APPLICATIONS OF TRIGONOMETRY...

DISTANCES AND HEIGHTS :

http://www.youtube.com/watch?v=W3C9unrSrRo

COORDINATE GEOMETRY : DISTANCE FORMULA...

COORDINATE GEOMETRY..

DISTANCE FORMULA...

http://www.youtube.com/results?search_query=coordinate+geometry+10&oq=coordinate+geometry+10&gs_l=youtube.3..33i21.29302.31351.0.32399.3.3.0.0.0.0.237.687.2-3.3.0...0.0...1ac.1.-jeVSBEbb1s

X ARITHMETIC PROGRESSION...

X

ARITHMETIC PROGRESSION

BASICS..

http://www.youtube.com/watch?v=zlsyrOcvq2k

TED-Ed | Scott Kennedy: How to prove a mathematical theory

Euclid of Alexandria revolutionized the way that mathematics is written, presented or thought about, and introduced the concept of mathematical proofs. Discover what it takes to move from a loose theory or idea to a universally convincing proof.

Source: Scott Kennedy Page at ed.ted.com

X QUADRATIC EQUATIONS.

Chapter 04 

QUADRATIC EQUATIONS

 a quadratic equation is a  polynomial equation of the second degree. A general quadratic equation can be written in the form




where x represents a variable or an unknown, and a, b, and c are constants with a ≠ 0. (If a = 0, the equation is a linear equation.)
The constants a, b, and c are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term. The term "quadratic" comes from quadratus, which is the Latin word for "square". Quadratic equations can be solved by factoring, completing the square  and using the quadratic formula
A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.
Having

http://en.wikipedia.org/wiki/Quadratic_equation

SYLLABUS FOR X FOR SEMESTER II

MATHEMATICS NCERT X :

1. QUADRATIC EQUATION
2. ARITHMETIC PROGRESSION
3. SOME APPLICATIONS OF TRIGONOMETRY
4. CIRCLES
5. CONSTRUCTIONS
6. SURFACE AREAS AND VOLUMES
7. AREAS RELATED TO CIRCLES
8. COORDINATE GEOMETRY
9. PROBABILITY

For more information related to syllabus please visit : quaddhelp

for any other queries, please feel free to ask..

Test decodes dolphins’ math skills

Nonlinear mathematics may be behind the mammals’ ability to see through bubbles

By Meghan Rosen

Web edition : Thursday, July 19th, 2012

Enlarge

By studying the clicks of Atlantic bottlenose dolphins, researchers have figured out a way to create sonar that sees through bubbles.Mike Price/Shutterstock

Dolphins could teach humans a thing or two about finding Nemo. The aquatic mammals may pinpoint prey hidden in bubbles by using mental math.

By adjusting the volume of sonar clicks, then processing the incoming echoes, dolphins might have solved a problem that still stymies man-made sonar: how to peer through frothy water. Using clicks that mimic an Atlantic bottlenose dolphin, scientists devised a system that weeds out sound clutter from underwater bubbles.

“It’s really ingenious, actually,” says oceanographer Grant Deane of the Scripps Institution of Oceanography in La Jolla, Calif. “I think it’s very clever work, and there are a number of significant applications for it.”

Using something like a fireman’s hose, researchers shot bubbles into a huge water tank set underground. The bubbles cloaked a submerged target: a steel ball slightly smaller than a baseball. Then, the researchers sent out short bursts of sound — the faux dolphin clicks — underwater, collected the echoes, and processed the data mathematically to figure out the steel ball’s location.

It worked “brilliantly,” says physicist and engineer Timothy Leighton who led the study, published online July 12 in the Proceedings of the Royal Society A.

The findings could improve man-made sonar, allowing mine-hunting submarines to see more clearly in murky shallow waters.

Normally, using sonar to locate targets in bubbly water is a bit like driving a car at night in thick fog, says Leighton, of the University of Southampton in England. Switching on the headlights doesn’t improve the driver’s vision; fog droplets just scatter the light.

Similarly, bubbles scatter sound. But the sound they reflect gets a sonic boost: Bubbles kick back sonar signals perhaps a million times more powerfully than equally sized steel balls in water, Leighton says. This peppy bounce back makes it tricky for sonar to look through foamy waters.

But scientists have observed dolphins blowing bubble nets — dense clouds of tiny bubbles — while hunting schools of fish.

“It seemed strange,” says Leighton. “Either dolphins are blinding their best sensory apparatus, or they’re doing something we can’t.” Compared to the best sonar technology, a dolphin’s hardware is less powerful. But dolphins still outperform their man-made counterparts.

Enlarge

Dolphins often send out a spray of bubbles to herd their lunch (in this case, sardines). Unlike man-made sonar, dolphins can see through the frothy clouds using echolocation.Courtesy of The Blue Planet, BBC

The researchers used nonlinear mathematics to process the dolphins’ sonar returns. This type of calculation exploits the energetic echoes of the bubbles, which unlike those from steel balls or fish, bounce back sound as the square of the pulse that hits them.
“I don’t think dolphins are sitting there playing with numbers in their heads,” he says, but their brains might be using mathematics to decipher bubble-scattered sonar signals.

There’s a chance dolphins aren’t underwater mathletes, but the research team isn’t trying to prove the animals use this method of sonar processing, says engineer Hugh Griffiths from University College London in England. The researchers are asking whether or not it’s feasible, he says. “And they conclude — quite rightly — that yes, in principle, it is.” 
source|sciencenews

MATHEMATICS PREPARATORY TEST SOLUTION..!!!

SECTION A

10 Easy Arithmetic Tricks!!.

Math can be terrifying for many people. This list will hopefully improve your general knowledge of mathematical tricks and your speed when you need to do math in your head.

1. The 11 Times Trick

We all know the trick when multiplying by ten – add 0 to the end of the number, but did you know there is an equally easy trick for multiplying a two digit number by 11? This is it:

Take the original number and imagine a space between the two digits (in this example we will use 52:

5_2

Now add the two numbers together and put them in the middle:

5_(5+2)_2

That is it – you have the answer: 572.

If the numbers in the middle add up to a 2 digit number, just insert the second number and add 1 to the first:

9_(9+9)_9

(9+1)_8_9

10_8_9

1089 – It works every time.

2. Quick Square

If you need to square a 2 digit number ending in 5, you can do so very easily with this trick. Multiply the first digit by itself + 1, and put 25 on the end. That is all!

25² = (2x(2+1)) & 25

2 x 3 = 6

625

3. Multiply by 5

Most people memorize the 5 times tables very easily, but when you get in to larger numbers it gets more complex – or does it? This trick is super easy.

Take any number, then divide it by 2 (in other words, halve the number). If the result is whole, add a 0 at the end. If it is not, ignore the remainder and add a 5 at the end. It works everytime:

2682 x 5 = (2682 / 2) & 5 or 0

2682 / 2 = 1341 (whole number so add 0)

13410

Let’s try another:

5887 x 5

2943.5 (fractional number (ignore remainder, add 5)

29435

3. Multiply by 5

Most people memorize the 5 times tables very easily, but when you get in to larger numbers it gets more complex – or does it? This trick is super easy.

Take any number, then divide it by 2 (in other words, halve the number). If the result is whole, add a 0 at the end. If it is not, ignore the remainder and add a 5 at the end. It works everytime:

2682 x 5 = (2682 / 2) & 5 or 0

2682 / 2 = 1341 (whole number so add 0)

13410

Let’s try another:

5887 x 5

2943.5 (fractional number (ignore remainder, add 5)

29435

4. Multiply by 9

This one is simple – to multiple any number between 1 and 9 by 9 hold both hands in front of your face – drop the finger that corresponds to the number you are multiplying (for example 9×3 – drop your third finger) – count the fingers before the dropped finger (in the case of 9×3 it is 2) then count the numbers after (in this case 7) – the answer is 27.

5. Multiply by 4

This is a very simple trick which may appear obvious to some, but to others it is not. The trick is to simply multiply by two, then multiply by two again:

58 x 4 = (58 x 2) + (58 x 2) = (116) + (116) = 232

6. Calculate a Tip

If you need to leave a 15% tip, here is the easy way to do it. Work out 10% (divide the number by 10) – then add that number to half its value and you have your answer:

15% of $25 = (10% of 25) + ((10% of 25) / 2)

$2.50 + $1.25 = $3.75

7. Tough Multiplication

If you have a large number to multiply and one of the numbers is even, you can easily subdivide to get to the answer:

32 x 125, is the same as:
16 x 250 is the same as:
8 x 500 is the same as:
4 x 1000 = 4,000

8. Dividing by 5

Dividing a large number by five is actually very simple. All you do is multiply by 2 and move the decimal point:

195 / 5

Step1: 195 * 2 = 390
Step2: Move the decimal: 39.0 or just 39

2978 / 5

step 1: 2978 * 2 = 5956
Step2: 595.6

9. Subtracting from 1,000

To subtract a large number from 1,000 you can use this basic rule: subtract all but the last number from 9, then subtract the last number from 10:

1000
-648

step1: subtract 6 from 9 = 3
step2: subtract 4 from 9 = 5
step3: subtract 8 from 10 = 2

answer: 352

10. Assorted Multiplication Rules

Multiply by 5: Multiply by 10 and divide by 2.
Multiply by 6: Sometimes multiplying by 3 and then 2 is easy.
Multiply by 9: Multiply by 10 and subtract the original number.
Multiply by 12: Multiply by 10 and add twice the original number.
Multiply by 13: Multiply by 3 and add 10 times original number.
Multiply by 14: Multiply by 7 and then multiply by 2
Multiply by 15: Multiply by 10 and add 5 times the original number, as above.
Multiply by 16: You can double four times, if you want to. Or you can multiply by 8 and then by 2.
Multiply by 17: Multiply by 7 and add 10 times original number.
Multiply by 18: Multiply by 20 and subtract twice the original number (which is obvious from the first step).
Multiply by 19: Multiply by 20 and subtract the original number.
Multiply by 24: Multiply by 8 and then multiply by 3.
Multiply by 27: Multiply by 30 and subtract 3 times the original number (which is obvious from the first step).
Multiply by 45: Multiply by 50 and subtract 5 times the original number (which is obvious from the first step).
Multiply by 90: Multiply by 9 (as above) and put a zero on the right.
Multiply by 98: Multiply by 100 and subtract twice the original number.
Multiply by 99: Multiply by 100 and subtract the original number.

Source :-

http://listverse.com/2007/09/17/10-easy-arithmetic-tricks/

MATHS TRICKS FOR FASTER AND ACCURATE CALCULATIONS ..!!!!

Visit the following link for amazing tricks for calculations :) :)
http://www.glad2teach.co.uk/fast_maths_calculation_tricks.htm

7 Tips To Finish Your Maths Paper Before Time !!..

Many students, even some really intelligent and talented ones, have a strange enemy. They often find it difficult to finish the paper within the allotted time. They are forced to leave a few questions just because they run out of time and often it has been found that the questions they leave are those which they otherwise could do very easily. It can be very disappointing if you are forced to skip such easy questions.

But how can you avoid a situation like this? Many people suggest a single tablet for this "Time management". But how to manage time and how to stop it from running out is a difficult proposition, especially for an average 14-15-year-old tenth grader.

1. Understand your exam
The most important thing is to understand the examination you are about to take. In the class X mathematics paper, there are 30 questions in four sections A, B, C and D and we have 180 minutes to answer these questions. Here, a rough calculation is that we get about six minutes to answer a question. But that is not the fact.

The question paper contains 'very short answer' type, 'short answer' type and 'long answer' type questions and the time requirement for each type is different. An ideal allotment for the four sections is as shown below:

2. Use the first 15 minutes effectively

You get a good 15 minutes in the beginning to read the question paper -- use this time to do just that, READ. Read all the 30 questions in 15 minutes. While reading, mark the questions into categories viz easy, manageable and tough. This is done to have an overall idea about the questions and make a rough plan.

3. Don't worry about the tough ones

The moment you find that there are a few tough questions; it is natural that you start worrying about them. This is not required and will only harm your performance.

The fact is that they may look a bit tough on the surface, but when you actually work on them you will find most of them to be much easier than they seemed. So be happy about the easy ones and don't get unduly worried about the tougher lot.

4. Prioritise your attempt

Always attempt the easy questions first and then move on to the manageable ones and ensure that you complete them before taking on the difficult ones. This will ensure that you are not leaving any question that you know.

Once you successfully attempt all the easy and manageable questions, your confidence will grow and you will be mentally ready to take on the more challenging questions.

5. Ensure speed and accuracy

Use quicker methods in calculations to ensure that you are not wasting time and your answers are correct. Mostly, we take a lot of time to solve a problem if we happen to make some error in the process.

For example, if you make an error in the sign of a term (+/-), you may not be able to solve questions involving quadratic equations or linear equations. Therefore avoiding silly mistakes is very important to save time. Always follow the tricks we discussed in speed strategies.

6. Keep an eye on your watch

Keeping an eye on your watch is of course not to increase your stress. This is just to see that you are broadly adhering to the time allocation we discussed in the beginning. A minor variation is not at all a reason to worry.

7. Avoid thinking too much about a question
Thinking about the questions before you attempt them is essential; but not to such an extent that you waste a lot of time on one question.

Also you need not write a very lengthy answer to a question just because the question is easy and you know it very well. Remember, you need to just answer the question and nothing more. Any over-attempt will be a mere waste of time.

Additionally, you must practice the habit of finishing samples papers in 140-150 minutes. This will help you simulate and exercise examination pressures better.

Source :- http://cbse-sample-papers.blogspot.in

Dear Readers this was just to increase your knowledge about attempting your Maths examination and helping you in sorting out the problem of time management in mathematics paper.

Fast Math Trick

Math Study Skills

Preparing for a Math Test

Your success on a math test can be maximized by proper preparation.

1. Practice good study techniques throughout the semester.

   Read the section on study techniques for mathematics. for details.
   

2. Begin studying for the test at least one week ahead of time.

3. Work out all practice tests in the textbook and those given by your instructor.

4. Review your study lists.

5. Find out from your instructor:

   • on which topics or objectives you will be tested.
   • what materials are needed for the test: calculators, rulers, etc.
   • what materials are prohibited from the tests: calculators, cell-phones, etc.

6. Prepare yourself physically

   
   
   • Get proper exercise weekly.
   • Eat properly prior to the test.
     • Avoid overeating just before the test.
     • Eat a good breakfast and/or lunch before the test.
     • Do not drink too much before the test: you do not want to have to use the restroom during the test.
     • Avoid too much caffinated beverages before the test; this may cause nervousness.
     • Do not use alcohol or recreational drugs before the test. These will impair your concentration and brain functioning.
     • If you are taking prescribed medications, be aware of their effects on your concentration and thinking. Adjust your intellectual activity accordingly.
   • Get a good night sleep before the test. Staying up late cramming is not productive and can reduce your mental sharpness.
7. Read the section on doing well on a math test so you know what to do once the test has started.
8. Do not study the day of the test. Relax and be confident that you have done your best to prepare. Additional studying will only make you more nervous and reducde your confidence. Before the test, take a nice walk around the campus and think positive thoughts.

Click here for more info. 

TRY THESE OUT !!.....

Q1. 17 cards numbered 1 to 17 are put in abox and mixed . Find the probability that anumber on card is odd , a prime , divisible by 3 .




Q2. The king , queen and jack of clubs are removed from a deck of 52 playing cards and well - shuffled . One card is selected from remaining cards . Find the probability of getting heart , king , and club.




Q3. A bag contains 5 red balls and some blue balls . If the probability of drawing a blue ball is double that of ared ball , find the number of blue balls in a bag.




Q4. Cards marked from numbers 2 to 101 are placed in a box and mixed thoroughly. Find the probability that a number on a card is an even number , a number less than 14 , a number which is a perfect square and a prime number less than 20.




Q5. A jar contains 54 marbles each of which is blue , green and white . The probability of selecting a blue marble at random from jar is 1 / 3 , and the probability of selecting a green marble at random is 4 / 9 . How many white marbles does the jar contain?

Try these questions and share your solutions..............
Feel free to ask about any query.............

SOME IMPORTANT QUESTIONS !!...

Q1. Find the probability that a leap year selected at random will contain 366 Sundays.

366 days = 52 weeks and 2 days.

therefore , leap year = 52 sundays.

the remaining two days can be :-

sunday and monday , monday and tuesday , tuesday and wednesday , wednesday and thursday ,
thursady and friday , friday and saturday , saturday and sunday.

total elementary events = 7 and favourable are 2

Therefore required probability = 2 / 7



Q2. Tickets numbered from 1 to 20 are mixed together . What is a probability that a ticket has
a number which is multiple of 3 or 7 ?

Here , total favourable outcomes = 3 , 6 , 7 , 9 , 12 , 14 , 15 and 18.

required probability = 8 / 20 = 2 / 5.



Q3. A bag contain 6oo pens out of which 12 are defective . What is probability for non - defective
pens ?

Total pens = 6oo
defective = 12
therefore , non - defective = 600 - 12 = 588

So , required probability = 588 / 600 = 49 / 50 = 0.98.



Q4. A letter is chosen at random from the letters of word 'ASSASSINATION' . Find the
probability that letter chosen is (i) vowel (ii) consonant.

Total letters = 13

(i) favourable for vowel = 6 (A , A , I , A , I , O )
SO , required probability = 6 / 13

(ii) probability for consonants = 1 - 6 / 13 = 7 / 13.



Q5. If x is chosen at random from the numbers -1 , 0 , 1 ,2 . What is a probability that x is
less than 2 ?

Total elementary outcomes = 5
favourable = 3 (-1 , 0 , 1)

So , required probability = 3 / 5.



Source :- R.D. Sharma and 100 % Success.

Feel free to ask about queries and clear your doubts !!!........

PROBABILITY

Probability : p = no.of favourable outcomes/Total number of possible outcomes
                   q(not happening) = 1 - p

Sample space when:
(i) A coin is tossed, S = {H,T}
(ii) When two coins are tossed simultaneously = {HH,HT,TH,TT}
(iii) When three coins are tossed simultaneously = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT)

Sample space of throwing a die:

(i) When a die is thrown once = {1,2,3,4,5,6}

(ii) When two dice are thrown simultaneously
             { (1,1) , (1,2) , (1,3) ,......, (1,6)
                (2,1) , (2,2) , (2,3), .......,(2,6)
                (3,1) , (3,2) , (3,3),........,(3,6)
                (4,1) , (4,2) , (4,3),.........,(4,6)
                (5,1) , (5,2) , (5,3),..........,(5,6)
                (6,1) , (6,2) , (6,3),...........(6,6)

Geometric Probability: If the total number of outcomes of a trial in a random experiment is infinite, then the definition of probability is not sufficient to find the probability of an event.In such cases, the definition of probability is modified and probability so obtained is called geometric probability. The geometric probability p of an event is given by
            p = measure of the specified part of the region/measure of the whole region
                      (measure means length,area,volume of the region etc.)

Playing cards consists of 52 cards which are divided into four suits of 13 cards each. Each suit consists of one king, one queen, one jack, one ace, and nine other cards numbered from 2 to 10. The four suits are named as spade, club, heart, and diamond.
       
                                       If you are having any doubt in this chapter then feel free to ask here..!!!

Source: 100% success in mathematics

MATHEMATICS PREPARATORY TEST..!!!!

Dear Readers,
How was your test????
If you are having any doubt in any question, then please feel free to discuss over here..!
                                                                                                                  Do share your views with us..!!!

Important Unsolved Questions!!....

Q1. If the eighth term of an A.P. is 31 and fifteenth term is 16 more than the eleventh term , find the A.P.


Q2. Which term of the A.P. 5 , 15 , 25 , ....... will be 130 more than its thirty first term ?


Q3. If the tenth term of an A.P. is 52 and seventeenth term is 20 more than the thirteenth term , find the A.P.


Q4. The sum of fifth and ninth terms of an A.P. is 72 and the sum of seventh and twelfth terms is 97 . Find the A.P.


Q5. Determine the general term of an A.P. whose seventh term is -1 and sixteenth term is 17.


Q6. If the sum of m terms of an A.P. is same as the sum of its n terms , show that the sum of its (m+n) terms is zero .


Q7. Find the sum of all odd integers between 2 and 100 divisible by 3.

Try out these Questions and feel free to ask the queries...................!!!!!

A FEW IMPORTANT QUESTIONS (A.P.)

Ques1: Find the sum of first two 40 positive integers divisible by 6.

Ques2:In an A.P., find n and S_n if a = 5, d = 3, a_n = 50

Ques3: Is -150 a term of 11, 8 , 5 ,2 .. ? Why?

Ques4: If the 8^th term of an A.P. Is 31 and the 15^th term is 16 more than the 11^th term, find the A.P.

                                     Please feel free to share your problems ...!!! 

Arithemetic Progressions ( A.P. ) !!......

Introduction :-

An Arithemetic Prgression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.

e.g:- 1 , 4 , 7 , 10 , 13..........

In this example , first term is 1 and all other terms are obtained by adding a fixed number to the preceding term.

First term is signified by :- a

Common Difference between each term is signified by :- d

Here , a = 1 i.e. first term and d = 3 i.e. common difference.

The fixed number i.e. common difference (d) in A.P. Can be positive , negative or zero.

General form an A.P. :-

An Arithemetic Prgression where a is the first term and d is the common difference is referred to as General form an A.P.

Finite A.P. :-

Some A.P.s have only finte number of terms . Such an A.P. is called as Finite A.P. .

Such Finite A.P.s have the last term (l) .

Infinite A.P. :-

The A.P.s having infinite number of terms is referred to as Infinte A.P.

Such A.P.s doesn't have the last term (l).

nth Term of an A.P. :-

Let a₁ , a₂ , a₃ , …. be an A.P. whose first term a₁ is a and the common difference is d .

Then ,

the second term (a₂) = a + d = a + (2 – 1) d

the third term (a₃) = a₂ + d = (a + d ) + d = a + 2d = a + (3 – 1 ) d

the fourth term (a₄) = a₃ + d = (a + 2d ) + d = a + 3d = a + (4 – 1 ) d

Looking at the pattern above , we can say that the nth term a_n = a + (n – 1 ) d .

So , the nth term a_n of the A.P. with first term a and common difference d is given by:-

a_n = a + ( n – 1 ) d .

an is also called the general term of an A.P. . If there are m terms in the A.P. , then a_m represents the last term which is sometimes donated by l .

Sum of first n terms of an A.P. :-

The sum of first n terms of an A.P. with first term ' a ' and common difference ' d ' is given by :-

S_n = n / 2 [ 2a + (n – 1 ) d ]

It can also be written as :-

S_n = n / 2 [ a + a ( n – 1 ) d ]

i.e. S_n = n / 2 (a + a_n )

So , if there are n terms in the A.P. , then an = l , the last term.

Therefore , the sum can also be given by :-

S_n = n / 2 ( a + l )

TRY THESE QUESTIONS!!........

Q1. Find the roots of the following equation 4x² + 4bx – (a² – b²) = 0 by the method of completing the square.

Q2. Solve the quadratic equation :- x² + 3x – (a² + a – 2) = 0 by factorization method.

Q3. By using the method of completing the square , show that the equation 4x² + 3x + 5 = 0 has no real roots.

Q4. Find the values of k for which the given equation i.e. 2x² – 10x + k = 0 has real and equal roots.

Q5. The sum of two numbers is 15. If the sum of their reciprocals is 3 / 10 , find the numbers.

Q6. The product of Rohan's age five years ago with his age 9 years later is 15. Find Rohan's present age.

Q7. One year ago , a man was 8 times as old as his son. Now his age age is equal to the square of his son's age. Find their present ages.

Try to sort out these questions and share your solutions..

Ask the queries in case it occurs in any question!!!!

A FEW IMPORTANT QUESTIONS (Quadratic Equations)

Ques1: For what values of  'k' does the quadratic equation (k-5)x² + 2(k-5)x + 2 = 0 have equal roots?

Ques2: A plane left 30minutes later than the schedule time and in order to reach its destination 1500km away in time, it has to increase its speed by 250km/hr from its usual speed. Find its usual speed.

  

 FEEL FREE TO SHARE YOUR PROBLEMS..!!!

DO COMMENT BELOW IF YOU ARE HAVING ANY DOUBT RELATED THESE QUESTIONS AND CHAPTER..!!

  

CHAPTER :- QUADRATIC EQUATIONS!!.....

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0 , where a , b , c are real numbers , a ≠ 0 .

In fact , any equation of the form p(x) = 0 , where p(x) is a polynomial of degree 2 , is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees , then we get the standard form of the equation .

Therefore the standard / general form of quadratic equation is ax² + bx + c = 0 , a 0 and a , b , c are real numbers. In this equation , variable is 'x' and the degree of the equation is '2'.

If x =  satisfies the equation ax² + bx + c = 0 , then is known as the root of the quadratic equation.

The roots and zeroes of the quadratic equation ax² + bx + c = 0 , a ≠ 0 are same.

If the given quadratic equation ax² + bx + c = 0 , a ≠ 0 can be factorised into two linear factors, then its roots can be found by equating each factor equal to zero.

We can solve quadratic equation by completing the square.

The roots of the quadratic equation ax² + bx + c = 0 , a ≠ 0 can be found by using the following formula, if its discriminate (D = b² – 4c) is greater than or equal to zero.

x = - b  b2 – 4ac / 2a

A quadratic equation ax² + bx + c = 0 , a ≠ 0 has:-

1. two distinct real roots , if D > 0.

2. two equal roots (i.e., repeated roots), if D = 0.

3. no real roots. if D < 0.

Questions related to this chapter will be posted soon!!..

TRY OUT THESE.....!! (CIRCLES)

Q1. Two tangents are drawn to a circle from an exterior point A, touching the at B and C. From another point K a third tangent is drawn to a circle at Q. If AB = 20cm, then find the perimeter of triangle APR.

Q2. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Q3. Find the length of a tangent drawn from a point whose distance from the centre of a circle is 25cm. Also the radius of a circle is 7cm.

Q4. From an external point P, two tangents PA and PB are drawn to the circle with centre O. Prove that OP is the perpendicular bisector of AB.

Q5. In two concentric circles, prove that a chord of larger circle which is tangent to smaller circle is bisected at the point of contact.

Q6. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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SOME IMPORTANT QUESTIONS (Chapter:Circles)

Q1. ABC is an isosceles triangle in which AB = AC, circumscribed about a circle. Prove that the base is bisected by the point of contact. i.e. if AB = AC, prove that BE = EC.                 CBSE 2008

Solution:-

Since the tangents drawn from any exterior point to a circle are equal in length, therefore:-

AD = AF - ( Tangents from A)

BD = BE - ( Tangents from B)

CE = CF - ( Tangents from C)

Now,

AB = AC

AB- AD = AC- AD - (Subtracting AD from both sides)

AB- AD = AC – AF - (Since, AD=AF)

BD = CF

BE = CF

BE = CE , Hence proved.

Q2. A circle touches all the four sides of a quadrilateral ABCD. Prove that AB + CD = BC + DA.

                                                                                                                               CBSE 2008,2009

Solution:-

Since the tangents drawn from any exterior point to a circle are equal in length, therefore:-

AP = AS - (Tangents from A) - (i)

BP = BQ - ( Tangents from B) - (ii)

CR = CQ - ( Tangents from C) - (iii)

DR = CQ - ( Tangents from D) - (iv)

Adding (i) , (ii) , (iii) , (iv) , we get :-

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

AB + CD = BC + DA , Hence proved.

Q3. Prove that a parallelogram circumscribing a circle ia a rhombus.                   CBSE 2002, 2008

Solution:-

Let ABCD be a parallelogram such that its sides touch a circle with centre O.

Since the tangents to a circle from an exterior point are equal in length , therefore:-

AP = AS - (Tangents from A) ….......(i)

BP = BQ - (Tangents from B) …........(ii).

CR = CQ - (Tangents from C) ….........(iii)

DR = DS - (Tangents from D) …...........(iv).

Adding (i) , (ii) , (iii) , (iv) , we get:-

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)

AB + CD = AD + BC

2AB = 2BC ….....................(Since, ABCD is a parallelogram, therefore AB = CD and BC = AD)

AB = BC

AB = BC = CD = AD

Therefore, ABCD is a rhombus, Hence proved.

Q4. A circle is inscribed in a triangle ABC having sides 8cm , 10cm and 12cm. Find AD , BE and CF.                                                                                                                                 CBSE 2001

Solution:-

Since, the tangents drawn from an external point to a circle are equal , therefore:-

AD = AF = say , x

BD = BE = say , y
CE = CF = say , z

Now,

AB = 12cm , BC = 8cm and CA = 10cm

Therefore,

x + y = 12 , y + z = 8 and z + x = 10

(x+y)+(y+z)+(z+x) = 12 + 8 + 10

2(x+y+z) = 30

x+y+z = 15

Now,

x+y = 12 and x+y+z = 15

12 + z = 15

z = 15 – 12 = 3

Similarly,

y+z = 8 and x+y+z = 15

x+8 = 15

x = 15 – 8 = 7

Similarly,

z+x = 10 and x+y+z = 15

y+10 = 15

y = 15 – 10 = 5

Therefore, x = 7cm , y = 5cm ad z = 3cm i.e. AD = 7cm , BE = 5cm and CF = 3cm.

Q5. Prove that the tangents at the extremities of any chord make equal angles with the chord.

                                                                                                                             CBSE 2000,2001,2002

Solution:-

Let AB be a chord of a circle with centre O , and let AP and BP be the tangents at A and B.

Suppose the tangents meet at P. Join OP. Suppose OP meets AB at C .

We need to prove that PAC = PBC.

In triangles PCA and PCB , we have:-

PA = PB ….......(Since, tangents from an external points are equal)

  APC =  BPC ….........(Since, PA and PB are equally inclined to OP)

PC = PC ….......(Common)

therefore, by SAS criterion of congruence , we have:-

triangle PAC = triangle PBC

PAC = PBC , Hence proved.

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