Category: Maths Notes PPT

Bernoulli trial, binomial distribution and Bernoulli distribution are briefly explained in this article. Let us first learn about Bernoulli trials. Bernoulli trials are also known as binomial trials as there are only possible outcomes in Bernoulli trials i.e success and failure whereas in a binomial distribution, we get a number of successes in a series of independent experiments. A Bernoulli distribution is the probability distribution for a series of Bernoulli trials where there are only two possible outcomes. In this article, we will discuss,bernoulli trial binomial distribution, bernoulli trial formula, bernoulli trial example, bernoulli distribution, bernoulli distribution examples, properties of bernoulli distribution, how bernoulli trial is related to binomial distribution etc. 

Bernoulli Trials

In the field of probability, the experimentation of different concepts led to major mathematical theories. Let us assume one experiment which will be finite in number. It should have only two results as outcome one will be termed as success and the other one as a failure. And in all the experiment terms the probability of events failure and success does not change. This setup of the experiment is known as the Bernoulli trial. This was created by Jacob Bernoulli, a Swiss mathematician and published in his book Ars Conjectandi in 1713. This helped us in understanding the nature of probability better.

Definition

A successive event in a sequence of independent experiments where there are only two possible outcomes and the probability of the outcomes remains the same in each event. When we conduct these events in a succession for a finite number of times then the series of experiments is called Bernoulli’s trials.

Binomial Distribution

The binomial distribution is a graphical representation of the results of Bernoulli’s trials. It gives us the idea of the probability of events throughout the experiment successions. This can also be said as the frequency distribution of the probability of a given number of successes in a random number of repeated independent Bernoulli trials. If Bernoulli’s trials are the experiment then the binomial theorem can be said as the result or finding of the experiment. 

The Use

Say we have to find the probability of heads in a coin toss then we can easily find it by using success occurrence divided by sample space formula and the same can be said for finding the probability of two heads in two consecutive coin tosses. If we have to find the probability of two heads and three tails in 5 consecutive coin tosses then it will be difficult but solvable by the above method but the complexity is very huge. But say we have to find the probability of even no heads counting in 20 consecutive coin tosses. It will be next to impossible for us to fund the probability by the above method. In these complex cases, we will have to use the binomial distribution formula which will help us to find the probability of complex predictable problems. 

Formula

[P_{r} = (frac{n}{r})p^{r} q^{n-r}]

n= number of trials

r= number of success

p= probability of success

q=probability of failure

More About Binomial Distribution

The binomial distribution is a kind of probability distribution that has two possible outcomes. In probability theory, binomial distributions come with two parameters such as n and p.

The probability distribution becomes a binomial probability distribution when it satisfies the below criteria.

1. The number of trials must be fixed.

2. The trials are independent of each other.

3.  The success of probability remains similar for every trial.

4. Each trial has only two outcomes namely success or failure.

()

Bernoulli Trial

A random experiment that has only two mutually exclusive outcomes such as “success “and “not success “is known as a Dichotomous experiment.

If a Dichotomous experiment is repeated many times and if in each trial you find the probability of success p (0< p <1) is constant, then all such trials are known Bernoulli trials.

As, Bernoulli trials has only two possible outcomes, it can easily frame as “yes” or “no” questions

Bernoulli Trials Example

The Bernoulli trial example will explain the concept of bernoulli trial in two different situation:

8 balls are drawn randomly including 10 white balls and 10 black balls. Examine whether the trials are Bernoulli trials if the balls are replaced and not replaced.

1. In the first trial, when the ball is drawn with replacement, the probability of success (say, the black ball) is 10/20 = ½ which is similar for all 8 trials. Hence, the trials including the drawing of balls with replacement are considered as Bernoulli trials.

2. In the second trial, when drawn without replacement, the probability of success (say, the black balls) changes with the number of trials =10/20 = ½ for second trials, the probability of success p =9/19 which is not similar to the first trial. Hence, the trails including the drawing of balls without replacement are not considered as Bernoulli trial

Bernoulli Trials Conditions

• The Bernoulli trial has only two possible outcomes i.e. success or failure.

• The probability of success and failure remain the same throughout the trials.

• The Bernoulli trials are independent of each other.

• The number of trials is fixed.

• If x is the probability of success then the probability of failure is 1-x.

Bernoulli Trials Formula

Here, you can see the Bernoulli trial formula in Bernoulli Math.

Let us take an example where n bernoulli trials are made then the probability of getting r successes in n trials can be derived by the below- given bernoulli trials formula.

P(r) = Cn pr qn-r

The term n! / r!(n!-r!)!  is known as a binomial coefficient.

Bernoulli Trials and Binomial Distribution

• The student will be able to design a  Bernoulli trial or experiment

• The student will be easily able to use binomial formula

• The student will be able to design a  binomial distributions

• The students will be able to compute applications including Bernoulli trials and binomial distribution’

Bernoulli Distribution

A Bernoulli distribution in Bernoulli Maths is the probability distribution for a series of Bernoulli trials where there are only two possible outcomes. It is a kind of discrete probability distribution where only specific values are possible. In such a case, only two values are possible;e ( n=0 for failure and n=1 for success). This makes the Bernoulli distribution the simplest form of the probability distribution that persists.

Bernoulli Distribution Examples

Some of the bernoulli distribution examples given in bernoulli Maths are stated below:

1. A newly born child is either a girl or a boy ( Here, the probability of a child being a  boy is roughly 0.5)

2. The student is either pass or fail in an exam

3. A tennis player either wins or losses a match

4. Flipping of a coin is either a head or a tail.

Properties of Bernoulli Distribution

Here, you can find some of the properties of bernoulli distribution in bernoulli Maths.

1. The expected value of the bernoulli distribution is given below.

E(X) = 0 * (1-P) + 1 * p = p

2. The variance of the bernoulli distribution is computed as 

Var (X) = E(X²) -E(X²) = 1² * p +0² * ( 1-p) – p² = p – p² = p (1-p)

3. The mode, the value with the highest probability of appearing, of a Bernoulli distribution is 1 if p > 0.5 and 0 if < 0.5, success and failure are equally likely and both 0 and 1 are considered as modes. 

4. The basic properties of Bernoulli Distribution can be computed by considering n=1 in probability

Quiz Time

1. How many possible outcomes can there be for Bernoulli trials?

1. More than 1

2. More than 2

3. Exactly 1

4. Exactly 2

2. What will be the variance of the Bernoulli trials, if the probability of success of the Bernoulli trial is  0.3.

1.  0.3

2. 0.7

3. 0.21

4. 0.09

3. The mean and variance are equal in binomial distributions.

1. True

2. False

Solved Examples

1. If the probability of the bulb being defective is 0.8, then find the probability of the bulb not being defective.

Solution: 

Probability of bulb being faulty, p = 0.8

Probability of bulb not being defective, q = 1-p = 1-0.8= 0.2

Hence, probability of bulb not being defective, q = 0.2 

2. In an exam, 10 multiple choice questions are asked where only one out of four questions are correct. Find the probability of getting 5 out of 10 questions correct in an answer sheet.

Solution: Probability of getting an answer correct, p = ¼

Probability of getting an answer incorrect , q = 1-p = 1

Probability of getting 5 answers correct, P(X=5) = (0.25)5 ( 0.75)5 = 0.5839920044

3. Toss a coin 12 times. What is the probability of getting 7 heads? 

Solutions:

Number of trials(N) = 12

Number of success (r)= 7

Probability of single trial  (P) = ½ = 0.5

nCr =

n!r!

n!r! * (n-r)!

12!7! (12-7)!

= 95040120

= 792

pr= 0.57=

0.0078125

To find (1-p)n-r,calculate (1-p) and (n-r)

Solving P (X=r) = nCr .pr. (1-p)n-r

= 792 * 0.0078125 *0.03125

= 0.193359375

The probability of getting 7 heads is 0.19

As the power increases the expansion of terms becomes very lengthy and tedious to calculate. It can be easily calculated with the help of the Binomial Theorem. 

What is Binomial Theorem?

The binomial theorem in mathematics is the process of expanding an expression that has been raised to any finite power. A binomial theorem is a powerful tool of expansion, which is widely used in Algebra, probability, etc.

Binomial Expression 

A binomial expression is an algebraic expression that contains two dissimilar terms such as a + b, a³ + b³, etc.

Binomial Theorem Expansion

According to the theorem, we can expand the power (x + y)[^{n}] into a sum involving terms of the form ax[^{b}]y[^{c}], where the exponents b and c are nonnegative integers with b+c=n and the coefficient a of each term is a specific positive integer depending on n and b.

The theorem is given by the formula:

(x + y)[^{n}] = [sum_{k=0}^{n}] ([_{k}^{n}]) x[^{n-k}]y[^{k}] = [sum_{k=0}^{n}] ([_{k}^{n}]) x[^{k}]y[^{n-k}]

(x + y)[^{n}] = [sum_{k=0}^{n}] (nk) x[^{n-k}]y[^{k}] = [sum_{k=0}^{n}] (nk) x[^{k}]y[^{n-k}]

Binomial Theorem Rules

The coefficients that appear in the binomial expansion are known as binomial coefficients. These are usually written ([_{k}^{n}]) or [ ^{n}C_{k}]. which means n choose k.

The coefficient of a term x[^{n-k}]y[^{k}] in a binomial expansion can be calculated using the combination formula. The formula consists of factorials:

([_{k}^{n}]) = [frac{n!}{k!(n-k)!}] 

Important Points to Remember While Solving Binomial Expansion:

• The total number of terms in the expansion of (x + y)[^{n}] is (n+1)

• The sum of exponents is always equal to n i.e (x + y) = n.

• ⁿC[_{0}], ⁿC[_{1}], ⁿC[_{2}], … .., ⁿC[_{n}] is called binomial coefficients and also represented by C[_{0}], ^CC[_{1}], ^CC[_{2}] ….., C[_{n}] respectively.

• The binomial coefficients which are equidistant from the beginning and from the ending are of equal value i.e. ⁿC₀= ⁿC_n,ⁿC₁= ⁿC_n-1 , ⁿC₂= ⁿC_n-2 ,….. etc.

• To find binomial coefficients we can also use Pascal’s Triangle.

Some Other Useful Expansions that Help in an Easy Way to Solve Binomial Theorem :

• (x + y)[^{n}] + (x – y)[^{n}] = 2[C[_{0}]x[^{n}] + C[_{2}]x[^{(n-1)}]y[^{2}] + C[_{4}]x[^{n-4}]y[^{4}]+ …]

• (x + y)[^{n}] – (x – y)[^{n}] = 2[C[_{1}] x[^{(n-1)}]y + C[_{3}] x[^{(n-3)}]y[^{3}] + C[_{5}]x[^{(n-5)}]y[^{5}]…]

• (1 + x)[^{n}] = [C[_{0}] + C[_{1}]x + C[_{2}]x[^{2}] + … C[_{n}]x[_{n}]]

• (1 + x)[^{n}] + (1 – x)[^{n}] = 2[C[_{0}] + C[_{2}]x[^{2}] + C[_{4}]x[^{4}] + …]

• (1 + x)[^{n}] – (1 – x)[^{n}] = 2[C[_{1}]x + C[_{3}]x[^{3}] + C[_{5}]x[^{5}] + …]

• The number of terms in the expansion of (x + a)[^{n}] + (x – a)[^{n}] is (n+2)/2 if “n” is even or (n+1)/2 if “n” is odd.

• The number of terms in the expansion of (x + a)[^{n}] – (x – a)[^{n}] is (n/2) if “n” is even or (n+1)/2 if “n” is odd.

Properties of Binomial Coefficients

Binomial coefficients refer to the integers that are coefficients in the binomial theorem. Some of the important properties of binomial coefficients are given below:

• C[_{0}] + C[_{1}] + C[_{2}] + … + C[_{n}] = 2n

• C[_{0}] + C[_{2}] + C[_{4}] + … = C[_{1}] + C[_{3}] + C[_{5}]… = 2n – 1

• C[_{0}] – C[_{1}] + C[_{2}] – C[_{3}] + … + (-1)[^{n}] . nC[_{n}] = 0

• ⁿC[_{1}] + 2ⁿC[_{2}] + 3.ⁿC[_{3}] + … +n.ⁿC[_{n}] = n.2[^{(n-1)}].

• C[_{1}] – ²C[_{2}] + ³C[_{3}] – ⁴C[_{4}]+ … +(-1)[^{(n-1)}]C[_{n}] = 0 for n ＞ 1

• C[_{0}] [^{2}] + C[_{1}] [^{2}] + C[_{2}] [^{2}] + …C[_{n}] [^{2}] = [frac{[(2n)!]}{(n!)^{2}}]

Terms in the Binomial Expansion

In binomial expansion, we generally find the middle term or the general term. The different Binomial Term involved in the binomial expansion is:

• General Term

• Middle Term

• Independent Term

• Determining a Particular Term

• Numerically greatest term

• The ratio of Consecutive Terms/Coefficients

General Term in Binomial Expansion:

We have (x + y)[^{n}] = ⁿC[_{0}]x[^{n}] + ⁿC[_{1}]x[^{(n-1)}]y + ⁿC[_{2}]x[^{n-2}]y[^{2}] + … + ⁿC[_{n}]Y[^{n}]

General Term = T[_{(r+1)}] = ⁿC[_{r}]x[^{n-r}] .y[^{r}]

General Term in (1+x)[^{n}] ⁿC[_{r}]x[^{r}]

In the binomial expansion of (x + y)[^{n}], the r[^{th}] term from the end is (n – r + 2)[^{th}].

Middle Term(S) in the expansion of (x + y)[^{n,n}] 

If n is even then (n/2 + 1) term is the middle term.

If n is odd then [(n+1)/2][^{th}] and [(n+3)/2)[^{th}] terms are the middle terms of the expansion.

Applications of Binomial Theorem

The binomial theorem has various applications in mathematics like finding the remainder, finding digits of a number, etc. The most common binomial theorem applications are:

• Finding Remainder using Binomial Theorem.

• Finding Digits of a Number.

• Relation Between two Numbers.

• Divisibility Test.

Binomial Theorem Problems are explained with the help of Binomial theorem formula examples which is given below:

1. Find the coefficient of x[^{9}] in the expansion of (1 + x) (1 + x[^{2}]) (1 + x[^{3}]) . . . . . . (1 + x[^{100}]).

Sol:
x[^{9}] can be formed in 8 ways.

i.e., x[^{9}] x[^{(1+8)}] x[^{(2+7)}] x[^{(3+6)}] x[^{(4+5)}], x[^{(1+3+5)}], x[^{(2+3+4)}]

∴ The coefficient of x[^{9}] = 1 + 1 + 1 + . . . . + 8 times = 8.

2.Find the number of terms in (1 + 2x + x[^{2}])[^{50}].

Sol: Use formula of the number of terms (1 + 2x + x[^{2}])[^{50}] = [(1 + x)[^{2}]][^{50}] = (1 + x)[^{100}]

Hence, the number of terms = (100 + 1) = 101

3. Find the value of $binom{6}{2}$ using a pascal triangle.

Solution:

Look at the 2nd element in the 6th row in pascal’s triangle. The value of $binom{6}{2}$ will be that element.

Hence, the value of $binom{6}{2}$ is $15$.

4. Expand $( a + 2)^6$ using binomial theorem.

Solution:

Let $a = x, y = 2$ and $n = 6$

Substituting the values on binomial formula, we get

$(a)^6+6(a)^5(2)+dfrac{6(5)}{2text{!}}(a)^4(2)^{2}+dfrac{6(5)(4)}{3text{!}}(a)^3(2)^3+dfrac{6(5)(4)(3)}{4text{!}}(a)^2(2)^4+dfrac{6(5)(4)(3)(2)}{5text{!}}(a)(2)^5+ 2^6$

$= a^6 + 12a^5 + 60a^4 +160a^3 + 240a^2 + 192a + 64$

Facts to Remember

• The binomial theorem was invented by Issac Newton.

• The Pascal triangle was invented by Blaise Pascal.

• The numbers in each row in the pascal triangle are known as the binomial coefficients.

• The numbers on the second diagonal and third diagonal in the pascal triangle form counting numbers and triangular numbers respectively.

• The sum of the numbers on each row are powers of $2$ whereas a series of diagonals of pascal’s triangle forms the Fibonacci Sequence.

Important Questions For Class 10 Maths

The question paper of 10th CBSE Maths will be for 80 marks. There will be four sections of the paper. Section-A a will have all the 1 mark questions and the number of questions in this section is 20. So, the total weightage of section-A is 20 marks. The marks weightage of Section-B is 12 marks where each question will be for 2 marks each. The number of questions in this section is 6. In Section-C, there will be 8 questions asked which carry 3 marks each and the total weightage of this section is 24. 

10th Maths Important Questions

There will be 40 questions in total asked in the question paper, The type of questions is different in each section. There will be objective-type questions from Section-A which will carry 1 mark. Section-B contains short-type questions which are for 2 marks each. The short type questions will be asked in section-C too which will be for 3 marks each. The section-D will have long-type questions which will carry 4 marks each.

The Important Question Of Maths Class 10 Short Type Questions

These questions will be mostly based on the formulae or describing the part of the formula. The questions will be helpful to find the surface area of the given shape, finding the decimal representation, completing the sequence of Arithmetic progression, the ratio of areas of two similar triangles when the ratio of sides are given, basic probability questions, finding if the pair of straight lines are parallel, nature of roots in a quadratic equation. 

CBSE Maths Important Questions of Section- C

The section-C will have short-type questions which will be mostly from the chapter Arithmetic progressions. These will include finding the sum to n terms. The next important topic is from applications of trigonometry like finding the height of the tree if they make a certain angle of elevation or deviation. Questions will also be asked from real numbers (which include finding) if the number is irrational, terminating or non-terminating, etc.

The statistics chapter will have good weightage and the questions can be related to mean median, mode. Coming to the chapter of coordinate geometry, the section formula, distance formula, midpoint of 2 points will be mostly asked in the exam.  

CBSE Maths Important Questions From Section D

There will be an internal choice in this section to answer the paper. In each question, two questions will be asked from which you can answer any one of your choices. 

The questions from quadratic equations will be to solve for the value of x. The practical questions from surface areas and volumes will be asked where the real-time examples will be asked and the area or volume should be calculated. The questions from statistics will be related to the mean, median, and mode of the given data. 

Solved Examples

1. If ΔAB[tilde{C}] ΔPQR and the Ratio of AB and PQ is 4 : 5. Find the Ratio of their Areas and Perimeters.

Given: ΔAB[tilde{C}] ΔPQR

AB : PQ = 4 : 5

[frac{AB}{PQ}] = [frac{4}{5}]

We know that the ratio of areas of 2 similar triangles is equal to the square of the ratio of their corresponding sides.

[frac{Ar(triangle ABC)}{Ar(triangle PQR)}] = ([frac{AB}{PQ}])[^{2}]

[frac{Ar(triangle ABC)}{Ar(triangle PQR)}] = ([frac{4}{5}])[^{2}]

[frac{Ar(triangle ABC)}{Ar(triangle PQR)}] = [frac{16}{25}]

Ar([triangle]ABC) : Ar([triangle]PQR) = 16 : 25  

Hence, the ratio of areas of two triangles is 16 : 25 .

We know that the ratio of areas of 2 similar triangles is equal to the square of their perimeters.

[frac{Ar(triangle ABC)}{Ar(triangle PQR)}] = ([frac{text{Perimeter of triangleABC}}{text{Perimeter of trianglePQR}}])[^{2}]

[sqrt{frac{Ar(triangle ABC)}{Ar(triangle PQR)}}] = ([frac{text{Perimeter of triangle ABC}}{text{Perimeter of triangle PQR}}])

[sqrt{frac{16}{25}}] = ([frac{text{Perimeter of triangle ABC}}{text{Perimeter of triangle PQR}}])

[sqrt{(frac{4}{5})^{2}}] = ([frac{text{Perimeter of triangle ABC}}{text{Perimeter of triangle PQR}}])

[frac{4}{5}] = ([frac{text{Perimeter of triangle ABC}}{text{Perimeter of triangle PQR}}])

Perimeter ([triangle]ABC) : Perimeter ([triangle]PQR) = 4 : 5

Hence, the ratio of the perimeters of two similar triangles is 4:5.

Q2. The Point P(x, y) Divides the Line Segment Joining the Points A(1, 4) and B(5, -1). If the x-coordinate of a Point P is 2, Find the Ratio in Which the Point P Divides Point A,B and also Find the y -coordinate of a Point P.

Given: A(1,4) and B(5,-1)

Let Point P(x,y) divides the points A, B in the ratio k : 1

x -coordinate is 2 ⇒ P(2,y) 

Section formula: If a point p(x,y) divides (x[_{1}],y[_{1}]) and (x[_{2}],y[_{2}]) in the ratio m : n, then

p(x,y) = ([frac{mx_{2} + nx_{1}}{m + n}] , [frac{my_{2} + ny_{1}}{m + n}]

p(2,y) = ([frac{5k+1}{k+1}] , [frac{-k+4}{k+1}])

On comparing the x -coordinate, we get

2 = [frac{5k + 1}{k + 1}]

2(k + 1) = 5k + 1

2k + 2 = 5k + 1

5k – 2k = 2 – 1

3k = 1

k = [frac{1}{3}]

So, the ratio of k : 1 = [frac{1}{3}] : 1 

On comparing the y -coordinates, we get

y = [frac{-k+4}{k+1}]

⇒ y = [frac{-frac{1}{3} + 4}{frac{1}{3} + 1}]

⇒ y = [frac{-frac{-1+12}{3}}{frac{1+3}{3}}]

⇒ y = [frac{frac{11}{3}}{frac{4}{3}}]

⇒ y = [frac{11}{4}]

Hence, the ratio in which point P divides points A, B is [frac{1}{3}] : 1 and the value of y -coordinate is [frac{11}{4}].

If we talk about the central limit theorem meaning, it means that the mean value of all the samples of a given population is the same as the mean of the population in approximate measures, if the sample size of the population is fairly large and has a finite variation. The central limit theorem is one of the important topics when it comes to statistics. In this article, we will be learning about the central limit theorem standard deviation, the central limit theorem probability, its definition, formula, and examples.

Central Limit Theorem Definition

Let us first define the central limit theorem. 

The Central Limit Theorem states that the overall distribution of a given sample mean is approximately the same as the normal distribution when the sample size gets bigger and we assume that all the samples are similar to each other, irrespective of the shape of the total population distribution.

Central Limit Theorem Statistics Example

To understand the Central Limit Theorem better, let us consider the following example.

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Assume that you have 10 different sports teams in your school and each team consists of 100 students. Now, we need to find out the average height of all these students across all the teams. How will we do it when there are so many teams and so many students? 

Well, the easiest way in which we can find the average height of all students is by determining the average of all their heights. To do so, we will first need to determine the height of each student and then add them all. Then, we will need to divide the total sum of the heights by the total number of the students and we will get the average height of the students. Well, this method to determine the average is too tedious and involves tiresome calculations. So, how do we calculate the average height of the students? We can do so by using the Central Limit Theorem for making the calculations easy.

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In this method of calculating the average, we will first pick the students randomly from different teams and determine a sample. Every sample would consist of 20 students. Then, we would follow the steps mentioned below:

1. First, we will take all the samples and determine the mean of each sample individually.

2. Then, we will determine the mean of these sample means.

3. This way, we can get the approximate mean height of all the students who are a part of the sports teams. 

If we find the histogram of all these sample mean heights, we will obtain a bell-shaped curve.

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Note: It is important to remember that the samples that are taken should be enough by size. When we take a larger sample size, the sample mean distribution becomes normal when we calculate it by repeated sampling.

Central Limit Theorem Formula

Now that we learned how to explain the central limit theorem and saw the example, let us take a look at what is the formula of the Central Limit Theorem.

We can apply the Central Limit Theorem for larger sample size, i.e., when n ≥ 30.

The formula of the Central Limit Theorem is given below.

μx = μ

𝜎x= 𝜎/√n

Here, 

μ is the population mean

𝜎 is the standard deviation of the population

μx is the sample mean

𝜎x is the sample standard deviation

n is the sample size

Circumcenter of triangle

The point of intersection of the perpendicular bisectors of the sides of a triangle is called its circumcenter.

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Note: The perpendicular bisectors of the sides of a triangle may not necessarily pass through the vertices of the triangle.

Properties of Circumcenter of Triangle

1. Circumcenter is equidistant to all the three vertices of a triangle.

2. The circumcenter is the centre of the circumcircle of that triangle.

3. Circumcenter is denoted by O (x, y).

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4. The circumcenter of an acute angled triangle lies inside the triangle.

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5. The circumcenter of the right-angled triangle lies at the midpoint of the hypotenuse of the triangle.

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6. The circumcenter of the obtuse angled triangle lies outside the triangle.

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Locating Circumcenter of Triangle Through Construction

The circumcenter of any triangle can be constructed by drawing the perpendicular bisector of any of the two sides of that triangle. 

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Following are the Steps to Locate the Circumcenter of the Triangle.

Step:1 Draw the perpendicular bisector of any two sides of the given triangle.

Step:2 Extend the perpendicular bisectors until they intersect each other.

Step:3 Mark the intersecting point as O which will be the circumcenter of the triangle.

Finding the third perpendicular bisector will ensure more accuracy of the location of the circumcenter.

Construction of Circumcircle to a Given Triangle

Following are the steps to construct the circumcircle to a given triangle.

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Step:1 Locate the circumcenter by constructing the perpendicular bisectors of at least two sides of the given triangle.

Step:2 Place the compass point on the circumcenter O and stretch to any one of the vertices of the given triangle.

Step:3 Rotate compass to draw a circumcircle.

Circumcenter formula

If A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the ∆ ABC and A, B, C are their respective angles. Then,

[Oleft( {x,y} right) = left( {frac{{{x_1}sin 2A + {x_2}sin 2B + {x_3}sin 2c}}{{sin 2A + sin 2B + sin 2C}},frac{{{y_1}sin 2A + {y_2}sin 2B + {y_3}sin 2c}}{{sin 2A + sin 2B + sin 2C}}} right)]

 

Methods to Calculate Circumcenter of Triangle

METHOD:1

Following are the steps to calculate the circumcenter of a given triangle

Step:1 Find the coordinates of midpoint (xm, ym), of sides AB, BC and AC, using the mid-point theorem.

Step:2 Calculate the slope of each side. If the slope of any side is ‘m1’ then, the slope of the line perpendicular to it will be 1/’m1’. Assume, m = 1/’m1’.

Step:3 Using coordinates of midpoint (xm, ym), and the slope of perpendicular line ‘m’. write the equation of line (y-ym) = m (x-xm).

Step:4 Similarly, find the equation of other lines.

Step:5 Solve them to find out their intersection point.

The obtained intersection point will be the circumcenter of the given triangle.

METHOD:2

Since, we know the property of circumcenter that, Circumcenter is equidistant to all the three vertices of a triangle.

Let O (x, y) be the circumcenter of ∆ ABC. Then, the distances to O from the vertices are all equal, we have AO = BO = CO = Circumradius.

Assume that D1 be the distance between the vertex A (x1, y1) and the circumcenter O (x, y), then

D₁² = (x – x1)2 + (y – y1)2 ( Using distance formula of two points in a coordinate)

Similarly,

[D_2^2 = {(x – {x_2})^2} + {(y – {y_2})^2}]

[D_3^2 = {(x – {x_3})^2} + {(y – {y_3})^2}]

Since, D1 = D2 and D2 = D3,  

By this we will get two linear equations:

Equation:1

(x – x1)2 + (y – y1)2 = (x – x2)2 + (y – y2)2

Equation:2

(x – x2)2 + (y – y2)2 = (x – x3)2 + (y – y3)2

By solving these two linear equations using a substitution or elimination method, the coordinates of the circumcenter O (x, y) can be obtained.

 

Solved Examples:

Q.1. Find the circumcenter of ∆ ABC with vertices A = (1, 4), B = (-2, 3), C = (5, 2). 

Ans. 

Since the distances to the circumcenter O from the vertices are all equal.

So, AO=BO=CO. 

From the first equality, we have AO2 = BO2

      (x – 1)2 + (y – 4)2 = (x + 2)2 + (y – 3)2

⇒ -2x + 1 – 8y + 16 = 4x + 4 – 6y + 9

⇒ 3x + y = 2          (1)

 

​Similarly, from the second equality, we have BO2 = CO2 

⇒ (x + 2)2 + (y – 3)2 = (x – 5)2 + (y – 2)2

⇒ 4x + 4 – 6y + 9 = -10x + 25 – 4y + 4

⇒ 7x – y = 8           (2) 

 

Solving equations (1) and (2)

Adding (1) + (2) gives:

 x = 1 which in turn gives y = −1

Therefore, the circumcenter of triangle ABC is O = (1, -1).

 

Q.2. Using the circumcenter formula, find the circumcenter of ∆ ABC whose vertices A (0, 2), B (0, 0) and C (2, 0) and respective measures of angles A, B and C are 450, 900 and 450.

 

Ans. 

We know that, if A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of the ∆ ABC and A, B, C are their respective angles. Then, 

 

Circumcenter =  [Oleft( {x,y} right) = left( {frac{{{x_1}sin 2A + {x_2}sin 2B + {x_3}sin 2c}}{{sin 2A + sin 2B + sin 2C}},frac{{{y_1}sin 2A + {y_2}sin 2B + {y_3}sin 2c}}{{sin 2A + sin 2B + sin 2C}}} right)]

On putting the corresponding values of coordinates of vertices and angle measures of the ∆ ABC in the above formula. We get:

      [Oleft( {x,y} right) = left( {frac{{0sin 2left( {45} right) + 0sin 2left( {90} right) + 2sin 2left( {45} right)}}{{sin 2left( {45} right) + sin 2left( {90} right) + sin 2left( {45} right)}},frac{{2sin 2left( {45} right) + 0sin 2left( {90} right) + 0sin 2left( {45} right)}}{{sin 2left( {45} right) + sin 2left( {90} right) + sin 2left( {45} right)}}} right)]

      [Oleft( {x,y} right) = left( {frac{{2left( {sin 90} right)}}{{sin left( {90} right) + sin left( {180} right) + sin left( {90} right)}},frac{{2left( {sin 90} right)}}{{sin left( {90} right) + sin left( {180} right) + sin left( {90} right)}}} right)]On putting the values of corresponding trigonometric ratios:

[sin {45^o}frac{1}{{sqrt 2 }}] Sin900 = 1     Sin1800 = 0

⇒ O (x, y) =[left( {frac{2}{{1 + 0 + 1}},frac{2}{{1 + 0 + 1}}} right)]

⇒ O (x, y) = [left( {frac{2}{2},frac{2}{2}} right)]

⇒ O (x, y) = (1, 1)

 

Q.3. Find the circumcenter of ∆ ABC with the vertices A= (1, 2), B= (3, 6) and C= (5, 4).

Ans. 

To calculate the coordinates of circumcenter of the ∆ ABC, we have to solve any two bisector equations and then, find out the intersection points that will give the coordinates of the circumcenter.

So, the midpoint of side AB =[left( {frac{{1 + 3}}{2},frac{{2 + 6}}{2}} right) = left( {2,4} right)]

And slope of AB = [frac{{6 – 2}}{{3 – 1}} = 2]

The slope of the perpendicular bisector of side AB is negative reciprocal of the slope of AB

So, slope of the perpendicular bisector of side AB = [ – frac{1}{2}]

The Equation of perpendicular bisector of AB with slope [ – frac{1}{2}]and the coordinates (2,4) is,

[left( {y – 4} right) =  – frac{1}{2}left( {x – 2} right)]

x +2y = 10            (1)

Similarly, we will proceed for side AC

The midpoint of side AC =[left( {frac{{1 + 5}}{2},frac{{2 + 4}}{2}} right) = left( {3,3} right)]

And slope of AC = [frac{{4 – 2}}{{5 – 1}} = frac{1}{2}]

So, slope of the perpendicular bisector of side AC = -2

The Equation of perpendicular bisector of AC with slope -2 and the coordinates (3, 3) is,

(y – 3) = -2 (x – 3)

2x + y = 9            (2)

On solving equations (1) and (2), we get:

x =[frac{8}{3}]  and 

y =  [frac{{11}}{3}]

So, the circumcenter of the ∆ ABC is O[left( {frac{8}{3},frac{{11}}{3}} right)].

A common denominator is useful for conducting a number of mathematical operations on numbers. A crucial concept in Mathematics is the addition and subtraction of fractions. Fractions or a fractional number has 2 parts that include a numerator (the number on top) and a denominator (the number below). That said, the fractions with the same denominators are called common denominators. Consider the following denominator examples: 3/7 + 5/7 = 3. In these cases, the denominators in the fraction are common, thus, it is also easy to calculate the answer.

Methods to Find a Common Denominator?

Want to know how to find the common denominator easily? There may be instances where you would be asked to add fractions with different denominators like 5/7 + 9/13. In such times, you are required to determine the common denominator and then solve the fractions. Following are the two methods to find the common denominator:

1. By cross multiplication

2. By finding the least common multiple (LCM)

When we solve for the common denominator using the LCM (least common multiple) method, you find the LCM of the given numbers. In this equation, the LCM is 6. Thus, the equation becomes 1/3 + 1/6 = (1 x 2 + 1)/6 = (2 + 1)/6 = 3/6 = 1/2. However, If you the cross multiplication method, you will find the solution as: 1/3 + 1/6 = 2/6 + 1/6 = (2 + 1)/6 = 3/6 = 1/2

How to Find the Least Common Denominator?

The least common denominator depends upon the type of denominator. For denominators having co-prime numbers, the least common denominator is the product of the two denominators. In addition, the least common denominator is the LCM of the two given denominators. Having said that, let us take two denominator values: 6 and 4.  As said, the least common denominator is the least common multiple of 4 and 6, which is the number 12.

Now, taking the case of numbers 3 and 4. In such an instance, neither 3 nor 4 are factors of each other. Therefore, you can calculate the value of the common denominator simply by multiplying both numbers and you obtain 12.

What is the Greatest Common Denominator?

In mathematics, the Greatest Common Denominator of two or more fractions, which are not zero (0), is the biggest positive integer which divides each of the given denominators.

Can a Common Denominator be 0 or 1?

For a fraction having a common denominator of zero (0), it becomes undefined. Whereas, for fractions with only whole numbers as numerators and 1 as a denominator, the common denominator will be 1. In the instance of whole numbers being taken into account as fractions, the common denominator is 1.

Fractions Without a Common Denominator

We can get the common denominators by multiplying both numerator (the top number) and denominator (the bottom number) by the same amount.

For example, take out the addition of two fractions which do not have a similar denominator:

Addition of 2/5 and 1/2.

Firstly, multiply 2/5 by 2/2  to obtain 4/10 .

Now, multiply the 2nd fraction 1/2 by 5/5  to obtain 5/10 .

The new fractions with a common denominator are 10.

Add the two fractions 4/10  + 5/10 , you will get  9/10.

Solved Examples on Common Denominator

Example:

Can you help Alex find the common denominator for the fractions 7/5 and 4/3?

Solution:

The denominators of the given fractions are 5 and 3 respectively. Thus, the LCM of 5 and 3 is 15 and 15 is the common denominator of the two given fractions.

Therefore, the common denominator for 7/5 and 4/3 2/3 is 15.

Example:

Find out if the rational numbers -9/12 and 21/-28 equal?

Solution:

The two given fractions can be simplified as:

-9/12 = -3/4

21/-28 = 3/-4 = -3/4.

Seeing that, the simplified value of the given two fractions is the same, thus, the two rational numbers are also equal.

∴ We can conclude that the two rational numbers -9/12 and 21/-28 are equal.

Fun Fact

The LCM of the denominators helps us in getting the common denominator.