yes, we can. Let us see how we can form the polynomial.

5 is a zero of the polynomial, it means that (x – 5) is a factor of the polynomial p(x). Similarly (x + 3) is also a factor of the polynomial p(x). We know that a quadratic polynomial can have only two linear factors. Thus, we can write the polynomial p(x) as follows:

p(x) = (x – 5) (x + 3)

= x² + (– 5 + 3) x + (– 5) (3)

[Using the identity (x + a) (x + b) = x² + (a + b) x + ab, where a = – 5 and b = 3]

Thus, p(x) = x² – 2x – 15

This is the required quadratic polynomial.

If we know the zeroes of the quadratic polynomial, then we can find the polynomial as above. On the other hand, if we know the sum and product of the zeroes of the polynomial, then we can use the following formula to find the polynomial.

For example: if the sum and the product of the zeroes of a polynomial are 2 and –8 respectively, then find the polynomial.

Using the above formula, we have

p(x) = x² – (2)x + (–8)

p(x) = x² – 2x – 8

In this way, we can find the polynomial if the relationship between the zeroes is known to us.

Let us discuss some more examples based on formation of quadratic polynomial using sum and products of zeroes.

Example 1: If the sum and product of zeroes of a quadratic polynomial are 13 and 36 respectively, find the quadratic polynomial.

Solution:

Let the polynomial be p(x).

It is given that the sum of the zeroes of the polynomial = 13 and the product of the zeroes of the polynomial = 36

But we know that the quadratic polynomial p(x) in terms of sum and product of zeroes is given by the formula

p(x) = x² – (sum of zeroes)x + product of zeroes

so, we have

p(x) = x² – 13x + 36

Therefore, x² – 13x + 36 is the quadratic polynomial whose sum and the product of zeroes are 13 and 36 respectively.

Example 2: If the sum and product of zeroes of a quadratic polynomial are and – 6 respectively, find the quadratic polynomial.

Solution:

Let the quadratic polynomial be p(x).

It is given that the sum of the zeroes of the polynomial is and the product of the zeroes of the polynomial is – 6.

We know that the quadratic polynomial p(x) in terms of sum and product of zeroes is given by the formula

p(x) = x² – (sum of zeroes)x + product of zeroes so, we have

p(x) = x² – ( )x + (– 6)

p(x) = x² x – 6

Therefore, x² x – 6 is the quadratic polynomial whose sum and product of zeroes are and – 6 respectively.

Example 3: If and are the zeroes of a quadratic polynomial, then find the polynomial.

Solution:

Let p(x) be the required quadratic polynomial. and are the zeroes of the polynomial p(x).

Now, the sum of zeroes =

and the product of zeroes =

Now, the polynomial is given by the formula

p(x) = x² – (sum of zeroes)x + product of zeroes

Hence, this is the required polynomial whose zeroes are and respectively.

Division Of Polynomials by Polynomials (Degree More than 1) Using Long Division Method

All of us have already studied the method of long division of polynomials by other polynomials in earlier classes. But there, our divisor was just of degree one. Now we will

revise long division method of polynomials with other polynomials and extend the discussion to do the division when the divisor is of a higher degree.

Let us discuss the method of division by taking an example. Let us divide (x³ − 7x + 6) by (x² + 2x − 3).

Now let us solve some examples to practice the above method.

Example 1: Divide the polynomial by monomial and write the quotient and remainder.

(i) 36n⁵ – 3n³ + 12n² ÷ 9n²

(ii) 7p⁷ + 3p⁵ – 42p³ + 10 ÷ 21p³

Solution:

(i)    36n⁵ – 3n³ + 12n² can be divided by 9n² using long division method as follows:

Therefore,

Quotient =

Remainder = 0

(ii)    7p⁷ + 3p⁵ – 42p³ + 10 can be divided by 21p³ using long division method as follows:

Therefore,

Quotient =

Remainder = 10

Example 2: Check whether (x² − 3x + 2) is a factor of (x³ − 7x² + 14x − 8) or not. Solution:

(x² − 3x + 2) will be a factor of (x³ − 7x² + 14x − 8), if the polynomial (x³ − 7x² + 14x − 8) on dividing by (x² − 3x + 2) gives remainder zero.

The division has been shown below:

The remainder is zero.

∴ (x² − 3x + 2) is a factor of (x³ − 7x² + 14x − 8).

Example 3: Find the quotient and remainder when (15x − 4 + x³ − 6x²) is divided by

(x² − 3x + 2).

Solution:

Firstly, we arrange the terms of the dividend and the divisor in the standard form.

The dividend in the standard form will be

x³ − 6x² + 15x − 4.

The divisor is already in standard form. The division has been shown below:

Therefore, Quotient = x – 3

Remainder = 4x + 2

Example 4: and are the zeroes of the polynomial 16x⁴ − 64x³ + 40x² + 80x − 75. Find the other zeroes of the polynomial.

Solution:

Let p(x) = 16 x⁴ − 64x³ + 40x² + 80x − 75

and  are two zeroes of p(x).

⇒ (x − ) and (x + ) are the factors of p(x)

⇒ x² − is a factor of p(x)

To find the other factors, we have to divide p(x) by x² − as follows:

Now,  is the factor of p(x).

So, we can write =

= (Splitting the middle term)

=

=

To find the zeroes, put  = 0

We have 

Thus, the remaining zeroes of the polynomial are x = and .

Division Algorithm Of Polynomials

We are aware of the process of division of numbers. Let us divide 487 by 3. We can divide 487 by 3 as follows:

From the above division, we can see that when we divide 487 by 3, we get 1 as remainder and 162 as quotient.

We can also write the dividend 487 in terms of divisor, quotient and remainder as follows: 487 = 3 × 162 + 1

Thus, for real numbers we can write

Dividend = Divisor × Quotient + Remainder

In the same way, we can also divide a polynomial p(x) by another polynomial g(x) when the degree of the polynomial p(x) is greater than or equal to the degree of polynomial g(x). The above relation is also true in case of polynomials and then it is known as division algorithm of polynomials.

Using the division algorithm of polynomials, we can also find any one of the values among p(x), g(x), q(x) and r(x) if other three are known.

Note that degree of q(x) = degree of p(x) – degree of g(x)

Let us discuss more examples based on division algorithm of polynomials.

Example 1: Find the quotient, if the polynomial −2x³ + 7x² − 7x − 2 when divided by 2x² − 5x + 2 gives remainder −4.

Solution:

Let the quotient be q(x).

It is given that when we divide the polynomial (−2x³ + 7x² − 7x − 2) by (2x² − 5x + 2), we get the q(x) as quotient and −4 as remainder.

Thus, we have

p(x) = −2x³ + 7x² − 7x − 2

g(x) = 2x² − 5x + 2

r(x) = −4

We know that the division algorithm for polynomials is:

p(x) = g(x) × q(x) + r(x)

Now, on putting the values in this equation, we have

−2x³ + 7x² − 7x − 2 = (2x² − 5x + 2) × q(x) − 4

⇒ −2x³ + 7x² − 7x − 2 + 4 = (2x² − 5x + 2) × q(x)

⇒ −2x³ + 7x² − 7x + 2 = (2x² − 5x + 2) × q(x)

We can divide −2x³ + 7x² − 7x + 2 by (2x² − 5x + 2) as follows:

Now, we have q(x) = 1 − x

Therefore, the quotient is 1 − x.

Example 2: Find the quotient, if the polynomial 3x³ − 3x² + 9x + 2 when divided by x² + 5x + 1 gives remainder 4x³.

Solution:

Let the quotient be q(x).

It is given that when we divide the polynomial (3x³ − 3x² + 9x + 2) by (x² + 5x + 1), we get the q(x) as quotient and 4x³ as remainder.

Thus, we have

p(x) = 3x³ − 3x² + 9x + 2

g(x) = x² + 5x + 1

r(x) = 4x³

We know that the division algorithm for polynomials is:

p(x) = g(x) × q(x) + r(x)

Now, on putting the values in this equation, we have

3x³ − 3x² + 9x + 2 = (x² + 5x + 1) × q(x) + 4x³

⇒ 3x³ − 3x² + 9x + 2 − 4x³ = (x² + 5x + 1) × q(x)

⇒ −x³ − 3x² + 9x + 2 = (x² + 5x + 1) × q(x)

We can divide −x³ − 3x² + 9x + 2 by (x² + 5x + 1) as follows:

Now, we have q(x) = 2 − x

Therefore, the quotient is 2 − x.

Example 3: By applying division algorithm, find the quotient and remainder when p(x)

= x⁴ + 3x³ + 2x² + 5x – is divided by g(x) = x³ + 2x – 1.

Solution:

p(x) = x⁴ + 3x³ + 2x² + 5x – , g(x) = x³ + 2x – 1

deg p(x) = 4, deg g(x) = 3

Degree of quotient q(x) = 4 – 3 = 1, and deg of remainder r(x) < deg g(x) = 3 Let q(x) = ax + b, r(x) = cx² + dx + e

By division algorithm,

p(x) = g(x) × q(x) + r(x)

⇒ x⁴ + 3x³ + 2x² + 5x – = (ax + b)(x³ + 2x – 1) + (cx² + dx + e)

= ax⁴ + 2ax² – ax + bx³ + 2bx – b + cx² + dx + e

= ax⁴ + bx³ + (2a + c)x² + (–a + 2b + d)x + (–b + e)

Equating the coefficients of respective powers, we obtain

Quotient, q (x) = x + 3

Remainder, r (x) =