Maths Concept Map_x Cbse

Concept Maps Class X Real Numbers Polynomials Linear equations in two variables Quadratic equation Airthmetic Progression Similar Triangles Height & Distance Tengent Coordinate Geometry Trignometry Construction Area related to circle Surface area & volume Probability Statistics

Real Numbers Real Numbers

p,

1,1/2,–7/5...

2

Irrational Numbers

Rational Numbers

89

Integers Non-Integers

Terminating

.... ...............

Non-Terminating

....

-3

-2

-1

0

1

2

3

.... Application H.C.F

3 22 ! 5

L.C.M

10 3

Euclid's division lemma

Definition

(p ! q ! r ) ! H.C.F(p, q, r ) H.C.F.(p, q) ! H.C.F.(q, r ) ! H.C.F(p, r ) H.C.F. (p,q,r) =

(p ! q ! r ) ! L.C.M.(p, q, r ) L.C.M.(p, q) ! L.C.M.(q, r ) ! L.C.M.(p, r )

Polynomials Polynomials 2

F(x) = ax + bx + c F(x) = ax3 + bx2 + cx + d Remainder theorem

Value of polynomial

Degree of polynomial

Factor theorem

Replace 'x' with 'k' ® P(k)

Highest power of x in polynomial

If P(k) = 0 Dividend = divisor ´ quotient + remainder

Then 'k' is a root or zero of the polynomial.

Geometrical interpretation of roots of zeroes of polynomial.

" coefficien t of x n"1

b a

coefficien t of xn cons tan t term coefficien t of x n

Y

Y

1 root

2 roots

Y

1 root

0

X

Y

No roots

0

X

Y

3 roots

0 X'

0

Y'

X

X'

0

Y'

X

X'

Y'

X'

Y'

X'

X

Y'

d a

Linear equations in two variables Linear equation in two variables a1x + b1y + c1 = 0 a2x + b2y + c2 = 0

..(1) ..(2)

Equation of a straight line

Equations reducible to a pair of linear equations

) a2 + b 2 * 0 ## ax + by + c = 0 ® ( a * 0,b * 0 # a, b, c are real nos. '#

Methods to solve

& # # % # # $

2 3 5 4 + , 13, " , "2 x y x y

Let

y = 2x – 4

Y

2p + 3q = 13 5p – 4q = –2

1 1 , p, , q x y

Solution (x,y) ® point lying on straight line. Algebric method

Graphical method a1 b1 * a2 b2

Intersecting (intersect at 1 point)

2 X'

0

X

-4

one solution (consistent)

a1 b1 c1 , , - Infinite a 2 b2 c 2

Coincident (Coincide)

solution

Y' a1 b1 c1 , * - No solution a2 b2 c 2

Parallel (No intersection)

Substitution method " c1 " b1y x, {Substitute in ( 2)} a1 " a2c1 " a2b1y + a1b 2 y , " c 2a1

x,

b1c 2 " b2c1 a1b2 " a2b1

y,

a2c1 " a1c 2 a1b2 " a2b1

Elimination method Multiply b2 in (1) & b1 in (2) a1b2x + b1b2y + c1 b2 = 0 ..(3) b1 a2x + b2b1y + c2b1 = 0 ..(4) (3) -(4) .... (a1b2-b1a2)x + (c1b2-c2b1) = 0 x,

b1c 2 " b2c1 a1b2 " a2b1

y,

a2c1 " a1c 2 a1b2 " a2b1

(inconsistent)

Cross multiplication method y x 1 b1 a1 c1

b1

b2

b2

a2

c2

x y 1 , , b1c 2 " b2c1 c1a2 " a1c 2 a1b2 " a2b1 x,

b1c 2 " b2c1 a1b2 " a2b1

y,

a2c1 " a1c 2 a1b2 " a2b1

Quadratic equation Methods to solve

Quadratic Formula

Completing the square

Factorization

Convert to (x+a)2-b2 = 0

e.g. x2 – 3x – 4 = 0 2 Þ x + (–4 + 1)x + (–4´1) = 0 Þ (x – 4) (x + 1) = 0 Product of roots = –4 Sum of roots = 3

e.g. x2 + 4x = 0 Þ x2 + 4x + 4 – 4 = 0 Þ (x + 2)2 – 22 = 0

. ,

" b + (b2 " 4ac ) 2a

/,

" b " (b2 " 4ac ) 2a

Y

Roots Standard form

X'

0

X

(b2– 4ac ³ 0 ® real roots)

Y' Y

2

ax + bx + c = 0 ; 2

ax +bx + c = 0 ; a ¹ 0 p(x) = 0 "where p(x) is polynomial of degree 2" e.g. (x + 2)3 = x3 – 4 Þ x2 + 4x + 2 = 0 (after simplification)

If a and b are the roots of the above equation then : aa2 + ba + c = 0 ab2 + bb + c = 0

X'

0

X

(b2– 4ac = 0 ® equal roots)

Y'

"b

Y

\a+ b = c a ab = a

X'

APPLICATIONS 1. Speed =

OF QUADRATIC EQUATIONS

Dis tan ce Time

2. Area of figures 3. Flow rate ´ time = volume of water 4. Number or ages

0

: Y'

X

(b2– 4ac < 0 ® Complex roots)

Airthmetic Progression Note : Taken 3 terms in A.P. : (a – d), a, (a + d)

Terms Properties

Numbers in the series are called Terms.

Common difference

Taken 5 terms in A.P. : (a – 2d), (a – d), a, (a + d), (a + 2d)

Common difference 'd' can be zero, positive, negative. a2 – a1 = a3 – a2 = a4 – a3 = a5 – a4 = an-an-1 = d. 5 – 2 = 8 – 5 = 11 – 8 = 14 – 11 = 17 – 14 = 3.

a1 ® a1 = 2 ® 1st Term a2 ® a1 + d = 5 ® 2nd Term nth Term

Arithmetic Progressions

Tn = a + (n–1) d.

a3 ® a1 + 2d = 5 ® 3rd Term . . an ® a1 + (n–1)d = 74 ® 25th Term

Sum of n Terms

st

Sum of 1 n terms of an A.P. ® Sn Sn= a1 + a2 + a3 + .... + an- 1 + an. n(n + 1) Sn = 2

Sum of n natural nos.

RESULTS

2

(n – m + 1)th term

mth term from end

th

Find A.P. whose n term is given ? a1, a2, a3,..... an-1, an. form an A.P.

n

e.g. Sum of first 7! 8 , 28 7 natural nos. Þ

Find Tn when Sn is given

e.g. 2, 5, 8, 11, ..... up to 25 terms.

e.g. Tn = 3n + 5 put n = 1,2,3...... T1 = 8, T2 = 11, T3 = 14......

Tn = Sn – Sn-1 e.g. Sn = n2 + 2n \ Tn = 2n + 1

Condition of an A.P. If a, b, c are 3 terms of an A.P. then : a + c = 2b.

n

Sn= 2 [2a + (n-1)d] = 2 [a1 + an]. e.g. 25 S25 = 2 [2(2) + (25-1)3] = 950

Similar Triangles Similar Triangles

Means 'Same shape' e.g. All circles are similar. All squares are similar.

Criteria

If Ds are similar

Q

P

Then :

A

Area theorem

R

Pythagoras theorem

C

B

Q AA criterion : If ÐA = ÐP and ÐB = ÐQ then DABC ~ DPQR.

P

B

R C

B

A

If DABC ~ DPQR : 2

2

2

Area( 6 ABC) 5 AB 2 5 AC 2 5 BC 2 ,3 0 ,3 0 ,3 0 Area( 6 PQR ) 4 PQ 1 4 PR 1 4 QR 1

SSS criterion : AB

AC

BC

If PQ , PR , QR then DABC ~ DPQR

D

AB AC BC , , PQ PR QR

C

DABC is right angled D (ÐB = 90°) DABC ~ DBDC ~ DADB : BD2 = AD ´ BC. BC2 = CD ´ AC. AB2 = CA ´ AD. AB2 + BC2 = AC2 ® P.Th.

AB

BC

ÐB = ÐQ then DABC ~ DPQR

and

Proved by Basic Proportionality theorem (THALES THEOREM) A

SAS criterion :

C

ÐA = ÐP ÐB = ÐQ ÐC = ÐR

E

D

If P Q , Q R and

R

P

B

A

A

Q

If DE | | BC then, C

B

AD AE , DB EC

NOTE : DABC ~ DPQR doesn't mean DABC ~ DQPR.

Converse of B.P.T. : If

AD AE , DB EC

DE | | BC.

then

Height & Distance

Heights & Distance C

A

Elevation Horizontal A

q

B

Height of tower BC = AB ´ tan q (given AB & q)

Horizontal

Height of tower AB = tan q ´ BC (given a & BC)

Applications

Depression a C B a = Angle of Depression

q = Angle of elevation @ Navigation @ Land surveys @ Buildings @ Optics @ Statics @ Crystallography

Definition : Angle formed by the line of sight with the horizontal when the point is above the horizontal.

Definition : Angle formed by the line of sight with the horizontal when the point is below the horizontal.

e.g.

e.g. Height of tower AB =

b a

CD ! tan . ! tan / (tan / " tan . )

PD =

(given CD, a, b) a

b

AB ! tan / (tan . " tan / )

(given AB, a, b) b

e.g. Height of tower BD =

(tan / " tan . ) ! AB tan .

(given AB, a, b) a b

Height of tower

a

Tengent

QT= 1/2 QR

NO TANGENT

ONE TANGENT

P

Q

P

O

P

R

of tangents from an external point to a circle are equal . * Length

In DOQP & ORP = OQ = OR (radii), OP = OP \ DOQP 7 DORP Þ PQ = PR AND

TANGENT

In DPQR PQ=PR, hence ÐPQR = ÐPRQ = 1/2 (180 – q) Since ÐOQR = 90º, ÐOQR = ÐOQP – ÐPQR ÐOQR = 90º – (90 – 1/2q) =1/2q Þ2 ÐOQR = ÐQPR.

&ÐQPR + ÐQOR = 180º ÐQPO = ÐRPO ÐQOP = ÐROP

P Tangent

B

Coordinate Geometry

Distance formula

Area of Triangle

Section formula

Q (x2,y2)

y

y P (x1,y1)

A (x1,y1)

R (x1,O)

S (x2,O)

x

RS = (x2–x1) = PT SQ = y2 ; ST = y1 ; QT = (y2 – y1) Applying Pythagoras in DPTQ PQ2 = PT2 + QT2 = (x2 – x1)2 + (y2 – y1)2

C

S

R

O

( x 2 " x1)2 + ( y 2 " y1)2

(x2,y2) B

Q

T

x

O

Q

P

R

A(x1y1) B(x2,y2) C(x3,y3)

(m : n Internally)

{ m mx ++ mm x 1 2

1

2 1

2

, m1y 2 + m2 y1 m1 + m2

{

Area = 1/2 ´ base ´ Altitude

{

m1x 2 " m2 x1 , m1y 2 " m2 y1 m1 " m2 m1 " m2

Area = 1/2 [x1(y2 –y3) + x2(y3 –y1) +x3(y1-y2)]

{

If Area = 0,

(Distance formula)

Application of distance formula Verifying collinearity A(x1,y1), B(x2,y2), C(x3,y3) ® Find AB,BC,CA using distance formula ® If AB + BC = CA or BC + CA = AB or AB + CA = BC ® Then 3 pointare collinear

x

Coordinates of P(x–xy) =

(m : n externally) PQ =

A (x,y) C(x3,y3)

(x,y)

O

y

B (x2,y2)

T (x2,y1)

Verifying collinearity A(x1,y1), B(x2,y2), C(x3,y3), D(x4,y4) ® Find AB,BC,CD,DA using distance formula ® Find diagonals CA & BD ® Check with different properties of quadrilaterals

then points are collinear Verifying triangle formations A(x1,y1), B(x2,y2), C(x3,y3) ® Find AB,BC,CA using distance formula Check if AB + BC > CA BC + CA > AB AB + CA > BC ® If yes, it is triangle, check for right D using pythagoras therom ® If no, it is not a D

Complimentry Angles

Graphs

T-ratios sinA =

AB

cosecA = BC = P

B

AB

cosA = AC

H

AC

cosA = AC = H

tanA =

H

AC

BC P = AC H

secA = AB = B

BC P = AB B

cotA =

secA =

cos(90-A) =

AB B = BC P

sinA =

BC AC

BC AC

tanA =

AC

AB BC

cot(90-A) =

y

AC cosec(90-A)= AB

AB tan(90-A)= BC

AB sin(90-A) = AC

secA = AB BC AB

sec(90-A) =

BC AB

AC BC

x'

" 38 2

"8

8 2

"8 0 2

8

38 2

AC BC

cosecA =

x

TAN

y'

y C ÐA of DABC = ÐA of DAMP P BC MP sin A = = AC AP A Note ; M B The values of the trignometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.

Interrelationship between T-ratios

sinA = T R I G O N O M E TRY

1 cos ecA

cosA =

1 sec A

tanA =

1 cot A

C

90° B

A

2

Divide both sides by Identities

AC2 AB

BC2 AC2

2

BC2 1+

+

+ 1,

BC2 AB2

0.5 0

8 2

0.5

8

38 2

28

2

,

2

2

, 1 =sin A + cos A = 1

AC

2

BC2 AC2 AB2

2

2

=cot A+1 = cosec A

AC2

BC

2

2

= 1+tan2A = sec2A

AB

2

0°

30°

45°

60°

90°

sin

0 0

1 4

2 4

3 4

4 4

cos

4 4

3 4

2 4

1 4

0 4

tan

0 4

1 3

2 2

3 1

9

cosec

9

4 1

4 2

4 3

4 4

sec

4 4

4 3

4 2

4 1

9

cot

9

3 1

2 2

1 3

0 4

Simplified trigonometric values

x

-1

SIN

y'

AB + BC = AC

AB2

1

y 1 0.5 0

8 2

0.5

8

38 2

28

x

-1 y'

COS

Construction

Triangle Scaling

Division of a line segment in m:n(3:2)ratio

Constructing Tangents

SCALING A

B2

B3 B

4

A' A C'

B1

B5 x

B

A

B

C B1

Q

B5C'||B3C C'A'||CA

B

B

A

1. Scaling up 5/3

P

O2

B4

X

B5 x

2. Scaling down 3/4

B1

B2

B3 B

C

B

4

x

Alternate-Method

3. 2 circles intersect at Q & R

A' C B2

X

AC 3 AA 3 = = A 3 A 5 CB 2

2. O2 as centre & O1O2 as radius , draw a circle

A

B1

Join A3C such that A3C is parallel to A5B

DAA3C ~ DAA5B hence 1. Join PO1 and bisect it (at O2)

C

B

R

B3

5 BC' BB 5 = = 3 BC BB 3

B

2 C

C B2

AB AC BC = = A' B A ' C' BC'

B3C'||B4C C'A'||CA

3

O1

DABC ~ A'BC'

A

A

B3

B4

DABC ~ A'BC' A ' B A ' C' BC' 3 = = = AC BC 4 AB

B

A

4. Join PQ & PR (Tangents) , Y

B2

x

Y

A

B Join A3B2

A1 A2

A3

B2

B1 A

3

B

A1 A2

X DAA3C ~ DBB2C AA 3 AC 3 = = BB2 BC 2

B1

2

A3

X

Area related to circle

Sector of a circle

Major sector

Area of combination of plane figures

Q

e.g. q = 60° r = 3m

e.g.

2 units

A

B

2

Area = pR r

q

r Generally sector implies minor sector

B

A

C

Sol.

P Length of sector AB =

r

r + r + r + r = length of square Þ 4r = ; r = 1/2 Area of shaded reg. = Area of square – Area of 4 circles 2 = (2´2) – 4 ´ 8 ( 1 )

Area of sector : AOBP = ´ pR2 360

: ´ 2pR 360

e.g.

e.g.

60 ´ p ´ 32 360 38 Þ m 2

: ´ 2p ´ 3 360 Þ pm

D

4

AOBP =

AB =

=4–p

Note: For rotating wheel 2pRn = speed ´ Time = Distance covered n = number of rotations

e.g.

2 units

A

B

I

Area of segment (APB) = Area of sector (AOB) – Area of DAOB

IV

II III

e.g. A

C 2 units

Sol.

B

I IV

II III

C

D

Sol. Area I + III = A(ABCD) – Area of semicircle with AD & BC as diameter =4–p

D

Area of shaded reg. = Area of square ABCD –Area of (1+2+3+4) = 4 – 2(4 – p) = 2p – 4

Surface area & volume

Cuboid

Frustum Cylinder

Cube Cone

Hemisphere Sphere

l1-l r2

h

r a

b

h

h

1. T.S.A ® 2[lb +bh + hl] 2. C.S.A ® 2[bh +hl] 3. Volume ® l ´ b ´ h 4. Diagonal ® 2 + b2 + h2

a 2 1. T.S.A ® 6a 2 2. C.S.A ® 4a 3 3. Volume ® a 4. Diagonal ® 3a

NOTE : 1. C.S.A.® Curved surface area 2. T.S.A.® Total surface area 3. l ® Length 4. b ® Breadth 5. h ® Height 6. l ® Slant height 7. r ® Radius 8. a ® Side of cube

h

l

r1

r

l

a

h1

r

r r

r

2

1. T.S.A ® pr(l + r) 2. C.S.A ® prl 3. Volume ® 1 8 r 2h 3

1. T.S.A ® 2pr(r+h) 2. C.S.A ® 2prh 2 3. Volume ® pr h

Volume = Volume of cylinder – Volume of hemisphere

1. C.S.A = T.S.A. = 4pr2 2. Volume = 4/3pr3

1. C.S.A = 2pr 3 2. T.S.A. = 3pr 3. Volume = 2/3pr3

1. C.S.A = p1r1l1 – pr2(l1–l) 2 2 2. T.S.A. = pl(r1+r2) + p(r1 + r2 ) 2 2 3. Volume = 1/3ph(r1 + r2 + r1r2) 4. Slant height = h2 + (r1 " r2 )2

Total surface area = C.S.A. of 2 hemispheres + C.S.A of cylinder

Probability PROBABILITY Elementary event (E) : Only one outcome sum of the probabilities of all elementary events is 1.

Applications Gambling, insurance & statistics control theory.

0 £ P(E) £ 1

Complimentary Events(E) Not E = E, P(E) = 1 – P (E) ln case 1 :

No. of outcomes favourable to E P(E) = No. of all possible outcomes

EVENT(E) 1 P(1) = ; P(1) = 1 – 1 = 5 6 6 6 P(1) = not getting 1 ; = getting 2,3,4, 5 and 6

In case 1 P(E) > 4 =P(5) + P(6) =

2 1 = 6 3

P(E) = 4 = P(1) + P(2) + P(3) + P(4) =

4 2 = 6 3

In case 2 In case 3 Sum = 8 for (2,6) (3,5) (4,4) (5,3) (6,2) 5 P(sum 8) = 36

Probability of picking up black P(B) = 2 9

Certain event P(E) = 1 Impossible event P(E) = 0 In case 2 : P(sum 13) is impossible. hence, P (sum 13) = 0

1

2

36 posibilities 3

4

5

6

1

1,1

1,2

1,3

1,4

1,5

1,6

2

2,1

2,2

2,3

2,4

2,5

2,6

3

3,1

3,2

3,3

3,4

3,5

3,6

4

4,1

4,2

4,3

4,4

4,5

4,6

5

3,1

5,2

5,3

5,4

5,5

5,6

Examples Case -1) One dice is rolled: P(1) = Probability of getting 1. Similarly P(2),P(3), P(4), P(5) & P(6) P(1) =

1 1 , 1,2,3,4,5,6 6

P(E) > 4 ® Probability of getting 5 & 6.

Case -2) Two dice are rolled: P(sum = 8) = Probability of getting two numbers whose sum is 8.

Case -3) Box with balls : A box contains 2 black, 4 green and 4 red balls.

Red

Red

Red

Black

Black

Red

Green

Green

Green

Statistics

Grouped Class of Intervals NO. of Students

Ungrouped 10-25

25-40

40-55

55-70

70-85

85-100

2

3

7

6

6

6

MODE

MEAN

The value of the observation having the max. frequency

; fx x= ;f

i i i

MODE

MEDIAN Intervals

No. of students

C.f.

n-C.f.

10-25

2

2

28

25-40

3

5

25

40-55

7

12

18

55-70

6

18

12

70-85

6

24

6

85-100

6

30

0

Class size (h) = 15 Max. frequency f1 = 7 Modal class = 40-55 Lower limit of modal class = 40 f0 = 3 (Previous class f value) f2 = 6 (next class f value)

n / 2 " c.f . ! h = 62.5 Median = l + f

x1, x2, ........ xn ® observations f1, f2, ........ fn ® frequencies th

5n 2 Average of 3 + 10 42 1

5 f1 " f0 2 Mode = l + 33 2f " f " f 00! h 4 1 0 21

Step deviation

x =

; fx ;f

i i

Assumed mean method

x = a + hu = 62

5 x = a + h 33 4

i

Class Intervel

fi

xi

10-25

2

25-40

3

40-55

; f d 20 = 62 ; f 01

i i i

di = x i – a

17.5

35

-30

32.5

97.5

-15

7

47.5

332.5

55-70

6

62.5

70-85

6

85-100

6 30

x = a + d 5 x = a + 3 3 4

; f x 20 =62 ; f 01

fi x i

xi " a h

i i i

fi ui

fi di

-2

-4

-60

-1

-3

-45

0

0

0

0

375

15

1

6

90

77.5

465

30

2

12

180

92.5

555

45

3

18

1860

ui =

n is even

th

3 Methods

Direct Method

th

5n 2 & 3 0 observation. 42 1 5 n + 12 3 0 observation. 4 2 1

Mode = 40 + 4/5 ´15 = 52 MEAN

n = 30 ; n/2 = 15 55-70 is median class lower limit of the median class (l) = 55 c.f. = cum frequency of median preceeding class

MEDIAN

;

f iu i = 29

270

;

fi di

= 435

n is odd