Maths Concept Map 9 CBSE

July 12, 2017 | Author: Nilesh Gupta | Category: Triangle, Euclid, Classical Geometry, Elementary Geometry, Euclidean Plane Geometry
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Concept Maps Class IX Number System Polynomials Lines & Angles Congruent Triangles Perimeter & Area of D Application of Parallelogram Area of parallogram & Triangle Circles

Surface Area and Volume Statistics Probability

Number System

a,b Î R ; a < b n rational no between integer a=a´

Number System

(n $ 1) (n $ 1) &b=b´ (n $ 1) (n $ 1)

Rational no between 2 integral no.

x=

Imaginary No. (Not represented on number line)

Real No. (Represented on no. line)

p q ¹0

q is not in 2m ´ 5n e.g:- 5

Rational No.(R) p X = q ; q ¹ 0 ; p, q Î I

11 p x= q¹0

Insertion of irrational between 2 real no. a,b Î Real Irrational = a ´ b

Irrational No. p

Cannot be q ! 0

; e.g . 2 , p Surds

q is in form 2m ´ 5n e.g:-

3 3 = 20 22 # 51

Integers (I or Z) {eg. -3,-2,-1,0,1,2,...}

Negative

Positive Whole No.(W) {eg:- 0,1,2,3.....}

Non-Integers {eg.

"1 3 , } 2 2

Natural No. (N) {eg:- 1,2,3.....} Prime No. {Exactly have 2 distinct factors 1 & itself. e.g. 2,3,5,7,11....}

Laws of Surds

Non-terminating & non-recuring (eg. 3.14285763)

(iii) n a ¸ n b =

Composite No. {More than 2 distinct factors e.g. 4,9,15....}

One (Exception) Neither Prime nor Composite

Non-terminating but recurring Conversion : Decimal (i) x = 0. 6 = 0.666.. (ii) Multiply by 10 10x = 6.666..... (iii) Subtract (ii),(i) 9x = 6

2 (iv) x = = 0. 6 3

na

(i) n an = a ; (ii)

Decimal No eg. 3.65, 4.67 Terminating eg. 3.65

Irrational of form n a a= Radicand n is radical sign n = order

Fraction

n

´ n b = n ab

a b

(iv)

n m

(v)

n

a = n´p ap

(vi)

n

am

a = nm a

= n´p am´p

(a) Addition & subtraction eg. 32 + 18 – 50 (b) Multiply & divide eg. 3 2 ´ 4 3 ¸ 5 10 (c) Comparison eg. 3 6 > or < 5 8 (d) Rationalization eg.

1 7$ 5

#

7" 5 7" 5

Polynomials

2

(i) Zero degree ; f(x) = ax°, a ¹ 0 (ii) Linear ; f(x) = ax + b ; a ¹ 0 2 (iii) Quadratic ; f(x) = ax + bx + c ; a ¹ 0 3 2 (iv) Cubic ; f(x) = ax + bx + cx + d ; a ¹ 0 4 3 2 (v) Biquadratic ; f(x) = ax + bx + cx + dx + e; a ¹ 0

(i) P(x) = x1; degree 1 3 2 (ii) P(x) = x - 5x ; degree 3 (iii) P(x) = x4 + 4x2; degree 4 Highest power of x in polynomial

3

(i) Monomial = 1 term; e g:- ax , 5x (ii) Binomial = 2 terms; e g:- 2x2 + 3x (iii) Trinomial = 3 terms; e g:- 2x2 + 3x + 5

Based on number of terms

Degree of Polynomial Types of polynomial Remainder Theorem {If P(x) ¸ (x - a) ; Rem = P(a)}

Based on degree Types of polynomial Value of polynomial . 2 e g. f(x) = x - 3x + 4, value at x = 2 2 f(2) = 2 - 3 ´ 2 + 4 = 2

Polynomials F(x) = a0 + a1x + a2x2 + ..... +anxn

Zero or root of polynomial . 2 2 e g. f(x) = 2x + x + 7x - 6, value at x = 2 f(2) = 0, x = 2 is a root or zero

Factor Theorem {If P(x) ¸ (x - a) ; Rem = P(a) = 0}

Algebraic Identities

Factors of Polynomials (i) By taking common factor eg. = 2a(x + y) - 3b(x + y) (ii) By grouping eg. (x2 + 3x)2 - 5(x2 + 3x) - y(x2 + 3x) + 5y = (x2 + 3x) (x2 + 3x - 5) - y (x2 + 3x -5) (iii) By making perfect square eg. 4a2 + 12ab + 9b2 (iv) Difference of 2 squares 8 8 eg. x - y = (x4)2- (y4)2 (v) By spliting middle term eg. x2 + 6 2x + 10 2 = x + 5 2x + 2x + 10

(i) (a + b)2 = a + 2ab + b2 2 2 2 (ii) (a - b) = a - 2ab + b (iii) a2 - b2 = (a + b) (a - b) 2 2 2 2 (iv) (a + b + c) = a + b + c + 2ab + 2bc + 2ca 3 3 2 2 (v) a + b = (a + b) (a - ab + b ) 3 3 2 (vi) a + b = (a - b) (a + ab + b2) 3 3 3 (vii) (a + b) = a + b + 3ab(a + b) (viii) (a - b)3 = a3 - b3 - 3ab(a + b) (ix) a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca) Special case : If a + b + c = 0, then 3 3 3 a + b + c = 3abc (a2 " b2 )3 $ (b2 " c 2 )3 $ (c 2 " a2 )3 (a " b)3 $ (b " c )3 $ (c " a)3 Here, (a2– b2) + (b2 – c2) + (c2 – a2) = 0 & (a – b) + (b – c) + (c – a) = 0

e.g.

Lines & Angles (vi) Vertically opp. angles :B

D

ÐAOC = ÐBOD ÐAOD = ÐBOC

(vii) Zero Ð :- q = 0º (vi) Complete Ð :- q = 360º (v) Reflex Ð : - 180º q < 360º (iv) Straight Ð : - q = 180º

C

A

(v) Linear Pair :- q1 + q2 = 180º (q1 & q2 are adjacent)

(iii) Obtuse Ð : - 90º < q < 180º

(iv) Adjacent angles :-

(ii) Acute Ð : - 0 < q < 90º

{

(a) they have the same vertex, (b) they have a common arm, (c) non common arms are on either

(i) Right Ð : - q = 90º

side of the common arm. B X

Types of Angles

(iii) Angle Bisector :-

A ÐAOX = ÐBOX

O

(ii) Supplementary angles :- q1 + q2 = 180º

Lines & Angles

(i) Complementary angles :- q1 + q2 = 90º Pair of Ð's (q1 & q2) Types of Triangles

Properties

Angles made by Transveral Basis of Sides

Q C A

6 5 7 8 2 1 3 4

D

(i) Scalene (all sides diff.)

(i) Right (q = 90º)

(i) Sum of interior angles of D = 180º 1 (ii) ÐBOC = 90 + ÐA 2

B

(ii) Isoceles (2 sides equal)

(ii) Acute (0 < q < 90º)

(iii) Ext. Ð = Sum of 2 interior opp. Ð's.

(iii) Equilateral (all sides equal)

(iii) Obtuse (90º < q < 180º)

P (i) Corresponding angles (Ð1 & Ð5, Ð2 & Ð6 , Ð3 & Ð7, Ð4 & Ð8) (ii) Alternate interior angles (Ð1 & Ð7, Ð2 & Ð8) (iii) Co-interior angles (Ð1 + Ð8 = 180°= Ð2 + Ð7) (iv) Alternate exterior angle (Ð4 & Ð6, Ð3 &Ð5)

Basis of angles

(iii) ÐBOC = 90 –

1 ÐA 2

Congruent Triangles

Properties

Criterion (i) SAS (Side Angle Side)

(i) Angle opposite to equal sides of an isosceles angle are equal.

(ii) ASA (Angle Side Angle)

(ii) Triangle is isosceles, if bisector of vertical angle bisects the base of triangle.

(iii) AAS (Angle Angle Side)

Congruent Triangles

(iv) SSS (Side Side Side) (v) RHS (Right Hypolenuese Side)

Inequalities (i) a + b > c, b + c > a, c + a > b

A

(ii) a – b < c, b – c < a, c– a < b b

c B

a

(iii) If a > b, then ÐA > ÐB & vice versa C

e.g. Any 2 side of D are greater than twice the median drawn to third side. In DABC and AD is median so, AB + AC > 2AD.

Perimeter & Area of D A Area of D = S ( S " a )( S " b )( S " c )

b

c

S(semi-perimeter) = B

a$b$c 2

C

a

Heron's Formula

Perimeter & Area of D (a) Right angled D A

(b) Isosceles right angled D A

(c) Equilateral D A

(b2 = a2 + c2) b

b

c

a B

B

a P=a+b+c 1 Area = a ´ c 2

a

a

(b2 = 2a2)

C

B

a P = 2a + b Area =

1 2 a 2

a/2 a

C

C P = 3a 1 a´h 2 = 3 a2 4

Area =

h=

3 a 2

Application of Parallelogram

Isosceles Trapezium

Trapezium

{

(i) One pair of opposite side is parallel.

(i) non-parallel sides are equal (ii) 2 pair of adjacent angles are equal.

}

Rhombus (i) Diagonals bisect each other at 90° (ii) All sides are equal

{ Each angle less than 180°} Convex quadrilaterals

Quadrilaterals

Application of Parallelogram Special Quadrilaterals

Parallelogram Both pair of opposite sides are llel e.g. AB||CD, AD||BC Properties (i) Diagonal divide it into 2 cong. D

Concave quadrilaterals angle is move } { One than 180°

Rectangle

{

(i) Diagonals are equal (ii) Each angle is equal to 90°

(ii) Opposite sides are equal (iii) Opposite angles are equal (vi) Diagonals bisect each other

}

{ {

(a) Mid point theorem A (i) DE || BC & 1 E D (ii) DE = BC 2 C

B

(b) Converse of mid point theorem A PQ||BC P

(i) Q is mid 1 point of AC i.e AQ = CQ 2 (ii) PQ = BC

Q C

B

Square Kite Two pair of adjacent side are equal

(i) All sides are equal

E.g:-.

A E

(ii) Each angle is equal is 90° (iii) Diagonals are equal & bisects each other at 90°

}

F B

D

BE is median BE || DF AE = CE

C

So, CF =

1 AC 4

}

Area of parallogram & Triangle Area of IIgm

Area of D Properties

Properties (i) A diagonal of II gm divides it into 2 D's of equal area

(i) Two D's on the same base and between same parallels are equal in area.

Area of IIgm & Triangle C

D

D Hint : DABD

A

(i) Area of trapezium is half the product of its height & sum of Ilel sides.

(ii) Parallelogram on same base and between same parallel are equal in area.

D

C Hint : DAFD

A

D

b

C

ar.(trap ADCD) 1 % h (a $ b ) 2

h a

N

(ii) Median of triangle divides it into 2 triangles of equal area. A

B ar (D ABD) = ar. (D ACD ) B

B

D

M

C

D

Eg : - ABC is D in which D is mid-pt of BC and E is mid-Point of AD. PT. or D BED = 1/4 ar. D ABC A

C Ar. (IIgm ABCD) = Ar(rect. ABML)

A

C

B

DBCE

(iii) Area of IIgm is product of its base & corresponding alt.

L

L

A

E

Q ar(DABC) = ar (DPBC)

DBDC

B

F

A

P

E

Hint : AD is median of D ADC & BE is median of DABD

B B

D

C

Circles

Region PQ is segment

Major

CIRCLES Segment

O

Chord

Q

P

P

(i) Collection of all points ina plane. (ii) At fixed distance from a fixed pt in plane

minor Major O

Region POQ is sector Q

P

PQ is chord Q

major PQ is arc

Arc

Sector

P

minor

Q minor

Cydic Quadrilteral D

C

A

B

ABCD is cyclic as all vertices touches the circle.

Properties (i) Equal chord of a circle subtend equal angles at the centre.

O Property (i) Sum of either pair of opposite angles of cydic quad. is 180º D

A

D B

(ii)

(i) Angle subtended at the centre is double the subtended on remaining put of circle. A

AB = CD So, Ð AOB = COD

Q

(ii) Angles is the same segment of a circle are equal. A

B m

B

OM AB So, AM=BM

O A

ÐPOQ = 2 ÐPAQ

O P

C

from centre of a circle bisectsthe chord.

C ÐA + ÐC = 180º ÐB + ÐD = 180º

A

Angle subtended by an arc

B

ÐPAQ = ÐPBQ Q

P

Eg :- Cyclic IIgm is rectangle (iii) Equal chord equi - distant from centre. A

D

B ÐA + ÐC = 180º ÐB + ÐD = 180º ÐA = ÐC & C ÐB = ÐD So, each angle is 90º

(iii) Angle in semi-cirde is right

C

A O M

N D

B

C

AB = CD So, OM = ON A

O

AB is diameter ÐACB = 90º B

Surface Area and Volume Length of diagonal = l2 + b2 + h2

h l

C.S.A = 2prh h

r 2

T.S.A = 2prh + 2pr = 2pr(r + h)

r

Volume = pr2h

b

T.S.A = 2(lb +bh + hl) C.S.A = 2h(l + b) Volume = l × b × h

T.S.A = C.S.A = 4pr2 Volume %

Cylinder

4 3 &r 3

Sphere

Cuboid

SURFACE AREA AND VOLUME

Hemi - Sphere r

Cube Cone

C.S.A = 2pr2 T.S.A = 2pr2 + pr2 = 3pr2

x

Volume %

x x T.S.A = 2(x2 + x2 + x2) = 6x2 2 C.S.A = 4x 3 Volume = x Length of diagonal = x 3

2 3 &r 3

l

h

C.S.A = prl Hollow hemi-sprere R

r 2

T.S.A = prl + pr = pr(l + r) 1 2 Volume % &r h 3

l = h2 + r2 (Slant height)

Volume %

C.S.A = 2p(R2 +r2) 2 2 T.S.A = 2p(R +r ) 2 2 + p(R – r )

2 & (R 3 " r 3 ) 3

y

Definitions (i) Range = U. L– L. L

CI

Range (ii) C I % No. of classes (iii) Discrete : Fixed magnitude eg :- 10, 20, 25 (iv) Continuous = Magnitude not fixed. eg :- 10 – 20 , 20 – 30

18 16 14 12 10 8 4 2

(iii) Frequency Polygon

Statistics

Freq.

20 - 30 30 - 40 40 - 50

8 12 17

50 - 60

9

Classification

Collection of Data (i) Primary Data : Data collected himself

0

Frequency Distribution

Cumulative Frequency

(i) Discrete freq.

(i)Discrete freq.

10

20 30 40 50 60

x

Graphical Representation y

(ii) Secondary Data : Data cellected by Person or society.

Measure of Centre Tendency

Wages

No.

Wages

4000 6000

10 8

4000 6000

10 8

10 18

8000 11000

5

8000 11000

5

23

7

30

7

No. CF

(i) Bar graph Heads

Erpenditue

Gracery Rent Medicine Fuel

4 5 2 2

7 6 5 4 3 2 1

(i) Mean %

Sum of observatio n No. of observatio n

(ii) Continuous freq. CI

(ii) Median = middle observation n = odd

n = even th

, n $ 1) Median % * ' + 2 (

th

th

,n ) , n $ 1) * ' $ * ' 2( 2 ( + + Median % 2

72 103 50

15 - 20

25

G

R

M

F

(a) Less than

No.

0-5 5 - 10 10 - 15

O

(ii) Continuous freq. Less Than

CF

0-5 5 - 10 10 - 15

5 10

72 175

15

225

15 - 20

20

250

CI

(b) More than CF

0-5 5 - 10 10 - 15

0 5 10

250 178

15 - 20

15

25

CI

Empirical Relation b/w mode, median, mean Mode = 3 median – 2 mean

CI

No.

0-5 5 - 10 10 - 15

72 103 50

15 - 20

25

125 100 75 50 25

More Than

(iii) Mode = observation with max. freq.

y

(ii) Histogram

75

0

5 10 15 20

x

x

Probability

Examples

Experiment

(i) Deterministic Exp : Produce same outcome.

Probability of any event A, i.e 0 - P (A) - 1

(ii) Random Exp :Don't produce same outcome.

If, P(A) = 0, Impossible event If, P(A) = 1, Sure event & P(A) + P(A) = 1

(i) 2 coins tossed 500 times, we get 2H = 105 times 1 H = 275 times 0 H = 120 times 105 P ( 2H ) % 500

P (1 H ) %

275 500

P (0 H ) %

120 500

Sample Space

Probability Set of all possible outcomes. eg :- Dice is tossed S = {1,2,3,4,5,6} Probability of happening an event A, i.e P( A ) %

Total Favorable. outcome Total possible outcome.

eg :- Probability of geting even no. on tossing a dice Event Subset of Sample Space eg :- Even no. when dice is tossed E = {2,4, 6}

P(Even) %

3 1 % 6 2

(ii) 1500 families with 2 children selected randomly No. of girls 2 1 0 No. of family 475 814 211 P( 2 girls) % P(1 girl) %

475 1500

814 1500

P ( a t le a st 1 g irl ) %

814 $ 475 1500

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