Kvpy Mathematics

Resonance Pre-foundation Programmes

(PCCP) Division

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COURSE : KVPY (STAGE-) I

MATHEMATICS

WORKSHOP TAPASYA

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Subject : Mathematics KVPY

S. No. Topics Page No.

1. Ratio & Proportion, Mixture & Alligation 1 - 4

2. Linear Inequalities 5 - 7

Sets 13 16

4. Trigonometry 17 - 22

5. Lines, Angles & 23 - 25

6. Triangles -

2. Quadratic equation & Progressions 8 - 12

3. -

26 28

7. Circles 29 - 30

8. Number System 31 - 34

9. Mensuration 35 - 38

Quadrilaterals

.13RPCCP


1PAGE # 1

Ratio :

The comparison of two quantities a and b of similar

kind is represented as a : b is called a ratio also it can

be represented as ba

.

In the ratio a : b, we call a as the first term or

antecedent and b, the second term or consequent.

e.g. The ratio 5 : 9 represents 95

, with antecedent = 5

and consequent = 9.

The multiplication or division of each term of a ratio by

the same non-zero number does not affect the ratio.

e.g. 4 : 5 = 8 : 10 = 12 : 15 etc. Also, 4 : 6 = 2 : 3.

Proportion :

The equality of two ratios is called proportion.

If a : b = c : d, we write, a : b : : c : d and we say that a, b,

c, d are in proportion.

where, a is called first proportional, b is called second

proportional, c is called third proportional and d is

called fourth proportional.

Law of Proportion :

Product of means = Product of extremesThus, if a : b : : c : d (b × c) = (a × d),

Here a and d are called extremes, while b and c are

called mean terms.

Mean proportional of two given numbers a and b

is ab .

Some other ratios :

Compounded Ratio : The compounded ratio of the

ratios (a : b), (c : d), (e : f) is (ace : bdf).

Duplicate ratio : The duplicate ratio of (a : b) is (a2 : b2).

Sub-duplicate ratio : The sub-duplicate ratio of (a : b)

is ( a : b ).

Triplicate ratio : The triplicate ratio of (a : b) is (a3 : b3).

Sub-triplicate ratio : The sub-triplicate ratio of (a : b) is

.b:a 31

31

Componendo : If dc

ba then, the componendo is

ddc

bba

.

RATIO-PROPORTION, PARTNERSHIP AND

MIXTURE & ALLIGATION

Dividendo : If dc

ba then, the dividendo is

dd�c

bb�a .

Componendo and Dividendo : If dc

ba , then the

componendo-dividendo is d�cdc

b�aba

.

VARIATION :

(i) We say that x is directly proportional to y, if x = ky for

some constant k and we write, x y.(ii) We say that x is inversely proportional to y, if xy = k

for some constant k and we write, y1

x .

Ex.1 If a : b = 5 : 9 and b : c = 4 : 7, find a : b : c.

Sol. a : b = 5 : 9 and b : c = 4 : 7 =

49

7:49

4 = 9 : 4

63

a : b : c = 5 : 9 : 4

63 = 20 : 36 : 63.

Ex.2 Find out :(i) the fourth proportional to 4, 9, 12;(ii) the third proportional to 16 and 36;(iii) the mean proportional between 0.08 and 0.18.

Sol. (i) Let the fourth proportional to 4, 9, 12 be x.Then, 4 : 9 : : 12 : x 4 × x = 9 × 12

x = 4129

= 27.

Fourth proportional to 4, 9, 12 is 27.

(ii) Let the third proportional to 16 and 36 is x.Then, 16 : 36 : : 36 : x 16 × x = 36 × 36

x = 16

3636 = 81.

Third proportional to 16 and 36 is 81.

(iii) Mean proportional between 0.08 and 0.18

= 10012

100100144

10018

1008

18.008.0

= 0.12

Ex.3 If x : y = 3 : 4, find (4x + 5y) : (5x � 2y).

Sol.43

yx

.732

47

)53(

2�43

5

543

4

2�yx

5

5yx

4

y2�x5y5x4


2PAGE # 2

Ex.4 Divide Rs. 1162 among A, B, C in the ratio 35 : 28 : 20.

Sol. Sum of ratio terms = (35 + 28 + 20) = 83.

A�s share = Rs.

8335

1162 = Rs. 490;

B�s share = Rs.

8328

1162 = Rs. 392;

C�s share = Rs.

8320

1162 = Rs. 280.

Ex.5 A bag contains 50 p, 25 p and 10 p coins in the ratio

5 : 9 : 4, amounting to Rs. 206. Find the number of

coins of each type.

Sol. Let the number of 50 p, 25 p and 10 p coins be 5x, 9x

and 4x respectively.

Then, 10

x44x9

2x5

= 206

50x + 45x + 8x = 4120

103x = 4120

x = 40.

Number of 50 p coins = (5 × 40) = 200;

Number of 25 p coins = (9 × 40) = 360;

Number of 10 p coins = (4 × 40) = 160.

Ex.6 If a man goes from a place A to another place B 100 m

apart in 4 hours at a certain speed. With the same

speed going from B to C 400 m apart, what time will he

take ?

Sol. d = st, where d is distance in m, s is speed in m/sec.,

t is time in seconds. Speed is same d t

New distance is 4 times, now the time will be 4 times

the time it takes from A to B .So, the time taken from B

to C is 4 × 4 = 16 hours.

Alligation : It is the rule that enables us to find the ratio

in which two or more ingredients at the given price

must be mixed to produce a mixture of a desired price.

Mean Price : The cost price of a unit quantity of mixture

is called the mean price.

Rule of Alligation : If two ingredients are mixed, then,

cheaper of C.P.�price Mean

price Mean�dearer of C.P.dearer ofQuantity

cheaper ofQuantity

We can also represent this thing as under

(m) Mean price

C.P. of a unit quantity of dearer C.P. of a unit quantity of cheaper

(m � c)(d � m)

(c) (d)

Suppose a container contains x units of liquid from

which y units are taken out and replaced by water. Aftern operations, the quantity of pure liquid

=

n

xy

�1x units.

Ex.7 The cost of Type 1 rice is Rs.15 per kg and Type 2 riceis Rs.20 per kg. If both type-1 and type-2 are mixed inratio of 2 : 3 , then find the price per kg of the mixedvariety of rice.

Sol. Let the price of the mixed variety be Rs. x per kg.By the rule of alligation, we have :

Rs. xMean price

Cost of 1 kg of Type 2 riceCost of 1 kg of Type 1 rice

Rs. 20

(x � 15)(20 � x)

Rs. 15

)15�x()x�20(

= 32

60 � 3x = 2x � 30

5x = 90 x = 18.So, price of the mixture is Rs.18 per kg.

Ex.8 A milk vendor has 2 cans of milk. The first contains25% water and the rest milk. The second contains50% water. How much milk should he mix from eachof the containers so as to get 12 litres of milk such thatthe ratio of water to milk is 3 : 5 ?

Sol. Let cost of 1 litre milk be Rs.1.

Milk in 1 litre mixture in 1st can = 43

litre,

C.P. of 1 litre mixture in 1st can = Rs.43

.

Milk in 1 litre mixture in 2nd can = 21

litre,

C.P. of 1 litre mixture in 2nd can = Rs.21

.

Milk in 1 litre of final mixture =85

litre,

Mean price = Rs. 85 .

By the rule of alligation, we have :

yx

= 2/18/58/53/4

yx

= 8/18/1

Mean price

C.P. of 1 litre mixture in 1st can C.P. of 1 litre mixture in 2nd can

1/81/8

3/4 1/2

5/8

We will mix 6 from each can.

3PAGE # 3

Ex.9 Tea worth Rs.126 per kg and Rs.135 per kg are mixedwith a third variety in the ratio 1 : 1 : 2. If the mixture isworth Rs. 153 per kg, then find the price of the thirdvariety per kg.

Sol. Since first and second varieties are mixed in equal

proportions, so their average price =

2135126

.Rs

= Rs.130.50So, the mixture is formed by mixing two varieties, oneat Rs. 130.50 per kg and the other at say, Rs. x per kgin the ratio 2 : 2, i.e., 1 : 1. We have to find x.By the rule of alligation, we have :

Rs. 153Mean price

Cost of 1 kg tea of 1st kind Cost of 1 kg tea of 2nd kind

22.50(x � 153)

130.50 Rs. x

1 = 5.22

153x

153 + 22.5 = xx = Rs.175.50

Ex.10 A jar full of whisky contains 40% alcohol . A part of thiswhisky is replaced by another containing 19% alcoholand now the percentage of alcohol was found to be26%. Find the quantity of whisky replaced.

Sol. By the rule of alligation, we have :

26%Mean strength

Strength of first jar Strength of 2nd jar

147

40% 19%

So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2.

Required quantity replaced = 32

.

Ex.11 A vessel is filled with liquid, 3 parts of which are waterand 5 parts syrup. How much of the mixture must bedrown off and replaced with water so that the mixturemay be half water and half syrup ?

Sol. Suppose the vessel initially contains 8 litres of liquid.Let x litres of this liquid be replaced with water.

Quantity of water in new mixture =

x

8x3

�3 litres.

Quantity of syrup in new mixture =

8x5

�5 litres.

x

8x3

�3 =

8x5

�5

5x + 24 = 40 � 5x

10x = 16 x = 58

.

So, part of the mixture replaced =

81

58

= 51

.

When two or more persons jointly start a business withan objective to earn money. This is called partnership.These persons are called partners and the moneyinvested in the business is known as capital.

Distribution of Profit/Loss when unequal capital isinvested for equal interval of time :When partners invest different amounts of money, forequal interval of time, then profit/loss is divided in theratio of their investment.

Ex.12 A and B invested Rs. 3600 and Rs. 4800 respectivelyto open a shop. At the end of the year B�s profit was

Rs. 1208. Find A�s profit.

Sol. Profit sharing ratio = 3600 : 4800 = 3 : 4

43

BofofitPrAofofitPr

Profit of A = 43

Profit of B

Profit of A = 43

× 1208 = Rs. 906

Distribution of P/L when equal capital is invested fordifferent intervals of time :

Ex.13 Govind & Murari started a business with equalcapitals. Govind terminated the partnership after7 months. At the end of the year, they earned a profit ofRs. 7600. Find the profit of each of them.

Sol. Govind invested for 7 month, Murari invested for12 month.Since investment is same for both (Let it be Rs. x) Profit sharing ratio = 7x : 12x = 7 : 12

Govind�s profit = 127

7

× 7600 = 2800

Murari�s profit = 127

12

× 7600 = 4800.

Ex.14 Ramesh started a business by investing Rs. 25000.3 months later Mahesh joined the business byinvesting Rs. 25000. At the end of the year Ramesh gotRs. 1000 more than Mahesh out of the profit. Find thetotal profit.

Sol. Ramesh invested for 12 month, Mahesh invested for 9month. Profit sharing ratio = 12x : 9x = 12 : 9 = 4 : 3.Let Capital be Rs P.

Profit of Ramesh = P74

Profit of Ramesh = P73

P74

= P73

+ 1000

P74 � P

73 = 1000

7P

= 1000 P = Rs.7000.

4PAGE # 4

Distribution of P/L when capital and time both are

unequal :

Ex.15 Suresh & Ramesh entered into a partnership by

investing Rs.14000 and Rs. 18000 respectively suresh

with drew his money after 4 months. If the total profit at

the end is Rs. 12240, find the profit of each.

Sol. Profit sharing ratio = 14000 × 4 : 18000 × 12 = 7 : 27

Suresh�s profit = 347

× 12240 = 2520

Ramesh�s profit = 3427

× 12240 = 9720.

Ex.16 David started a business establishment by investing

Rs.15000. After 4 months william entered into a

partnership by investing a certain amount. At the end

of the year; the profit was shared in the ratio 9 : 8. Find

how much money was invested by william.

Sol. Let William invested Rs. x.

Profit sharing ratio = 15000 × 12 : 8x = 1,80,000 : 8x

Also profit ratio = 9 : 8

ATP 180000 : 8x = 9 : 8

89

8180000

= x

x = Rs. 20,000

Working and Sleeping partner :

Active Partner : A partner who manages the business

is known as active or working partner.

Sleeping Partner : A partner who only invests the money

is known as sleeping partner.

Ex.17 Nitesh & Jitesh invested Rs.15000 and Rs.18000

respectively in a business. If the total profit at the end

of the year is Rs. 8800 and Nitesh, being an active

partner, gets an additional 12.5% of the profit, find the

total profit of Nitesh.


Total profit = 8800

Nitesh gets 12.5% of the profit = 1000

5.12× 8800

= Rs. 1100

Net profit = 8800 � 1100 = Rs. 7700

Nitesh share in profit = 65

5

× 7700 = 3500

Total profit of Nitesh = 3500 + 1100 = Rs. 4600.

Ex.18 Tanoj & Manoj started a business by investing

Rs. 75000 and Rs. 90000 respectively. It was decided

to pay Tanoj a monthly salary of Rs. 1875 as he was

the active partner. At the end of the year if the total profit

is Rs. 39000, find the profit of each.


Total profit = Rs. 39000

Salary of Tanoj = 12 × 1875 = Rs. 22500

Profit left = Rs.39000 � Rs. 22500 = Rs.16500.

Tanoj�s profit = 115

× 16500 = 7500.

Total profit of Tanoj = 22500 + 7,500 = Rs. 30,000

Manoj�s profit = 116

× 16500 = Rs. 9,000

Change in invested capital :

Ex.19 Rajeev & Sanjeev entered into a partnership and

invested Rs. 36000 and Rs. 40000 respectively. After 8

months Rajeev invested an additional capital of

Rs. 4000, Sanjeev withdrew Rs. 4000 after 9 months.

At the end of the year of their profit was Rs. 45800. Find

the profit of each.

Sol. Rajeev�s capital = 36000 × 8 + (36000 + 4000) × 4

= Rs. 448000 for 1 month

Sanjeev�s capital = 40000 × 9 + (40000 � 4000) × 3

= Rs. 468000 for 1 month

Profit sharing ratio = 448000 : 468000 = 112 : 117

Rajeev�s profit = 229112

× 45800 = Rs. 22400

Sanjeev�s profit = 229117

× 45800 = Rs. 23400.

Ex.20 A, B and C start a business each investing Rs. 20000.

After 5 months A withdrew Rs. 5000, B withdrew Rs.

4000 and C invests Rs. 6000 more. At the end of the

year, a total profit of Rs. 69900 was recorded. Find the

share of each.

Sol. Ratio of the capitals of A, B and C

= 20000 × 5 + 15000 × 7 : 20000 x 5 + 16000 × 7 : 20000

× 5 + 26000 × 7

= 205000 : 212000 : 282000 = 205 : 212 : 282.

A�s share = Rs. (69900 × 699205

) = Rs. 20500 ;

B's share = Rs. (69900 × 699215

) = Rs. 21200 ;

C�s share = Rs. (69900 x 699282

) = Rs. 28200.

5PAGE # 5

A statement involving variable (s) and the sign of inequality

viz, >,< or is called an inequation.

An inequation may contain one or more variables.

Also, it may be linear or quadratic or cubic etc.

(i) 3x � 2 < 0 (ii) 2x2 + 3x + 4 > 0 (iii) 2x + 5y 4

(a) Properties of inequalities

(i) If �a� is a positive no. i.e. a > 0 then for x < y

ay

ax & ax < ay..

(ii) If �a� is �ve i.e. a < 0 then for x < y

ay

ax & ax > ay..

(iii) If �a� is a +ve no. i.e. a > 0 then for x > y

ay

ax & ax > ay..

(iv) If �a� is a �ve no. i.e. a < 0 then for x > y

ay

ax & ax < ay..

(i) Closed interval : Let a and b be two given real

numbers such that a < b. Then the set of all real

numbers x such that a x b is called closed interval

and is denoted by [a, b] may be graphed as :

a b

(ii) Open interval : If a and b are two real numbers

such that a < b, then the set of all real numbers

x satisfying a < x < b is called an open interval and is

denoted by (a, b) or ]a, b[ and may be graphed as :

a b

(iii) Semi-closed or semi-open interval : If a and b are

two real numbers such that a < b, then the sets

(a, b] = {xR : a < x b} and [a, b) = {xR : a x < b} are

known as semi-open or semi-closed intervals. (a, b]

and [a, b) are also denoted by ]a, b] and [a, b[

respectively.

INEQUATIONS

Let a be a non-zero real number and x be a variable.

Then inequations of the form ax + b < 0, ax + b 0,

ax + b > 0 and ax + b 0 are known as linear inequations

in one variable x.

For example, 9x � 15 > 0, 5x � 4 0, 3x + 2 < 0 and

2x � 3 0 are linear inequations in one variable.

(a) Solving linear inequations in one variable

Rule 1: Same number may be added to (or subtracted

from) both side of an inequation without changing the

sign of inequality.

Rule 2 : Both sides of an inequation can be multiplied

(or divided) by the same positive real number without

changing the sign of inequality. However, the sign of

inequality is reversed when both sides of an inequation

are multiplied or divided by a negative number.

Rule 3 : Any term of an inequation may be taken to the

other side with its sign changed without affecting the

sign of inequality.

Ex. 1 Solve the inequality ax > a.

Sol. This inequality has the parameter a that needs to be

investigated further.

If a > 0, then x > 1

If a < 0, then x < 0

Ex.2 Solve : 24 x < 100 when

(i) x is a natural number (ii) x is an integer.

Sol. We are given 24 x < 100

24

x24<

24100

x < 6

25

(i) When x is a natural number, the following values of

x make the statement true. 1, 2, 3, 4.

(ii) When x is an integer, the solutions of the given

equations are ....,� 3, � 2, � 1, 0, 1, 2, 3, 4.

The solution set of the equation is :

{ ...,� 3,� 2,� 1, 0,1, 2, 3, 4}.

Ex.3 Solve & graph the solution set of 3x + 6 9

and � 5x > �15, xR.

Sol. 3x + 6 9 and �5x > �15

3x 3 � x > � 3

x 1 x < 3

Combining the solution

�3 �2 �1 0 1 2 3

So, the solution is x [1,3).


6PAGE # 6

Ex.4 Solve & graph the solution set of � 2 < 2x � 4

and �2x + 5 13, xR

Sol.�4 �3 �2 �1 1 2 30

� 2 < 2x � 4

2x � 4 > �2 and �2x + 5 13

2x > 2 and �2x 13 � 5

x > 1 and �2x 8

x > 1 and �x 4

x > 1 and x �4

x > 1 or x � 4 or x (� , � 4] (1, ).

Ex.5 Solve the following equation :

2(2x + 3) � 10 < 6 (x � 2)

Sol. We have,

2(2x + 3) � 10 6 (x � 2)

4x + 6 � 10 6x � 12

4x � 4 6x � 12

4x � 6x � 12 + 4

[Transposing � 4 to RHS and 6x to LHS]

� 2x � 8

2�x2�

2�8�

x 4

x [4, )

Hence, the solution set of the given inequation is [4, )

which can be graphed on real line as shown in Figure.

0 4

Ex.6 Solve the following inequations : 4x

53�x7

�3

2�x5 .

Sol.4x

53�x7

�3

2�x5

15)3�x7(3�)2�x5(5

> 4x

159x21�10�x25

> 4x

151�x4

> 4x

4 (4x � 1) > 15 x

[Multiplying both sides by 60 i.e. Lcm of 15 and 4]

16x � 4 > 15x

16x � 15x > 4

[Transposing 15 x to LHS and � 4 to RHS]

x > 4

x (4, )

Hence, the solution set of the given inequation is

(4, ). This can be graphed on the real number line as

shown in figure.

0 4

Ex.7 Solve the following inequations :1�x4x2

> 5.

Sol. We have,

1�x4x2

5

1�x4x2

� 5 0

1�x

)1�x(5�4x2 0

1�x

5x5�4x2 0

1�x9x3�

0 [Multiplying both sides by � 1]

1�x9�x3

0

1�x

)3�x(3 0 [Dividing both sides by 3]

1�x3�x

0

1 < x 3

x (1, 3]

+ 1 3

� +

Hence, the solution set of the given inequations is

(1, 3].

Ex. 8 Solve : � 5 4

x3�2 < 9.

Sol. � 5 × 4 4

x3�2 × 4 9 × 4

[Multiplying throughout by 4]

� 20 2 � 3x 36

� 20 � 2 � 3x 36 � 2 [Subtracting 2 throughout]

� 22 � 3x 34

3�22�

3�x3�

3�

34 [Dividing throughout by � 3]

322

x 334�

334�

x 3

22

x [� 34/3, 22/3]

Hence, the interval [� 34/3, 22/3] is the solution set of

the given system of inequations.

7PAGE # 7

Ex.9 Solve 5x - 3 < 3x + 1 when

(i) x is an integer, (ii) x is a real number.

Sol. We have, 5x � 3 < 3x + 1

5x � 3 + 3 < 3x + 1 + 3

5x < 3x + 4

5x � 3 x < 3x + 4 � 3x

2x < 4

x < 2

(i) When x is an integer, the solutions of the given

inequality are ..........., � 4, �3, �2, �1, 0,1.

(ii) When x is a real number, the solutions of the

inequality are given by x < 2, i.e., all real number x

which are less than 2. Therefore, the solution set of

the inequality is x (- , 2).

The function f(x) defined by

f(x) =

0xwhen,x�0xwhen,xx

is called the modulus function. It is also called the

absolute value function.

x' x

y

y'

o

f(x) = �x f(x) =

x

The distance between two real numbers x and y is

defined as y�x .

Ex.10 Solve : 4�x = 7

Sol.

4x04�xwhere),4�x(�4x04�xwhere,4�x4�x

x � 4 = 7 and � (x � 4) = 7

x = 11 and � x + 4 = 7

� x = 3

x = � 3 Ans.

11x3�x

Ex. 11 Evaluate 7��3�3�2�3

Sol. 7��3�3�2�3

= 3 + 5 � 3 � {�(�7)}

= 3 + {�(�5)} � 3 � 7

= 3 + 5 � 10 = 8 � 10 = �2.

(a) Inequations involving absolute value

Result 1. If a is a positive real number, then

(i) | x | < a � a < x < a i.e. x (� a, a)

�a a

(ii) | x | a � a x a i.e. x [�a, a]

�a a

Result 2. If a is a positive real number, then

(i) | x | > a x < � a or x > a

�a a

(ii) | x | a x � a or x a

�a a

Result 3. Let r be a positive real number and a be a

fixed real number. Then,

(i) | x � a | < r a � r < x < a + r i.e. x (a � r, a + r)

(ii) | x � a | r a � r x a + r i.e. x [a � r, a + r]

(iii) | x � a | > r x < a � r, or x > a + r

(iv) | x � a | r x a � r, or x a + r

Ex.12 Find x from 2x1 and represent it on number line.

Sol. 1xx1 x > 1 or x < �1

x ),1()1,( ...(i)

also 2x x < 2 or x > � 2

x lies between � 2 & 2

)2,2(�x ...(ii)

Combining the two results, we get

2x1 {� 2 < x < �1} {1 < x < 2}

i.e. x (� 2, �1) (1, 2)

�2 �1 1 20

Ex.13 Find x satisfying 35�x .

Sol. as ]ra,r�a[x.e.iraxr�ara�x

]8,2[x.e.i8x2.e.i35x3�535�x

2 8

8PAGE # 8

QUADRATIC EQUATION & PROGRESSION

If P(x) is quadratic expression in variable x, thenP(x) = 0 is known as a quadratic equation.

General form of a Quadratic Equation :

The general form of a quadratic equation isax2 + bx + c = 0, where a, b, c are real numbers and a 0.

The value of x which satisfies the given quadraticequation is known as its root. The roots of the givenequation are known as its solution.For quadratic equation ax2 + bx + c = 0, the roots are

a2ac4bb 2

and

a2ac4bb 2

.

Consider the quadratic equation, a x2 + b

x + c = 0

having and as its roots and b2 4ac is calleddiscriminant of roots of quadratic equation. It isdenoted by D or .

Roots of the given quadratic equation may be(i) Real and unequal (ii) Real and equal(iii) Imaginary and unequal.Let the roots of the quadratic equation ax2 + bx + c = 0(where a 0, b, c R) be and then

= a2

ac4bb 2

... (i)

and = a2

ac4bb 2

... (ii)

The nature of roots depends upon the value ofexpression �b2 � 4ac� with in the square root sign. Thisis known as discriminant of the given quadraticequation.

Consider the Following Cases :

Case-1 When b2 � 4ac > 0, (D > 0)

In this case roots of the given equation are real anddistinct and are as follows

= a2

ac4bb 2

and = a2

ac4bb 2

(i) When a( 0), b, c Q and b2 � 4ac is a perfect

squareIn this case both the roots are rational and distinct.

(ii) When a( 0), b, c Q and b2 � 4ac is not a perfect

square

In this case both the roots are irrational and distinct. [See remarks also]

Case-2 When b2 � 4ac = 0, (D = 0)

In this case both the roots are real and equal to � a2

b.

Case-3 When b2 � 4ac < 0, (D < 0)

In this case b2 � 4ac < 0, then 4ac � b2 > 0

= a2

)bac4(b 2

and = a2

)bac4(b 2

or = a2

bac4ib 2

and = a2

bac4ib 2

[ 1 = i ]

i.e. in this case both the roots are imaginary and distinct.

REMARKS :

If a, b, c Q and b2 � 4ac is positive (D > 0) but not a

perfect square, then the roots are irrational and they

always occur in conjugate pairs like 2 + 3

and 2 � 3 . However, if a, b, c are irrational numbers

and b2 � 4ac is positive but not a perfect square, then

the roots may not occur in conjugate pairs.

If b2 � 4ac is negative (D < 0), then the roots are complex

conjugate of each other. In fact, complex roots of an

equation with real coefficients always occur in conjugate

pairs like 2 + 3i and 2 � 3i. However, this may not be true

in case of equations with complex coefficients.

For example, x2 � 2ix � 1 = 0 has both roots equal to i.

If a and c are of the same sign and b has a sign opposite

to that of a as well as c, then both the roots are positive,

the sum as well as the product of roots is positive

(D 0).

If a, b, c are of the same sign then both the roots are

negative, the sum of the roots is negative but the product

of roots is positive (D 0).

(a) Relation Between Roots & Coefficients :

(i) The solutions of quadratic equation a x2 + b

x + c = 0

are given by

x = a2

ca4bb 2

(ii) The expression b2 4 a c D is called discriminant

of the quadratic equation a x2 + b

x + c = 0.

If , are the roots of the quadratic equation a x2 + b

x + c = 0, then


9PAGE # 9

(a) Sum of the roots = � 2xoftcoefficien

xoftcoefficien

+ = ab

(b) Product of the roots = 2xoftcoefficien

termttancons

= ac

(iii) A quadratic equation whose roots are and is(x ) (x ) = 0i.e. x2 (sum of roots) x + (product of roots) = 0.

(b) Symmetric functions of roots of a quadraticequation

(i) � = ± 4)( 2

(ii) 2 � 2 = ± ( + ) 4)( 2

(iii) 2 + 2 = ( + )2 � 2

(iv) 3 + 3 = ( + )3 � 3(+)(v) 3 � 3 = ( � )3 + 3(�)

= ± 23

2 4)( ± 3 4)( 2

(vi) 4 + 4 = (2 + 2)2 � 222= {(+)2 �2 }2 � 222

(vii) 4 � 4 = ( + ) ( � )(2 + 2)

= ± (+) 4)( 2{(+)2 �2 }

(a) By Factorisation :

ALGORITHM :

Step (i) Factorise the constant term of the given

quadratic equation.

Step (ii) Express the coefficient of middle term as the

sum or difference of the factors obtained in step 1.

Clearly, the product of these two factors will be equal to

the product of the coefficient of x2 and constant term.

Step (iii) Split the middle term in two parts obtained in

step 2.

Step (iv) Factorise the quadratic equation obtained in

step 3.

(b) By the Method of Completion of Square :

ALGORITHM :

Step-(i) Obtain the quadratic equation. Let the quadratic

equation be ax2 + bx + c = 0, a 0.

Step-(ii) Make the coefficient of x2 unity, if it is not unity.

i.e., obtain x2 + ab

x + ac

= 0.

Step-(iii) Shift the constant term ac

on R.H.S. to get

x2 + ab

x = � ac

Step-(iv) Add square of half of the coefficient of x.i.e.2

a2

b

on both sides to obtain

x2 + 2

a2b

x + 2

a2b

=

2

a2b

�

ac

Step-(v) Write L.H.S. as the perfect square of a binomial

expression and simplify R.H.S. to get2

a2

bx

= 2

2

a4

ac4b .

Step-(vi) Take square root of both sides to get

x + a2

b = ± 2

2

a4

ac4b

Step (vii) Obtain the values of x by shifting the constant

term a2

b on RHS.

(c) By Using Quadratic Formula :

Solve the quadratic equation in general form viz. ax2 +bx + c = 0.

Step (i) By comparison with general quadratic equation,find the value of a, b and c.

Step (ii) Find the discriminant of the quadratic equation.D = b2 � 4ac

Step (iii) Now find the roots of the equation by givenequation

x = a2

Db ,

a2Db

REMARK :

If b2 � 4ac < 0, i.e., negative, then ac4�b2 is not real

and therefore, the equation does not have any real roots.

Common Roots :

Consider two quadratic equations, a1 x2 + b

1 x + c

1 = 0

& a2 x2 + b

2 x + c

2 = 0.

(i) If two quadratic equations have both rootscommon, then the equations are identical and theircoefficients are in proportion. i.e.

2

1

aa

=2

1

bb

=2

1

cc

.

(ii) If only one root is common, then the commonroot '

' will be :

=1221

1221

babaacac

=

1221

1221

acaccbcb

Hence the condition for one common root is:

21221 acac = 1221 baba 1221 cbcb

10PAGE # 10

SEQUENCE

A sequence is an arrangement of numbers in adefinite order according to some rule.e.g. (i) 2, 5, 8, 11, ... (ii) 4, 1, � 2, � 5, ...

(iii) 3, �9, 27, � 81, ...

Types of Sequence

On the basis of the number of terms there are twotypes of sequence :

(i) Finite sequences : A sequence is said to befinite if it has finite number of terms.

(ii) Infinite sequences : A sequence is said to beinfinite if it has infinite number of terms.

PROGRESSIONS

Those sequence whose terms follow certain patternsare called progressions. Generally there are three typesof progressions.

(i) Arithmetic Progression (A.P.)(ii) Geometric Progression (G.P.)(iii) Harmonic Progression (H.P.)

ARITHMETIC PROGRESSION

A sequence is called an A.P., if the difference of a termand the previous term is always same.i.e. d = tn + 1 � tn = Constant for all n N. The constantdifference, generally denoted by �d� is called thecommon difference.

GENERAL FORM OF AN A.P.

If we denote the starting number i.e. the 1st number by�a� and a fixed number to be added is �d� thena, a + d, a + 2d, a + 3d, a + 4d,........... forms an A.P.

n FORM OF AN A.P. th

Let A.P. be a, a + d, a + 2d, a + 3d,...........Then, First term (a1) = a + 0.d

Second term (a2) = a + 1.dThird term (a3) = a + 2.d. .. .. .nth term (an) = a + (n � 1) d

an = a + (n � 1) d is called the nth term.

mth

TERM OF AN A.P. FROM THE END

Let �a� be the 1st term and �d� be the common difference

of an A.P. having n terms. Then mth term from the endis (n � m + 1)th term from beginning or {n � (m � 1)}th

term from beginning.

SELECTION OF TERMS IN AN A.P.

Sometimes we require certain number of terms in A.P.The following ways of selecting terms are generallyvery convenient.

No. of Terms Terms Common Difference

For 3 terms a � d, a, a + d d

For 4 terms a � 3d, a � d, a + d, a + 3d 2d

For 5 terms a � 2d, a � d, a, a + d, a + 2d d

For 6 terms a � 5d, a � 3d, a � d, a + d, a + 3d, a + 5d 2d

SUM OF n TERMS OF AN A.P.

Let A.P. be a, a + d, a + 2d, a + 3d,............., a + (n � 1)d

Then, S

n = a + (a + d) +...+ {a + (n � 2) d} + {a + (n � 1) d} ..(i)

alsoS

n= {a + (n � 1) d} + {a + (n � 2) d} +....+ (a + d) + a ..(ii)

Add (i) & (ii)

2Sn = 2a + (n � 1)d + 2a + (n � 1)d +....+ 2a + (n � 1)d

2Sn = n [2a + (n � 1) d]

1) d( n2a2

nSn

Sn =

2n

[a + a + (n � 1)d] = 2n

[a + ]

a2

nSn where, is the last term.

rth term of an A.P. when sum of first r terms is Sr isgiven by, tr = Sr � Sr � 1

.

PROPERTIES OF A.P.

(i) For any real number a and b, the sequence whose

nth term is an = an + b is always an A.P. with common

difference �a� (i.e. coefficient of term containing n).

(ii) If a constant term is added to or subtracted from

each term of an A.P. then the resulting sequence is

also an A.P. with the same common difference.

(iii) If each term of a given A.P. is multiplied or divided

by a non-zero constant K, then the resulting

sequence is also an A.P. with common difference

Kd or K

d respectively. Where d is the common

difference of the given A.P.

(iv) In a finite A.P. the sum of the terms equidistant

from the beginning and end is always same and is

equal to the sum of 1st and last term.

(v) If three numbers a, b, c are in A.P. , then 2b = a + c.

Arithmetic Mean (Mean or Average) (A.M.)

If three terms are in A.P. then the middle term is

called the A.M. between the other two, so if a, b, c

are in A.P., b is A.M. of a & c.

A.M. for any n number a1, a2,..., an is;

A =n

a.....aaa n321 .

11PAGE # 11

n - Arit hmetic Means Between Two

Numbe rs :

If a, b are any two given numbers & a, A1, A2,...., An, b arein A.P. then A1, A2,... An are the n-A.M.�s between a & b.

Total terms are n + 2.

Last term b = a + (n+2�1)d.Now, d = 1nab

.

A1 = a +1nab

, AA2 = a +

1n)ab(2

, . . . . . . . . . . . . ,

An = a + 1n

)ab(n

.

NOTE :Sum of all n-A.M.�s inserted between a & b is equal

to n times the single A.M. between a & b.

i.e. r

n

1

Ar = nA where A is the single A.M. between a & b.

G.P. is a sequence of numbers whose first term isnon zero & each of the succeeding terms is equalto the preceding term multiplied by a constant. Thusin a G.P. the ratio of successive terms is constant.This constant factor is called the common ratio ofthe series & is obtained by dividing any term by thatwhich immediately precedes it.

Example : 2, 4, 8, 16 ... & 31

,91

,271

,811

...are in G.P..P.

(i) Therefore a, ar, ar2, ar3, ar4,...... is a G.P. with �a� as

the first term and �r� as common ratio.

nth term = a rn1

(ii) Sum of the first n terms.

Sn =

1r,r1r1a

1r,1r1ra

n

n

(iii) Sum of an infinite G.P. when r < 1. When

n rn 0 if r < 1 therefore,

S = )1|r(|r�1

a .

(i) If a, b, c are in G.P. then b2 = ac, in general if

a1, a

2, a

3, a

4,......... a

n � 1 , a

n are in G.P.,

then a1a

n = a

2a

n � 1 = a

3 a

n � 2 = ..............

(ii) Any three consecutive terms of a G.P. can be

taken as ra

, a , ar..

(iii) Any four consecutive terms of a G.P. can be taken

as 3r

a,

ra

, ar, ar3.

(iv) If each term of a G.P. be multiplied or divided or

raised to power by the some nonzero quantity, the

resulting sequence is also a G.P.

If a, b, c are in G.P., b is the G.M. between a & c.

b² = ac, therefore b = ac ; a > 0, c > 0.

n-Geometric Means Between a, b :

If a, b are two given numbers & a, G1, G2,....., Gn, bare in G.P.. Then,G1, G2, G3,...., Gn are n-G.M.s between a & b.

G1 = a(b/a)1/n+1, G2 = a(b/a)2/n+1,......, Gn = a(b/a)n/n+1

NOTE :The product of n G.M.s between a & b is equal to thenth power of the single G.M. between a & b

i.e.

n

1r

Gr = nab = Gn where G is the single G.M.

between a & b.

A sequence is said to be H.P. if the reciprocals of itsterms are in A.P. If the sequence a1, a2, a3,...., an isan H.P. then 1/a1, 1/a2,...., 1/an is an A.P. Here we donot have the formula for the sum of the n terms of aH.P. For H.P. whose first term is a and second termis b, the nth term is tn

tn = )ba)(1n(bab

.

If a, b, c are in H.P. b =ca

ac2

or ca

=cbba

.

NOTE :

(i) If a, b, c are in A.P. cbba

=

aa

(ii) If a, b, c are in G.P. cbba

=

ba

If a, b, c are in H.P., b is the H.M. between a & c, then

b =ca

ac2

.

If a1, a2 , ........ an are �n� non-zero numbers then H.M.

H of these numbers is given by :

H1

= n1

n21 a1

.......a1

a1

.

12PAGE # 12

Relation between means :

If A, G, H are respectively A.M., G.M., H.M. betweena & b both being unequal & positive then, G² = AH

i.e. A, G, H are in G.P. A.M. G.M. H.M.

Let a1, a

2, a

3, .......a

n be n positive real numbers, then

we define their

A.M. = n

a.......aaa n321 ,

G.M. = (a1 a

2 a

3 .........a

n)1/n and

H.M. =

n21 a1

.......a1

a1

n

.

It can be shown that A.M. G.M. H.M. and equalityholds at either places iff a

1 = a

2 = a

3 = ..............= a

n.

13PAGE # 13

A well defined collection of objects is known as sets.

If a is an element of a set A, then we write a A and

say a belongs to A. If a does not belong to A, then a Ais written.

For example : The collection of all states in the Indianunion is a set but collection of good cricket players ofIndia is not a set, since the term �good player is vague

and it is not well defined.

Some letters are reserved for the sets as listedbelow :

N : For the set of Natural numbers.

Z : For the set of Integers.

Z+ :For the set of all positive Integers.

Q : For the set of all Rational numbers.

Q+ : For the set of all positive Rational numbers.

R : For the set of all Real numbers.

R+ : For the set of all Positive real numbers.

C : For the set of all Complex numbers.

A set is often described in the following two ways :

(a) Roster Method :

In this method a set is described by listing elements,separated by commas, within braces { }.

Ex.1 Write the set of vowels of English alphabet in rosterform.

Sol. A = {a, e, i, o, u}.

Ex.2 Write the set of even natural numbers in roster form.Sol. B = {2,4,6,.....}.

Ex.3 Write the set of all prime numbers less than 11 inroster form.

Sol. C = {2,3,5,7}.

Ex.4 Write the set A = {x z, x2 < 20} in the roster from.Sol. We observe that the squares of integers 0, ± 1, ± 2, ± 3,

± 4 are less than 20. Therefore, the set A in roster form

i s A = { � 4, � 3, � 2, �1, 0, 1, 2, 3, 4}.

� NOTE :The order in which the element are written in a setmakes no difference.(b) Set Builder Method :

In this method, a set is described by a characterizingproperty P(x) of its elements x. In such a case the setis described by {x : P(x) holds } or, {x | P(x) holds,}which is read as �the set of all x such that P(x) holds�.The symbol �|� or �:� is read as �such that�.

SETS

Ex. 5 Write the set A = {0,1,4,9,16,........} in set builder form.Sol. A = {x2 : x Z).

Ex.6 Write the set X = {1, 4

1,

9

1,

16

1,

25

1, ........} in the set

builder form.Sol. We observe that the elements of set X are the

reciprocals of the squares of all natural numbers. So,

the set X in set builder form is X =

Nn ; 2n

1.

(a) Empty Set :

A set is said to be empty or null or void set if it has noelement and it is denoted by or { }.

Ex. 7 Write {x N : 5 < x < 6} in roster form.Sol. A = { }.

(b) Singleton Set :

A set consisting of a single element is called asingleton set.

Ex. 8 Write the set {x : x N and x2 = 9} in roster form.Sol. Let B is the set. So B is a singleton set equal to {3}.

(c) Finite Set :

A set is called a finite set if it is either void set or itselement can be listed (counted labelled) by naturalnumbers 1, 2, 3 ....... and the process of listingterminates at a certain natural number n (say).For example : Set of all persons on the earth is a finiteset.

(d) Infinite Set :

A set whose elements cannot be listed by naturalnumbers 1, 2, 3,...... for any natural number n is calledan infinite set.For example : Set of all points in a plane is an infiniteset.

Ex. 9 Which of the following sets are finite and which areinfinite ?(a) Set of concentric circle in a plane.(b) Set of letters of English alphabets.

(c) { xN, x > 5 }

(d) { xR, 0 < x < 1 }

(e) { xN, x < 200}Sol. (a) Infinite set (b) Finite set (c) Infinite set

(d) Infinite set (e) Finite set

(e) Cardinal Number of a Finite Set :

The number n in the above definition is called thecardinal number or order of a finite set A and is denotedby n(A).


14PAGE # 14

(f) Equivalents Set :

Two finite sets A and B are equivalent if their cardinalnumbers are same. i.e. n(A) = n(B).For example : A = {1,2,3} and B = {a,b,c} are equivalentsets.

(g) Equal Set :

Two sets A and B are said to be equal if every elementof A is a member of B, and every element of B ismember of A.

NOTE :Equal sets are equivalents but equivalent sets neednot be equal.

Ex. 10 Are the following sets equal ?A = { x : x is a letter in the word reap }B = { x : x is a letter in the word paper }.

Sol. A = { r, e, a, p}B = { p, a, e, r }So, A and B are equal sets.

(h) Subset :

Let A and B be two sets. If every element of A is anelement of B, then A is called a subset of B. If A is a

subset of B, we write A B, which is read as �A is asubset of B� or �A is contained in B�. Thus,

A B if a A a B. The symbol �� stands for�implies�. If A is not a subset of B, we write

A B.

NOTE :Every set is a subset of itself and the empty set issubset of every set. These two subsets are calledimproper subsets. A subsets A of a set B is called a

proper subset of B if A B and we write A B.

SOME RESULTS ON SUBSET :(i) Every set is a subset of itself(ii) The empty set is a subset of every set.(iii) The total number of subsets of a finite setcontaining n element is 2n.

(i) Universal Set :

A set that contains all sets in a given context is calledthe Universal Set.

Ex.11 If A = {1,2,3}, B = {2,4,5,6} and C = {1,3,5,7}, then findthe universal set.

Sol. U = {1, 2, 3, 4, 5, 6, 7} can be taken as the universalset.

( j) Power Set :

Let A be a set. Then the collection or family of allsubsets of A is called the power set of A and is denotedby P(A).

Ex.12 Let A = {1,2,3}. Then find the power set of A.Sol. Subset of A are : , {1}, {2}, {3}, {1,2}, {1,3}, {2,3} and

{1,2,3}.Hence, total number of subset are = 23 = 8.Hence, P(A) ={ {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }.

Ex.13 Let A = {1,2,3,4}, B = {1,2,3} and C = {2,4}. Find sets Xsatisfying each pair of conditions.

(i) X B and X C

(ii) X B, X B and X C

(iii) X A, X B and X CSol. We have

(i) X B and X C X is subset of B but X is not a subset of C

X P (B) but X P (C) X = {1}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}.

(ii) We have,

X B, X B and X C X is a subset of B other than B itself and X is not asubset of C

X P(B), X P(C) and X B X = {1}, {3}, {1,2}, {1,3}, {2,3}.

(iii) We have,

X A, X B and X C

X P(A), X P(B) and X X (C) X is a subset of A, B and C X = , {2}

Diagram drawn to represent sets are called Venn-Eulerdiagram or simply Venn diagram. In Venn-diagramthe universal set U is represented by points within arectangle and its subsets are represented by points inclosed curves (usually circles) within the rectangle.

(a) Union of Sets :

Let A and B be two sets. The union of A and B is the setof all those elements which belong either to A or to Bor to both A and B.

Thus, A B = { x : x A or x B}.

Ex.14 If A {1, 2, 3} and B = {1, 3, 5, 7}, then find A B.Sol. A B = {1, 2, 3, 5, 7}.

(b) Intersection of Sets :

Let A and B be two sets. The intersection of A and B isthe set of all those elements that belong to both A and B.

Thus, A B = {x : x A and x B}.

15PAGE # 15

Ex.15 If A = { 1, 2, 3, 4, 5 } and B = { 1, 3, 9, 12 }, thenfind A

Sol. A = { 1, 3 }.

(c) Disjoint Sets :

Two sets A and B are said to be disjoint, if A B = .If A B , then A and B are said to be intersecting oroverlapping sets.(d) Difference of Sets :

Let A and B be two sets. The difference of A and B,written as A � B, is the set of all those elements ofA which do not belong to B.

u

A�B

A B

Thus, A � B = {x : x A and x B}or, A � B = {x A : x B}.Clearly, x A � B x A and x B.Similarly, the difference B � A is the set of al lthose elements of B that do not belong to Ai .e. B � A = {x B : x A}.

u

A B

Ex.16 If A = { 2, 3, 4, 5, 6, 7 } and B = { 3, 5, 7, 9, 11, 13 }, thenfind A � B and B � A.

Sol. A � B = { 2, 4, 6 } and B � A = { 9, 11, 13 }.

(e) Symmetric Difference of Two Sets :Let A and B be two sets. The symmetric difference ofsets A and B is the set (A � B) (B � A) and is denotedby A B.

Ex.17 If A = { x R : 0 < x < 3 }, B = { x R : 1 x 5 }, then

find A B.Sol. A � B = { x R : 0 < x < 1 }, B � A = { x R : 3 x 5 }

A B = { x R : 0 < x < 1 } { x R : 3 x 5 }

A B = { x R : 0 < x < 1 or 3 x 5 }.

(f) Complement of a Set :

Let U be the universal set and let A be a set such thatA U. Then, the complement of A with respect to U isdenoted by A� or Ac or U� A and is defined the set of allthose elements of U which are not in A.Thus, A� = {x U : x A}.

Clearly, x A� x A.

Ex.18 Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4 },B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find(i) Ac (ii) ( A C )c (iii) ( A B )c (iv) (B � C)c

Sol. (i) Ac = U � A = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } � { 1, 2, 3, 4 }

= { 5, 6, 7, 8, 9 }.(ii) ( A C )c = U � ( A C )

= { 1, 2, 3, 4, 5, 6, 7, 8, 9 } � { 3, 4 }

= { 1, 2, 5, 6, 7, 8, 9 }.(iii) ( A B )c = U � ( A B )= { 1, 2, 3, 4, 5, 6, 7, 8, 9 } � { 1, 2, 3, 4, 6, 8 } = { 5, 7, 9 }.

(iv) (B � C)c = U � ( B � C )

= { 1, 2, 3, 4, 5, 6, 7, 8, 9 } � { 2, 8 } = { 1, 3, 4, 5, 6, 7, 9 }.

If A, B and C are finite sets and U be the finite universalset, then(i) A � B = A B� = A � (AB)(ii) A � B = A AB = (iii) (A � B) ( B � A ) = ( A B ) � (AB)(vi) n (A B) = n(A) + n(B) � n (A B)(vii) n (A B) = n(A) + n(B) A, B are disjoint non-voidsets.(viii) n (A � B) = n(A) � n(A B)(ix) n (A B) = No. of elements which belong to exactlyone of A or B

= n ((A � B) (B � A)) = n (A � B) + n (B � A)

[ (A � B) and (B � A) are disjoint ]

= n(A) � n (A B) + n(B) � n (A B) = n(A) + n(B) � 2n (A B)

(x) n(A B C) = n (A) + n(B) + n(C) � n (A B)� n (B C) � n(A C) + n(A B C)(xi) Number of elements in exactly two of the sets A, B, C= n(A B) + n(B C) + n(C A) � 3 n(A B C)(xii) Number of elements in exactly one of thesets A, B, C= n(A) + n(B) + n(C) � 2n (A B) � 2n (B C) � 2n (A C)+ 3n (A B C)(xiii) n(A� B�) = n((A B)�) = n(u) � n (A B)(xiv) n(A� B�) = n((A B)�) = n(u) � n (A B).

Ex. 19 In a group of 800 people, 550 can speak Hindi and450 can speak English. How many can speak bothHindi and English ?

Sol. Let H denote the set of people speaking Hindi and Edenote the set of people speaking English.n(H) = 550, n(E) = 450 and n(H E) = 800.

n(H E) = n(H) + n(E) � n (H E) n(H E) = n(H) + n(E) � n(H E) n(H E) = 550 + 450 � 800 = 200.

Hence, 200 persons can speak both Hindi and English.

Ex.20 There are 40 students in a chemistry class and 60students in a physics class. Find the number ofstudents which are either in physics class or chemistryclass in the following cases:(i) the two classes meet at the same hour.(ii) the two classes meet at different hours and 20students are enrolled in both the subjects.

16PAGE # 16

Sol. Let A be the set of students in chemistry class and B bethe set of students in physics class.It is given that n (A) = 40 and n(B) = 60.(i) If two classes meet at the same hour, then there willnot be a common student sitting in both the classes.Therefore, n (A B) = 0 n (A B) = n(A) + n(B) � n(A B) n (A B) = 40 + 60 � 0 = 100

(ii) If two classes meet at different timings then therecan be some student sitting in both the classes.Therefore, n (A B) = 20 n (A B) = n(A) + n(B) � n(A B)

= 40 + 60 � 20 = 80.

Ex. 21 In a school there are 200 students 100 play cricket,50 play hockey and 60 play basket ball. 30 studentsplay both circket and Hockey, 35 play both hockey andbasket ball and 45 play both basket ball and cricket.(i) What is the maximum number of students who playat least one game?(ii) What is the maximum number of student who playall the 3 games ?(iii) What is the minimum number of students playingat least one game?(iv) What is the minimum number of students playingall the three games.

Sol. Converting all values in terms of variable x, now thenumber of students cannot be negative

020x015x

For minimum number students playing all threegames i.e. x = 20For maximum value of x, again none of the categoriesshould have � ve number of students

30 � x 0x 30

If x is more than 30, 30 �x would be � ve which is not

possible. Total number of students playing at leastone game= 100 + x � 15 + 35 � x + x � 20

= 100 + x Minimum number of students playing at least onegame = 100 + 20 = 120.Maximum number of students playing at least onegame = 100 + 30 = 130.

17PAGE # 17

Pre-requisite : Before going through this chapter,

you should be thorough with the basic concepts of

the chapter explained in X NCERT.

An angle is the amount of rotation of a revolving line

with respect to a fixed line. If the rotation is in

anticlock-wise sense, then the angle measured is

positive and if the rotation is in clock-wise sense,

then the angle measured is negative.

Let X�OX and YOY� be two lines at right angles in a

plane. These lines divide the plane into four equal

parts are known as quadrants. The lines X�OX and

YOY� are known as X-axis and Y-axis respectively.

These two lines taken together are known as the

co-ordinate axes. The regions XOY, YOX�, X�OY� and

Y�OX are known as first, second, third and fourth

quadrants respectively.

(a) Systems of measurement of angles :

(i) Sexagesimal system

(ii) Centesimal system

(iii) Circular system

(i) Sexagesimal system : In this system a right angle

is divided into 90 equal parts called degrees. Each

degree is divided into 60 equal parts called minutes

and each minute is divided into 60 equal parts called

seconds.

Thus, 1 right angle = 90 degrees ( 90º)

1º = 60 minutes (60�)

1� = 60 seconds (60�)

TRIGONOMETRY

(ii) Centesimal system : In this system a right angle is

divided into 100 equal parts, called grades. Each grade

is sub divided into 100 minutes, and each minute into

100 seconds.

Thus, 1 right angle = 100 grades (100g)

1 grade = 100 minutes (100�)

1 minute = 100 seconds (100�)

(iii) Circular system : In this system the unit of

measurement is radian. One radian, written as 1c, is

the measure of an angle subtended at the centre of a

circle by an arc of length equal to the radius of the

circle.

The number of radians in an angle subtended by an

arc of a circle at the centre is equal to radius

arc of length.

= rs

Where, = angle in radian, s = arc length and r = radius.

(b) Relation Between Three System of

Measurement of Angles :

2R

100

G

90

D

Where, D = number of degrees,

G = number of grades,

and R = number of radians.

NOTE :

(i) The angle between two consecutive digits in a

clock = 30º = (/6 radians).

(ii) The hour hand rotates through an angle of 30º in

one hour, i.e. (1/2)º in one minute.

(iii) The minute hand rotates through an angle of 6º in

one minute.

Ex.1 Express 110º 30� in radians.

Sol. 110º 30� = 110º

º

60

30

=

º

2

1110

= 1802

221 c

= 360

221 c

.


18PAGE # 18

Ex.2 Express in radians 47º 25� 36�.

Sol. 47º 25�

'

6036

= 47º

'

53

25

= 47º

'

5128

= 47º

º

601

5128

= º

75

3247

=

º

75

3557

= 18075

3557 c

= 135003557

c.

Ex.3 Express in degrees :

(a) c

15

2

(b) (� 2)c .

Sol. (a) ºc

180

15

2

15

2

= 24º.

(b) (� 2)c =

º180

2

= º

)2(722

180

=

º

116

114

= � 114º

'

60

116

= � 114º

'

11

832

= � 114º 32�

''

60

118

= � 114º 32� 44�.

Ex.4 Express in radians 345g 25� 36�.

Sol. 345g 25� 36� = 345.2536g

= 20000003452536

c

= 1.726268 c

Ex.5 One angle of a triangle is 3x2

grades another is 2x3

degrees, whilst the third is 75

x radians ; express them

all in degrees.

Sol.32

xg = 32

x 53

10º9

xº

And 75x

75xc

180º = 5

ºx12

But53

xº + 23

xº + 5

12 xº = 180º

6xº + 15xº + 24xº = 1800º

45xº = 1800º

x = 40º

Hence, three angles of the triangle are 24º, 60º and

96º.

Ex.6 The angles of a triangle are in A.P. and the number of

degrees in the least is to the number of radians in the

greatest as 60 to c. Find the angles in degrees.

Sol. The three angles in A.P. ; if y is common difference, let

these angles be (x + y)º, xº and (x � y)º

x + y + x + x � y= 180º

x = 60º

According to the question

60

180)yx(

)yx(c

or (x � y) = (x + y) 180

× 60

or 3 (x � y) = x + y

or 4y = 2x

or y = x/2

y = 2

º60 = 30º

Hence three angles are 30º, 60º and 90º.

Ex.7 The angles in one regular polygon is to that in another

as 3 : 2, also the number of sides in the first is twice

that in the second ; how many sides do the polygons

have ?

Sol. Suppose the second regular polygon has number of

side = x.

The first regular polygon will have number of side = 2x.

Each angle of the first polygon = x2

)4x4( right angle.

And each angle of the second polygon = x

)4x2( right angle.

x

4x2:

x24x4

= 3 : 2

orx

12x6x

4x4

or 4x � 4 = 6x � 12

or 2x = 8.

The number of sides in the first and second polygons

is respectively is 8 and 4.

19PAGE # 19

Ex.8 The radius of a certain circle is 30 cm, find the

approximately length of an arc of this circle ; if the length

of the chord of the arc be 30 cm.

Sol. Let ABC be the circle whose centre is O and AC is

chord.

In AOC, AO = OC = AC = 30 cm.

AOC = 60º = 3

Hence, arc AC = radius × 3

= 30 × 3

= 10 = 31.4159 cm.

Two angles are said to allied when their sum or

difference is either zero or a multiple of 90º. If is any

angle, then 90 ± , 180 ± , 270 ± , 360 ±

etc. are called allied angles.

sin ( ) = sin cos ( ) = cos

sin (90 + ) = cos cos (90 + ) = � sin

tan (90 + ) = � cot cot (90 + ) = � tan

sec (90 + ) = � cosec cosec (90 + ) = sec

sin (180 � ) = + sin sin (180 + ) = � sin

cos (180 � ) = � cos cos (180 + ) = � cos

tan (180 � ) = � tan tan (180 + ) = tan

cot (180 � ) = � cot cot (180 + ) = cot

sec (180 � ) = � sec sec (180 + ) = � sec

cosec (180 � ) = cosec cosec (180 + ) = � cosec

sin (270 � ) = � cos sin (270+ ) = � cos

cos (270 � ) = � sin cos (270 + ) = sin

tan (270 � ) = cot tan (270 + ) = � cot

cot (270 � ) = tan cot (270 + ) = � tan

sec (270 � ) = � cosec sec (270 + ) = cosec

cosec (270 � ) = � sec cosec (270 + ) = � sec

sin (360 � ) = � sin sin (360 + ) = sin

cos (360 � ) = cos cos (360 + ) = cos

tan (360 � ) = � tan tan (360 + ) = tan

cot (360 � ) = � cot cot (360 + ) = cot

sec (360 � ) = sec sec (360 + ) = sec

cosec (360 � ) = � cosec cosec (360 + ) = cosec

Ex.9 Prove that :

cot A + tan (180º + A) + tan (90º + A) + tan (360º � A) = 0.

Sol. cot A + tan (180º + A) + tan (90º + A) + tan (360º � A)

= cot A + tan A � cot A � tan A = 0.

Ex.10 Prove that : sec (270º � A) sec (90º � A) �

tan (270º � A) tan (90º + A) + 1 = 0

Sol. sec (270º � A) sec (90º � A) � tan (270º � A)

tan (90º + A) + 1

= � cosec2A + cot2A + 1 = 0.

Ex.11 Prove that : sin 420º cos 390º + cos (�300º)

sin (�330º) = 1.

Sol. sin 420º cos 390º + cos(� 300º) sin (� 330º)

= sin(360º + 60º) cos (360º + 30º) � cos(270º + 30º)

sin (270º + 60º)

= sin 60º cos 30º + sin 30º cos 60º

= 21

21

23

23

= 41

43 = 1.

Ex.12 tan 225º cot 405º + tan 765º cot 675º = 0.

Sol. tan 225º cot 405º + tan 765º cot 675º

= tan(180º + 45º) cot (360º + 45º) + tan (720 + 45º)

cot (630+ 45º)

= tan 45º cot 45º + tan 45º (� tan 45º)

= 1 � 1 = 0.

(i) y = sin x x R; y [�1, 1]

(ii) y = cos x x R; y [ � 1, 1]

(iii) y = tan x, x R � (2n + 1)/2, n ; y R

(iv) y = cot x, x R � n , n ; y R

20PAGE # 20

(v) y = cosec x,x R � n , n ; y (, 1] [1, )

(vi) y = sec x, x R � (2n + 1)/2, n ; y (, 1] [1, )

(i) sin (A ± B) = sinA cosB ± cosA sinB

(ii) cos (A ± B) = cosA cosB sinA sinB

(iii) sin²A sin²B = cos²B cos²A

= sin (A+B). sin (A B)

(iv) cos²A sin²B = cos²B sin²A

= cos (A+B). cos (A B)

(v) tan (A ± B) = BtanAtan1BtanAtan

(vi) cot (A ± B) = AcotBcot1BcotAcot

(vii) tan (A + B + C)

= AtanCtanCtanBtanBtanAtan1CtanBtanAtanCtanBtanAtan

.

Ex.13 Prove that : sin (45º + A) cos (45º � B) +

cos (45º + A) sin (45º � B) = cos (A � B).

Sol. sin (45º + A) cos (45º � B) + cos (45º + A) sin (45º � B)

= sin (45º + A + 45º � B)

= sin (90º + A � B)

= cos (A � B).

Ex.14 Prove : tan

4 tan

43

= �1.

Sol. tan

4 × tan

43

=

tan1tan1

×

tan1tan1

= � 1

Ex.15 If sin = 53

, cos = 135

, then find sin ( + ).

Sol. cos = 2sin1 =

2

53

1

cos = 54

and sin = 2cos1 = 2

135

1

sin = 1312

.

sin ( + ) = sin cos + cos sin

= 1312

54

135

53

` = 6548

6515

= 6563

Ex.16 Find the value of sin 105º.

Sol. sin 105º = sin (60º + 45º)

= sin 60º cos45º + cos 60º sin 45º

= 2

121

2

123

= 22

13 .

(i) sin (A + B) + sin (A � B) = 2 sinA cosB

(ii) sin(A+B) sin(A B) = 2 cosA sinB

(iii) cos(A+B) + cos(A B) = 2 cosA cosB

(iv) cos(A B) cos(A+B) = 2 sinA sinB

(v) sinC + sinD = 2 sin2DC

cos2DC

(vi) sinC sinD = 2 cos2DC

sin2DC

(vii) cosC + cosD = 2 cos2DC

cos2DC

(viii) cosC

cosD = 2 sin

2DC

sin

2DC

Ex.17 Prove that : sin 5A + sin 3A = 2sin 4A cos A.

Sol. L.H.S. sin 5A + sin 3A = 2sin 4A cos A = R.H.S.

[ sin C + sin D = 2 sin 2

DC cos

2DC

]

21PAGE # 21

Ex.18 Find the value of 2 sin 3 cos � sin 4 � sin 2

Sol. 2 sin 3 cos � sin 4 � sin 2

= 2 sin 3 cos � [2 sin 3 cos ]

= 0.

Ex.19 Prove that :

4sin3sincos2cos3cos6sincos8sin

= tan 2

Sol.

4sin3sin2cos2cos23cos6sin2cos8sin2

=

7coscoscos3cos3sin9sin7sin9sin

=

7cos3cos3sin7sin

=

2cos5cos25cos2sin2

= tan 2.

Ex.20 Prove that :

3tan5tan3tan5tan

= 4 cos 2 cos 4 .

Sol.

3tan5tan3tan5tan

=

5cos3sin3cos5sin5cos3sin3cos5sin

=

2sin8sin

=

2sin4cos2cos2sin4

2sin4cos4sin2

= 4 cos2 cos 4 .

Ex.21 Prove that : A7sinA5sin2A3sinA5sinA3sin2Asin

=

A5sinA3sin

.

Sol.A7sinA5sin2A3sinA5sinA3sin2Asin

= A5sin2)A7sinA3(sinA3sin2)A5sinA(sin

=A5sin2A2cosA5sin2A3sin2A2cosA3sin2

= )1A2(cosA5sin2)1A2(cosA3sin2

=

A5sinA3sin

.

Ex.22 Prove that : 2 cos 13

cos139

+ cos 133

+ cos 135

= 0.

Sol. 2 cos 13

cos 139

+ cos 133

+ cos 135

= cos

139

13 + cos

139

13 + cos133

+ cos135

= cos

1310

+ cos

138

+ cos133

+ cos135

= cos

133

+ cos

135

+ cos133

+ cos135

= � cos133

� cos135

+ cos133

+ cos135

= 0.

(i) sin 2A = 2 sinA cosA ; sinA = 2 sin2A

cos2A

(ii) cos 2A = cos2A sin2A = 2cos2A 1 = 1 2 sin2A;

2 cos2

2A

= 1 + cosA, 2 sin2

2A

= 1 cosA.

(iii) tan 2A =Atan1

Atan22

, tan =

222

tan1

tan2

(iv) sin 2A =Atan1

Atan22

, cos 2A =

Atan 1

Atan 12

2

(v) sin 3A = 3 sinA 4 sin3A

(vi) cos 3A = 4 cos3A 3 cosA

(vii) tan 3A = Atan31

AtanAtan32

3

(viii) sin 15° or sin12

= 22

13 = cos 75° or cos

125

(ix) cos 15° or cos12

= 22

13 = sin 75° or sin

125

(x) tan 15° =13

13

= 3 2 = cot 75° ;

tan 75° =13

13

= 3 2 = cot 15°

(xi) sin10

or sin 18° =4

15 & cos 36° or cos

5

= 4

15.

22PAGE # 22

Ex.23 Prove that : A2cos1

A2sin

= tan A .

Sol. L.H.S.A2cos1

A2sin

= Acos2

AcosAsin22 = tan A.A.

Ex.24 Prove that : tan A + cot A = 2 cosec 2 A.

Sol. L.H.S. tan A + cot A

= Atan

Atan1 2

= 2

Atan2Atan1 2

= A2sin

2

= 2 cosec 2 A.

Ex.25 Prove that :

)BAcos(BcosAcos1)BAcos(BcosAcos1

= tan

2A

cot 2B

.

Sol. L.H.S. )BAcos(BcosAcos1)BAcos(BcosAcos1

=

B2A

cos2A

cos22A

cos2

B2A

sin2A

sin22A

sin2

2

2

= tan 2A

B2A

cos2A

cos

B2A

sin2A

sin

= tan 2A

2B

sin2

BAsin2

2B

cos2

BAsin2

= tan 2A

cot 2B

.

Ex.26 Prove that : sin 20º sin 40º sin 60º sin 80º = 163

.

Sol. sin 20º sin 40º sin 60º sin 80º

= sin 20º sin 40º × 23

sin 80º

=23

(sin 20º sin 40º) sin 80º

=43

(2sin 20º sin 40º) sin 80º

=43

[cos 20º � cos 60º] sin 80º

=43

cos 20º sin 80º � 43

cos 60º sin 80º

=83

[2 sin 80º cos 20º] � 83

sin 80º

=83

[sin 100º cos 60º] � 83

sin 80º

=83

sin 100º + 83

sin 60º � 83

sin 80º

=83

sin (180º � 80º) + 83

× 23

� 83

sin 80º

=83

sin 80º + 163�

83

sin 80º

=163

.

E = a sin + b cos

E = 22 ba sin ( + ), where tan =ab

= 22 ba cos (), where tan =ba

Hence for any real value of ,

2222 baEba

So, the Maximum value = 22 ba

And Minimum value = � 22 ba

Ex.27 Find maximum and minimum values of following :(i) 3sinx + 4cosx(ii) 1 + 2sinx + 3cos2x

Sol. (i) We know

� 22 43 3sinx + 4cosx 22 43

� 5 3sinx + 4cosx 5(ii) 1+ 2sinx + 3cos2x= � 3sin2x + 2sinx + 4

= � 3

3xsin2

xsin2 + 4

= � 3

2

31

xsin

+

313

Now 0 2

31

xsin

916

� 3

16 � 3

2

31

xsin

0

� 1 � 3

2

3

1xsin

+

313

3

13

23PAGE # 23

LINES AND ANGLES & QUADRILATERAL

DEFINITIONS

LINE : A line has length but no width and no thickness.

ANGLE :An angle is the union of two non-collinear rays witha common initial point. The common initial point iscalled the �vertex� of the angle and two rays arecalled the �arms� of the angles.

REMARK :

Every angle has a measure and unit of measurementis degree.One right angle = 90º, 1º = 60� (min.), 1� =

60� (sec.)

Types of Angles :

(i) Right angle : An angle whose measure is 90º iscalled a right angle.

(ii) Acute angle : An angle whose measure is lessthan 90º is called an acute angle.

0º < BOA < 90º

AO

B

(iii) Obtuse angle : An angle whose measure ismore than 900 but less than 180º is called an obtuseangle.90º < AOB < 180º.

(iv) Straight angle : An angle whose measure is180º is called a straight angle.

(v) Reflex angle : An angle whose measure is morethan 1800 is called a reflex angle. 180º < AOB <360º.

(vi) Complementary angles : Two angles, the sumof whose measures is 90 º are calledcomplementary angles. AOC + BOC = 90º.

(vii) Supplementary angles : Two angles, the sumof whose measures is 180º, are called thesupplementary angles. AOC +BOC = 180º.

Angles Made by a Transversal with two Lines:

(i) Transversal : A line which intersects two or moregiven parallel lines at distinct points is called atransversal of the given lines.

(ii) Corresponding angles : Two angles on thesame side of a transversal are known as thecorresponding angles if both lie either above the twolines or below the two lines, in figure 1 & 5, 4& 8, 2 & 6, 3 & 7 are the pairs ofcorresponding angles.If a transversal intersects two parallel lines then thecorresponding angles are equal i.e. 1 = 5,4 = 8, 2 = 6 and 3 = 7.

(iii) Alternate interior angles : 3 & 5, 2 &8, are the pairs of alternate interior angles.If a transversal intersects two parallel lines then theeach pair of alternate interior angles are equal i.e.3 = 5 and 2 = 8.


24PAGE # 24

(iv) Consecutive interior angles : The pair ofinterior angles on the same side of the transversalare called pairs of consecutive interior angles. Infigure 2 & 5, 3 & 8, are the pairs ofconsecutive interior angles.If a transversal intersects two parallel lines then eachpair of consecutive interior angles are supplementaryi.e. 2 + 5 = 180º and 3 + 8 = 180º.

(v) Vertically opposite angles : In figure,1 = 3, 2 = 4, 5 = 7 & 6 = 8.

POLYGON

A closed plane figure bounded by line segments iscalled a polygon.

A polygon is named according to the number of sidesit has :

No. of sides 3 4 5 6 7 8 10

Figure Triangle Quadrilateral Pentagon Hexagon Heptagon Octagon Decagon

In general, a polygon having n sides is called 'n'sided polygon.

Diagonal of Polygon :

Line segment joining any two non-consecutivevertices of a polygon is called its diagonal.

Convex Polygon :

If all the interior angles of a polygon are less than1800, it is called a convex polygon.

Concave Polygon :

If one or more of the interior angles of a polygon isgreater than 1800 i.e. reflex, it is called a concavepolygon.

Regular Polygon :

A polygon is called a regular polygon if all its sideshave equal length and all the angles have equalmeasure.

REMARKS :

(1) The sum of the interior angles of a convex polygonof n sides is (2n � 4) right angles or (2n � 4) 90º.

(2) The sum of the exterior angles of a convex polygonis 4 right angles or 360º.

(3) Each interior angle of a n-sided regular polygon is

n

º904n2

(4) Each exterior angle of a regular polygon of n sides

=

n3600

(5) If a polygon has n sides, then the number of

diagonals of the polygon =

23nn

.

QUADRILATERAL

A quadrilateral is a four sided closed figure.

A

D

C

B

Let A, B, C and D be four points in a plane suchthat :

(i) No three of them are collinear.

(ii) The line segments AB, BC, CD and DA do notintersect except at their end points, then figureobtained by joining A, B, C & D is called aquadrilateral.

Convex and Concave Quadrilaterals :

(i) A quadrilateral in which the measure of eachinterior angle is less than 180° is called a convex

quadrilateral. In fig., PQRS is convexquadrilateral.

S

P Q

R

(ii) A quadrilateral in which the measure of one ofthe interior angles is more than 180° is called a

concave quadrilateral. In fig., ABCD is concavequadrilateral.

A

B

C

D

25PAGE # 25

Special Quadrilaterals :

(i) Parallelogram : A parallelogram is aquadrilateral in which both pairs of opposite sidesare parallel. In fig., AB || DC, AD || BC therefore,ABCD is a parallelogram.

A B

CD

(ii) Rectangle : A rectangle is a parallelogram, inwhich each of its angle is a right angle. If ABCD is arectangle then A = B = C = D = 90°.

A B

CD

900

(iii) Rhombus : A rhombus is a parallelogram inwhich all its sides are equal in length. If ABCD is arhombus then AB = BC = CD = DA.

(iv) Square : A square is a parallelogram havingall sides equal and each angle equal to right angle.If ABCD is a square then AB = BC = CD = DA andA = B = C = D = 90°.

(v) Trapezium : A trapezium is a quadrilateral withonly one pair of opposite sides parallel. In fig., ABCDis a trapezium with AB || DC.

A B

CD

(vi) Kite : A kite is a quadrilateral in which twopairs of adjacent sides are equal. If ABCD is a kitethen AB = AD and BC = CD.

A

B

C

D

(vii) Isosceles trapezium : A trapezium is said tobe an isosceles trapezium, if its non-parallel sides

are equal. Thus a quadrilateral ABCD is an isoscelestrapezium, if AB || DC and AD = BC.

PROPERTIES

Theorem 1 : The sum of the four angles of aquadrilateral is 360°.

Theorem 2 : A diagonal of a parallelogram dividesit into two congruent triangles.

Theorem 3 : In a parallelogram, opposite sides areequal.

Theorem 4 : The opposite angles of a parallelogramare equal.

Theorem 5 : The diagonals of a parallelogram bisecteach other.

Theorem 6 : Each of the four angles of a rectangleis a right angle.

Theorem 7 : Each of the four sides of a rhombus isof the same length.

Theorem 8 : Each of the angles of a square is aright angle and each of the four sides is of the samelength.

Theorem 9 : The diagonals of a rectangle are ofequal length.

Theorem 10 : The diagonals of a rhombus areperpendicular to each other.

Theorem 11 : The diagonals of a square are equaland perpendicular to each other.

Theorem 12 : Parallelograms on the same baseand between the same parallels are equal in area.

Theorem 13 : Two triangles on the same base(or equal bases) and between the same parallelsare equal in area.

Theorem 14 : Parallelogram and Triangles on thesame base (or equal bases) and between the sameparallels, then area of parallelogram is twice thearea of triangle.

Theorem 15 : Median of a triangle divides it intotwo triangles of equal area.

Theorem 16 : A diagonal of a parallelogram dividesit into two triangles of equal area.

A quadrilateral become a parallelogram when:

(i) Opposite angles are equal.

(ii) Both the pair of opposite sides are equal

(iii) A pair of opposite side is equal as well asparallel

(iv) Diagonals of quadrilateral bisect each other.

26PAGE # 26

TRIANGLE

TRIANGLE

A plane figure bounded by three lines in a plane iscalled a triangle. Every triangle have three sides andthree angles. If ABC is any triangle then AB, BC & CAare three sides and A,B and C are three angles.

Types of triangles :

A. On the basis of sides we have three types of triangle.1. Scalene triangle � A triangle in which no two sides

are equal is called a scalene triangle.2. Isosceles triangle � A triangle having two sides equal

is called an isosceles triangle.

3. Equilateral triangle � A triangle in which all sides

are equal is called an equilateral triangle.

B. On the basis of angles we have three types :

1. Right triangle � A triangle in which any one angle is

right angle is called right triangle.

2. Acute triangle � A triangle in which all angles are

acute is called an acute triangle.

3. Obtuse triangle � A triangle in which any one angle

is obtuse is called an obtuse triangle.

CONGRUENT TRIANGLES

Two triangles are congruent if and only if one of themcan be made to superimposed on the other, so as tocover it exactly.

If two triangles ABC and DEF are congruent thenthere exist a one to one correspondence between theirvertices and sides i.e. we get following six equalitiesA = D, B = E, C = F and AB = DE, BC = EF, AC= DF.

Sufficient Conditions for Congruence of two Triangles :

(i) SAS Congruence Criterion :A

B

A

C

P

RQ

If, AC = PR

ACB = PRQ

and BC = QR

then,ABC PQR

(ii) ASA Congruence Criterion :

If, ABC = PQR

BC = QR

and ACB = PRQ

then, ABC PQR

(iii) AAS Congruence Criterion :

If, ABC = PQR

BCA= QRP

and AC = PR

then, ABC PQR

(iv) SSS Congruence Criterion :

If AB = PQ BC = QR

and AC = PRthen, ABC PQR

(v) RHS Congruence Criterion :

If, AB= PQ BC = QR

and ACB = PRQ = 90º

then, ABC PQR


27PAGE # 27

NOTE : If two triangles are congruent then theircorresponding sides and angles are also congruentby cpctc (corresponding parts of congruent trianglesare also congruent).

In triangle ABC, AB = c, BC = a & CA = b then,

A

B C

c

a

b

(i) The sum of any two sides of a triangle is greater

than its third side. i.e. in ABC,

(A) a + b > c (B) b + c > a (C) a + c > b

(ii) If two sides of a triangle are unequal, then the longer

side has greater angle opposite to it

i.e. in ABC, if AB > AC then C > B.

(iii) In a right angle triangle the sum of squares of two

smaller sides is equal to the square of its third side.

i.e. in ABC, a2 + b2 = c2.

If sum of squares of two smaller sides is greater than

the square of its third side then that triangle is acute

angled triangle.

i.e in ABC, a2 + b2 > c2.

If sum of squares of two smaller sides is lesser than

the square of its third side then that triangle is obtuse

angled triangle.

i.e in ABC, a2 + b2 < c2.

Theorem-1 (Mid-Point Theorem) :

Statement : In a triangle, the line segment joining the

mid-points of any two sides is parallel to the third side

and is half of it.

Theorem-2 : The sum of the three angles of a triangle

is 1800.

Theorem-3 : If a side of a triangle is produced, the

exterior angle so formed is equal to the sum of the two

interior opposite angles.

Theorem-4 : Angles opposite to equal sides of an

isosceles triangle are equal.

Theorem-5 : If two angles of a triangle are equal, then

sides opposite to them are also equal.

Theorem-6 : If the bisector of the vertical angle bisects

the base of the triangle, then the triangle is isosceles.

Basic Proportionality Theorem (Thales Theorem) :

Statement : If a line is drawn parallel to one side of a

triangle to intersect the other two sides in distinct points,

then the other two sides are divided in the same ratio.

Converse of Basic Proportionality Theorem :

Statement : If a line divides any two sides of a triangle

in the same ratio, then the line must be parallel to

the third side.

Two triangles ABC and DEF are said to be similar if

their

(i) Corresponding angles are equal.

i.e. A = D, B = E, C = F

And,

(ii) Corresponding sides are proportional.

i.e. DE

AB = EF

BC = DF

AC.

(a) Characteristic Properties of Similar Triangles :

(i) (AAA Similarity) If two triangles are equiangular,

then they are similar.

(ii) (SSS Similarity) If the corresponding sides of two

triangles are proportional, then they are similar.

(iii) (SAS Similarity) If in two triangles, one pair of

corresponding sides are proportional and the included

angles are equal then the two triangles are similar.

(b) Results Based Upon Characteristic Properties

of Similar Triangles :

(i) If two triangles are equiangular, then the ratio of the

corresponding sides is the same as the ratio of the

corresponding medians.

(ii) If two triangles are equiangular, then the ratio of the

corresponding sides is same as the ratio of the

corresponding angle bisector segments.

28PAGE # 28

(iii) If two triangles are equiangular then the ratio of the

corresponding sides is same as the ratio of the

corresponding altitudes.(iv) If one angle of a triangle is equal to one angle ofanother triangle and the bisectors of these equalangles divide the opposite side in the same ratio, thenthe triangles are similar.

(v) If two sides and a median bisecting the third side ofa triangle are respectively proportional to thecorresponding sides and the median of anothertriangle, then two triangles are similar.

(vi) If two sides and a median bisecting one of thesesides of a triangle are respectively proportional to thetwo sides and the corresponding median of anothertriangle, then the triangles are similar.

Statement :The ratio of the areas of two similar

triangles is equal to the square of the ratio of their

corresponding sides.

Two triangles ABC and PQR such that ABC ~ PQR

then )PQR(ar)ABC(ar

= 2

PQAB

=

2

QRBC

=

2

RPCA

.

(a) Properties of Areas of Similar Triangles :

(i) The areas of two similar triangles are in the ratio of

the squares of corresponding altitudes.

(ii) The areas of two similar triangles are in the ratio of

the squares of the corresponding medians.

(iii) The area of two similar triangles are in the ratio of

the squares of the corresponding angle bisector

segments.

PYTHAGOR S THEOREMA

Statement : In a right triangle, the square of the

hypotenuse is equal to the sum of the squares of the

other two sides.

If in right triangle ABC, right angled at B then AC2 = AB2

+ BC2

Converse of Pythagoras Theorem :

Statement : In a triangle, if the square of one side is

equal to the sum of the squares of the other two sides,

then the angle opposite to the first side is a right angle.

If in triangle ABC such that AC2 = AB2 + BC2 then

angle opposite to AC is right angle

Some Results Deduced From Pythagoras Theorem :

(i) In the given figure ABC is an obtuse triangle,

obtuse angled at B. If AD CB,

then AC2 = AB2 + BC2 + 2BC. BDA

D CB

(ii) In the given figure, if B of ABC is an acute angle

and AD BC, then AC2 = AB2 + BC2 � 2BC . BD

(iii) In any triangle, the sum of the squares of any two

sides is equal to twice the square of half of the third

side together with twice the square of the median which

bisects the third side.

(iv) Three times the sum of the squares of the sides of

a triangle is equal to four times the sum of the

squares of the medians of the triangle.

29PAGE # 29

CIRCLES

DEFINITIONS

Circle :

The collection of all the points in a plane, which are ata fixed distance from a fixed point in the plane, is calleda circle.The fixed point is called the centre of the circle and thefixed distance is called the radius of the circle.

In figure, O is the centre and the length OP is the radiusof the circle. So the line segment joining the centreand any point on the circle is called a radius of thecircle.

Chord :

If we take two points P and Q on a circle, then the linesegment PQ is called a chord of the circle.

QO

P

If the chord which passes through the centre of thecircle, is called a diameter of the circle.

Arc :

A piece of a circle between two points is called an arc.

P Q

R

The longer one is called the major arc and the

shorter one is called the minor arc

Secant :Secant to a circle is a line which intersects the circle intwo distinct points.

Tangent :A tangent to a circle is a line that intersects the circle inexactly one point.

Theorem-1 : Equal chords of a circle subtend equalangles at the centre.If AB = CD then AOB = COD

Theorem-2 : The perpendicular from the centre of acircle to a chord bisects the chord.If OM AB then AM = BM

A M B

O

Theorem-3 : There is one and only one circle passingthrough three given non-collinear points.

Theorem-4 : The angle subtended by an arc at thecentre is double the angle subtended by it at any pointon the remaining part of the circle.POQ = 2PAQ

BO

A

PQ

Theorem-5 : Angles in the same segment of a circleare equal. PAQ = PCQ

O

P

A

Q

C

Theorem-6 : Angle in the semicircle is a right angle.PCQ = PAQ = 90º

Theorem-7 : The sum of either pair of opposite anglesof a cyclic quadrilateral is 180º.BCD + BAD = 180º

ABC + ADC = 180º

Theorem-8 : Equal chords of a circle (or of congruentcircles) are equidistant from the centre (or centres).If AB = CD then ON = OM

ON

C

DB

M

A


30PAGE # 30

Theorem-9 : A tangent to a circle is perpendicular to

the radius through the point of contact.

OP AB

Theorem-10 : Lengths of two tangents drawn from an

external point to a circle are equal.

If AP and AQ are two tangents then AP = AQ.

Theorem-11 : If two chords of a circle intersect inside

or outside the circle when produced, then the area

of rectangle formed by the two segments of one chord

is equal in area to the rectangle formed by the two

segments of the other chord. (PA × PB = PC × PD)

A

BC

D

P

Theorem-12 : If PAB is a secant to a circle intersecting

the circle at A and B and PT is tangent segment, then

PA × PB = PT2

P

AB

T

Theorem-13 : A line touches a circle and from the point

of contact a chord is drawn. Prove that the angles which

the chord makes with the given line are equal

respectively to angles formed in the corresponding

alternate segments.

BAT = ACB and BAP = ADB.

B

C

A

O

TP

D

Definition : A line which touches the two given circles iscalled common tangent to the two circles. Let C(O1, r1),C(O2, r2) be two given circles. Let the distance betweencentres O1 and O2 be d i.e., O1O2 = d.

(a) In fig. (i) d > r1 + r2 i.e. two circles do not intersect.In this case, four common tangents are possible.The tangent lines l and m are called direct commontangents and the tangent lines p and q are calledindirect (transverse) common tangents.

(b) In fig. (ii), d = r1 + r2. In this case, two circles touchexternally and there are three common tangents.

(c) In fig.(iii) d < r1 + r2. In this case two circles intersectin two distinct points and there are only two commontangents.

(d) In fig. (iv), d = r1 � r2 (r1 > r2), in this case, two circlestouch internally and there is only one common tangent.

(e) In fig. (v), the circle C(O2, r2) lies wholly in the circleC(O1, r1) and there is no common tangent.

31PAGE # 31

NUMBER SYSTEM

(i) Natural numbers :Counting numbers are known as natural numbers.N = { 1, 2, 3, 4, ... }.

(ii) Whole numbers :

All natural numbers together with 0 form the collectionof all whole numbers. W = { 0, 1, 2, 3, 4, ... }.

(iii) Integers :All natural numbers, 0 and negative of natural numbersform the collection of all integers.I or Z = { ..., � 3, � 2, � 1, 0, 1, 2, 3, ... }.

(iv) Rational numbers :These are real numbers which can be expressed in the

form of qp

, where p and q are integers and 0q .

e.g. 2/3, 37/15, -17/19.

All natural numbers, whole numbers and integers arerational.

Rational numbers include all Integers (without anydecimal part to it), terminating fractions ( fractions inwhich the decimal parts are terminating e.g. 0.75,� 0.02 etc.) and also non-terminating but recurringdecimals e.g. 0.666....., � 2.333...., etc.

Fractions :

(a) Common fraction : Fractions whose denominatoris not 10.

(b) Decimal fraction : Fractions whose denominator is10 or any power of 10.

(c) Proper fraction : Numerator < Denominator i.e.5

3.

(d) Improper fraction : Numerator > Denominator i.e.35 .

(e) Mixed fraction : Consists of integral as well as

fractional part i.e.72

3 .

(f) Compound fraction : Fraction whose numerator and

denominator themselves are fractions. i.e.5/72/3

.

Improper fraction can be written in the form of mixedfraction.

(v) Irrational Numbers :

All real number which are not rational are irrationalnumbers. These are non-recurring as well asnon-terminating type of decimal numbers.

For Ex. : 2 , 3 4 , 32 , 32 , 4 7 3 etc.

(vi) Real numbers : Numbers which can represent

actual physical quantities in a meaningful way areknown as real numbers. These can be representedon the number line. Number line is geometrical straightline with arbitrarily defined zero (origin).

(vii) Prime numbers : All natural numbers that have

one and itself only as their factors are called primenumbers i.e. prime numbers are exactly divisible by1 and themselves. e.g. 2, 3, 5, 7, 11, 13, 17, 19, 23,...etc.If P is the set of prime number then P = {2, 3, 5, 7,...}.

(viii) Composite numbers : All natural numbers, whichare not prime are composite numbers. If C is the setof composite number then C = {4, 6, 8, 9, 10, 12,...}.

1 is neither prime nor composite number.

(ix) Co-prime Numbers : If the H.C.F. of the given

numbers (not necessarily prime) is 1 then they areknown as co-prime numbers. e.g. 4, 9 are co-primeas H.C.F. of (4, 9) = 1.

Any two consecutive numbers will always be co-prime.

(x) Even Numbers : All integers which are divisible by 2are called even numbers. Even numbers are denotedby the expression 2n, where n is any integer. So, if E isa set of even numbers, then E = { ..., � 4, �2, 0, 2, 4,...}.

(xi) Odd Numbers : All integers which are notdivisible by 2 are called odd numbers . Oddnumbers are denoted by the general expression2n �1 where n is any integer. If O is a set of odd

numbers, then O = {..., �5, �3, �1, 1, 3, 5,...}.

(xii) Imaginary Numbers : All the numbers whosesquare is negative are called imaginary numbers.

e.g. 3i, -4i, i, ... ; where i = 1- .

(xiii) Complex Numbers : The combined form of realand imaginary numbers is known as complexnumbers. It is denoted by Z = A + iB where A is real part

and B is imaginary part of Z and A, B R.

The set of complex number is the super set of all thesets of numbers.

Squares : When a number is multiplied by itself thenthe product is called the square of that number.

Perfect Square : A natural number is called a perfectsquare if it is the square of any other natural numbere.g. 1, 4, 9,... are the squares of 1, 2, 3,... respectively.


32PAGE # 32

Cube : If any number is multiplied by itself three times

then the result is called the cube of that number.

Perfect cube : A natural number is said to be a perfect

cube if it is the cube of any other natural number.

Any irrational number of the form n a is given a special

name Surd. Where �a� is called radicand, rational. Also

the symbol n is called the radical sign and the index

n is called order of the surd.

n a is read as nth root of �a� and can also be written

as n

1

a .

Identification of Surds :

(i) 3 4 is a surd as radicand is a rational number..

Similar examples : ...,12,7,12,5 543

(ii) 2 + 3 is a surd (as surd + rational number will

give a surd)

Similar examples : ...,13,13,2�3 3

(iii) 34�7 is a surd as 7 � 4 3 is a perfect square

of 3�2 .

Similar examples : ...,549,54�9,347

(iv) 3 3 is a surd as 6613

1

21

3 3333

Similar examples : ...,6,5 4 53 3

(v) These are not a surds :

(A) 3 8 , because 3 33 28 which is a rational

number.

(B) 32 , because 2 + 3 is not a perfect square.

(C) 3 31 , because radicand is an irrational number..

Laws of Surds :

(i) n nnn aa = a

(ii) nnn abba [Here order should be same]

(iii) nnn

ba

ba

(iv) m nnmn m aaa

(v) pn pn aa

or, pn pmn m aa

[Important for changing order of surds]

Factors : �a� is a factor of �b� if there exists a relationsuch that a × n = b, where �n� is any natural number.

1 is a factor of all numbers as 1 × b = b.

Factor of a number cannot be greater than the number(infact the largest factor will be the number itself). Thusfactors of any number will lie between 1 and the numberitself (both inclusive) and they are limited.

Multiples : �a� is a multiple of �b� if there exists a relationof the type b × n = a. Thus the multiples of 6 are6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 = 24, and so on.

The smallest multiple will be the number itself and thenumber of multiples would be infinite.

NOTE :To understand what multiples are, let�s just take an

example of multiples of 3. The multiples are 3, 6, 9,12,.... so on. We find that every successive multiplesappears as the third number after the previous.So if one wishes to find the number of multiples of 6less than 255, we could arrive at the number through

6255

= 42 (and the remainder 3). The remainder is of

no consequence to us. So in all there are 42 multiples.

If one wishes to find the multiples of 36, find 36

255 = 7

(and the remainder is 3). Hence, there are 7 multiplesof 36.

Factorisation : It is the process of splitting any number

into form where it is expressed only in terms of the

most basic prime factors.

For example, 36 = 22 × 32. 36 is expressed in the

factorised form in terms of its basic prime factors.

Number of factors : For any composite number C,

which can be expressed as C = ap × bq × cr ×....., where

a, b, c ..... are all prime factors and p, q, r are positive

integers, then the number of factors is equal to

(p + 1) × (q + 1) × (r + 1).... e.g. 36 = 22 × 32. So the

factors of 36 = (2 +1) × (2 + 1) = 3 × 3 = 9.

LCM (least Common Multiple) : The LCM of given

numbers, as the name suggests is the smallestpositive number which is a multiple of each of the givennumbers.

HCF (Highest Common factor) : The HCF of givennumbers, as the name suggests is the largest factorof the given set of numbers.Consider the numbers 12, 20 and 30. The factors andthe multiples are :

FactorsGiven

numbersMultiples

1, 2, 3, 4, 6, 12 12 12, 24, 36, 48, 60, 72, 84, 96, 108, 120....

1, 2, 4, 5, 10, 20 20 20, 40, 60, 80, 100, 120.....

1, 2, 3, 5, 6, 10, 15, 30 30 30, 60, 90, 120....

33PAGE # 33

The common factors are 1 and 2 and the commonmultiples are 60, 120...Thus the highest common factor is 2 and the leastcommon multiple is 60. Meaning of HCF is that theHCF is the largest number that divides all the givennumbers.Also since a number divides its multiple, the meaningof LCM is that it is smallest number which can bedivided by the given numbers.HCF will be lesser than or equal to the least of thenumbers and LCM will be greater than or equal to thegreatest of the numbers.

For any two numbers x and y :x × y = HCF (x, y) × LCM (x, y).

HCF and LCM of fractions :

LCM of fractions = atorsmindenoofHCFnumeratorsofLCM

HCF of fractions = atorsmindenoofLCMnumeratorsofHCF

Make sure the fractions are in the most reducible form.

Division Algorithm : General representation of result is,

DivisormainderRe

QuotientDivisor

Dividend

Dividend = (Divisor × Quotient ) + Remainder

NOTE :(i) (xn � an) is divisible by (x � a) for all the values of n.

(ii) (xn � an) is divisible by (x + a) and (x � a) for all the

even values of n.(iii) (xn + an) is divisible by (x + a) for all the odd values of n.

Test of Divisibility :

No. Divisiblity Test

2 Unit digit should be 0 or even

3 The sum of digits of no. should be divisible by 3

4 The no formed by last 2 digits of given no. should be divisible by 4.

5 Unit digit should be 0 or 5.

6 No should be divisible by 2 & 3 both

8 The number formed by last 3 digits of given no. should be divisible by 8.

9 Sum of digits of given no. should be divisible by 9

11The difference between sums of the digits at even & at odd places should be zero or multiple of 11.

25 Last 2 digits of the number should be 00, 25, 50 or 75.

Rule for 7 : Double the last digit of given number andsubtract from remaining number the result should bezero or divisible by 7.

Rule for 13 : Four times the last digit and add toremaining number the result should be divisible by13.

Rule for 17 : Five times the last digit of the number andsubtract from previous number the result obtainedshould be either 0 or divisible by 17.

Rule for 19 : Double the last digit of given number andadd to remaining number The result obtained shouldbe divisible by 19.

The method of finding the remainder without actuallyperforming the process of division is termed asremainder theorem.

Remainder should always be positive. For example ifwe divide �22 by 7, generally we get �3 as quotient and

�1 as remainder. But this is wrong because remainder

is never be negative hence the quotient should be �4

and remainder is +6. We can also get remainder 6 byadding �1 to divisor 7 ( 7�1 = 6).

To find the remainder of big number

NOTE :

(i) Binomial Expansion :

(a + b)n = an + 1!

nan�1b +

2!

1)n(n an � 2b2 + .... + bn, or

(a � b)n = an � 1!

nan�1b +

2!

1)n(n an� 2b2 � ... + (� 1)nbn.

Hence, first term is pure of a i.e an and last digit is pureof b, i.e. bn.

(ii) Total number of terms in the expansion of (a + b)n is(n + 1).

Ex. What is the remainder when 738 is divided by 48.

Sol.48738

=

487

192

=

4849 19

=

48148 19

so by using

binomial expansion, we can say that 18 terms arecompletely divisible by 48 but the last term which is

481 19

is not divisible. So, 191 = 1 is the remainder..

We are having 10 digits in our number systems andsome of them shows special characterstics like they,repeat their unit digit after a cycle, for example 1 repeatits unit digit after every consecutive power. So, itscyclicity is 1 on the other hand digit 2 repeat its unitdigit after every four power, hence the cyclicity of 2 isfour. The cyclicity of digits are as follows :

Digit Cyclicity

0, 1, 5 and 6 1

4 and 9 2

2, 3, 7 and 8 4

So, if we want to find the last digit of 245, divide 45 by 4.The remainder is 1 so the last digit of 245 would besame as the last digit of 21 which is 2.

34PAGE # 34

Ex. In (32)33 unit digit is equal to the unit digit of 21 i.e. 2.

Factorial n : Product of n consecutive natural numbers

is known as �factorial n� it is denoted by �n!�.

So, n! = n(n � 1)(n � 2).............321. e.g. 5! = 5 × 4 × 3 × 2 × 1 = 120.

The value of factorial zero is equal to the value of

factorial one. Hence 0! = 1 = 1!

The approach to finding the highest power of x dividing

y! is

32 x

y

x

yxy

......., where [ ] represents just

the integral part of the answer and ignoring the fractionalpart.

Ex. What is the highest power of 5 that divides of x = 100!= 100 × 99 × 98 × ...... × 3 × 2 × 1.

Sol. Calculating contributions of the different powers of 5,

we have 15

100= 20, 25

100 = 4.

Hence, the total contributions to the power of 5 is 24, orthe number 100! is divisible by 524.

The number system that we work in is called the�decimal system�. This is because there are 10 digitsin the system 0-9. There can be alternative system thatcan be used for arithmetic operations. Some of themost commonly used systems are : binary, octal andhexadecimal.These systems find applications in computing.Binary system has 2 digits : 0, 1.Octal system has 8 digits : 0, 1, 2..., 7.Hexadecimal system has 16 digits : 0, 1, 2,..., 9, A , B,C, D, E, F.

After 9, we use the letters to indicate digits. For instance,A has a value 10, B has a value 11, C has a value 12,...so on in all base systems.The counting sequences in each of the systems wouldbe different though they follow the same principle.Conversion : conversion of numbers from (i) decimalsystem to other base system. (ii) other base system todecimal system.(i) Conversion from base 10 to any other base :

Ex. Convert (122)10

to base 8 system.

Sol.

10

718

2158

1228

The number in decimal is consecutively divided by thenumber of the base to which we are converting thedecimal number. Then list down all the remainders inthe reverse sequence to get the number in that base.So, here (122)

10 = (172)

8.

(ii) Conversion from any other base to decimal system

Ex. Convert (231)8 into decimal system.

Sol. (231)8 , the value of the position of each of the numbers

( as in decimal system) is :

1 = 80 × 1

3 = 81 × 3

2 = 82 × 2

Hence, (231)8 = (80 × 1 + 81 × 3 + 82 × 2)

10

(231)8

= (1 + 24 + 128)10

(231)8

= (153)10

35PAGE # 35

MENSURATION

MENSURATION

Figure lying in a plane is called a plane figure.Closed figure in a plane covers some part of theplane, then magnitude of that part of the plane iscalled the area of that closed figure. The unit ofmeasurement of area is square unit (i.e. squarecentimeter, square metre etc.).

Triangle :

D

Perimeter = a + b + c

Area = 21

× Base × Height = 21

ah

Heron�s formula :

Area = )cs)(bs)(as(s

Where, s = semi-perimeter = 2

cba

Area of equilateral triangle = 4

3 (side)2

Rectangle :

Perimeter = 2 ( + b)Area = × b

Length of diagonal = 22 b

Square :

Perimeter = 4aArea = a2

Length of diagonal = a 2

Parallelogram :

Perimeter = 2 (a + b)

Area = ah1 = bh2

Rhombus :

Perimeter = 4a = 2 22

21 dd

Area = 21

d1d2

Quadrilateral :

Let AC = d

Area = 21

d (h1 + h2)

Trapezium :

h

b

a

D

A B

C

Area = 21

h (a + b)


36PAGE # 36

AREA RELATED TO CIRCLE

Circle :Circle is a path of a point, which moves in such amanner that its distance from a fixed point is alwaysequal. The fixed point is called centre of the circleand the fixed distance is called radius of the circle.

Circle

r

C

Area of circle (A) = r2

Circumference (C) = 2 rDiameter (D) = 2r

Results :(i) If two circles touch internally, then the distancebetween their centres is equal to the difference oftheir radii.

(ii) If two circles touch externally, then the distancebetween their centres is equal to the sum of theirradii.

(iii) Distance moved by a rotating wheel in onerevolution is equal to the circumference of the wheel.

(iv) Number of revolutions completed by a rotatingwheel in one minute

=nceCircumfere

uteminoneinmovedcenatDis.

(v) Angle described by minute hand in one minute =6º.

(vi) Angle described by hour hand in one hour =30º.

Semicircle :

Semi-Circle

rC

r

Perimeter = r + 2r = ( + 2) r

Area (A) = 2r2

Sector :

Area (A) = º360

r2

Length of arc () = º360

r2

Area (A) = 21

× × r

Perimeter = + 2r

MENSURATION (SOLID FIGURES)

If any figure such as cuboid, which has three

dimensions length, width and height are known as

three dimensional figures.

Some of the main solid figures are :

Cuboid :

Total Surface Area (T.S.A.) : The area of surface

from which cuboid is formed. There are six faces

(rectangular), eight vertices and twelve edges in a

cuboid.

(i) Total Surface Area (T.S.A.)

= 2 [ × b + b × h + h × ]

(ii) Lateral Surface Area (L.S.A.)

(or Area of 4 walls)

= 2 [b × h + h × ]

= 2 h [ + b]

(iii) Volume of Cuboid

= (Area of base) × height

= × b × h

(iv) Length of Diagonal = 222 hb

Cube :

Cube has six faces. Each face is a square.

(i) T.S.A. = 2 [x x + x x + x x]

= 2 [x2 + x2 + x2] = 2 (3x2) = 6x2

(ii) L.S.A. = 2 [x2 + x2] = 4x2

(iii) Volume = (Area of base) × Height

= (x2) x = x3

(iv) Length of Diagonal = x 3

37PAGE # 37

Cylinder :

(i) C.S.A. of Cylinder = (2 r) × h = 2 rh.

(ii) Total Surface Area (T.S.A.) : T.S.A. = C.S.A. + Area of circular top & bottom

= 2 rh + 2 r2

= 2 r (h + r)

(iii) Volume of Cylinder :Volume = Area of base × height

= ( r2) × h

= r2hCone :

r

h

(i) C.S.A. = r

(ii) T.S.A. = C.S.A. + base area= r + r2

= r ( + r)

(iii) Volume = 31

r2h

Where, h = height r = radius of base = slant height

Sphere :

T.S.A. = S.A. = 4 r2

Volume =34

r3

Hemisphere :

C.S.A. = 2 r2

T.S.A. = C.S.A. + base area = 2 r2 + r2

= 3 r2

Volume = 32

r3

Volume of frustum= )rRRr(h3

22

Lateral Surface area= )Rr(l

Total surface area= 22 Rr)Rr(l

222 )rR(hl

Volume = Base area × height

Lateral surface area = perimeter of the base × height

Volume = 31

× base area × height

Lateral surface area

= 21

× perimeter of the base × slant height

38PAGE # 38

Total surface area = lateral surface area + base area

Kvpy Mathematics

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