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
Irrational Numbers Rational Numbers
Terminating Non-Integers
Non-Terminating
Integers
Euclid's division lemma Definition
Application H.C.F L.C.M
Real Numbers
p,
2
1,1/2,
–
7/5...
5 2 3 2 ! 103 0 1 -3 -2 -1 2 3 .... .... ... .... 89 ) r , p ( F . C . H ) r , q .( F . C . H ) q , p .( F . C . H ) r , q , p ( F . C . H ) r q p ( ! ! ! ! ! H.C.F. (p,q,r) = ) r , p .( M . C . L ) r , q .( M . C . L ) q , p .( M . C . L ) r , q , p .( M . C . L ) r q p ( ! ! ! ! !Polynomials
Polynomials
F(x) = ax + bx + c F(x) = ax + bx + cx + d 2 3 2Value of polynomial Remainder theorem
Factor theorem Degree of polynomial
Highest power of x in polynomial
Dividend = divisor´quotient + remainder
Replace 'x' with 'k'®P(k)
Then 'k' is a root or zero of the polynomial.
Geometrical interpretation of roots of zeroes of polynomial. If P(k) = 0 n 1 n x of t coefficien x of t coefficien " " a b n x of t coefficien term t tan cons a d
1 root 2 roots 1 root No roots 3 roots
0 0 0 0 0 X' X' X' X' X' X X X X X Y Y Y Y Y Y' Y' Y' Y' Y'
Linear equations in two variables
Methods to solve Linear equation in two variables a x + b y + c = 0 ..(1) a x + b y + c = 0 ..(2) 1 1 1 2 2 2 Equation of a straight lineAlgebric method Graphical method
Substitution method Elimination method
Intersecting
Coincident
Parallel
Cross multiplication method Equations reducible to a pair of linear equations ax + by + c = 0 ® Let Multiply b in (1) & b in (2) a b x + b b y + c b = 0 ..(3) b a x + b b y + c b = 0 ..(4) (3) -(4) .... (a b -b a )x + (c b -c b ) = 0 2 1 1 2 1 2 1 2 1 2 2 1 2 1 1 2 1 2 1 2 2 1 (intersect at 1 point) one solution (consistent) Infinite solution No solution (inconsistent) (Coincide) (No intersection) b1 c1 a1 b1 b2 c2 a2 b2
Solution (x,y) point lying on straight line. ® # # $ # # % & # # ' # # ( ) * * * + . nos real are c , b , a 0 b , 0 a 0 b a2 2 0 X' X Y Y' -4 2 y = 2x– 4 1 2 2 1 1 2 1 2 1 1 1 a c y b a y b a c a )} 2 ( in Substitute { a y b c x " , + " " " " , 1 2 2 1 1 2 2 1 b a b a c b c b x " " , 1 2 2 1 1 2 2 1 b a b a c b c b x " " , 1 2 2 1 1 2 2 1 b a b a c b c b x " " , 1 2 2 1 2 1 1 2 b a b a c a c a y " " , 1 2 2 1 2 1 1 2 b a b a c a c a y " " , 1 2 2 1 2 1 1 2 b a b a c a c a y " " , 2p + 3q = 13 5p – 4q = –2 1 2 2 1 2 1 2 1 1 2 2 1 ab ab 1 c a a c y c b c b x " , " , " x y 1 2 y 4 x 5 , 13 y 3 x 2 " , " , + q y 1 , p x 1 , , -* 2 1 2 1 b b a a -, , 2 1 2 1 2 1 c c b b a a -* , 2 1 2 1 2 1 c c b b a a
Quadratic equation
Methods to solve Factorization Quadratic Formula Completing the square Standard form Roots Convert to (x+a) -b = 02 2 Product of roots = –4 Sum of roots = 3 e.g. x + 4x = 0 x + 4x + 4 – 4 = 0 (x + 2) – 2 = 0 2 2 2 2 Þ Þ e.g. x – 3x – 4 = 0 x + (–4 + 1)x + (–4 1) = 0 (x – 4) (x + 1) = 0 2 2 Þ ´ Þ a 2 ) ac 4 b ( b+ 2" " , . a 2 ) ac 4 b ( b" 2" " , / ax +bx + c = 0 ; a p(x) = 0"where p(x) is polynomial of degree 2" 2 ¹ 0 ax + bx + c = 0 ;2 \ a+ b = ab= e.g. (x + 2) = x – 4 x + 4x + 2 = 0 (after simplification) 3 3 2 Þ
If and are the roots of the above equation then : a + b + c = 0 a + b + c = 0 a b a a b b 2 2 a b " a c 0 0 0
PPLICATIONS OF QUADRATIC EQUATIONS
A :
1. Speed =
2. Area of figures
3. Flow rate time = volume of water
4. Number or ages ´
X' (b – 4ac 0 real roots)
2
³ ®
(b – 4ac = 02 ® equal roots)
(b – 4ac < 02 ® Complex roots) X' X' X X X Y Y Y Y' Y' Y' Time ce tan Dis
Arithmetic
Progressions
Note :
Taken 3 terms in A.P. :
Taken terms in A.P. :
(a – d), a, (a + d) 5
(a – 2d), (a – d), a, (a + d), (a + 2d)
Numbers in the series are called Terms.
T = a + (n–1) d.n Sum of 1 n terms of an A.P. S S = a + a + a + .... + a + a . S = [2a + (n-1)d] = [a + a ]. [2 + ( -1) ] = st ® n n 1 2 3 n- 1 n n 1 n e.g. S25 = (2) 25 3 950 a1®a = 21 ®1 Term st a2®a + d = 51 ®2 Term nd a a + 2d = 5 3 Term . . a a + (n–1)d = 74 25 Term 3 1 n 1 ® ® ® ® rd th
Common difference 'd' can be zero, positive, negative. a – a = a – a = a – a = a – a = a -a = d. 5 – 2 = 8 – 5 = 11 – 8 = 14 – 11 = 17 – 14 = 3. 2 1 3 2 4 3 5 4 n n-1 Properties n Termth Sum of n Terms RESULTS Terms
Sum of n natural nos.
S =n
e.g. Sum of first 7 natural nos. Þ (n–m + 1) termth e.g. T = 3n + 5 put n = 1,2,3... T = 8, T = 11, T = 14... n 1 2 3 T = Sn n–Sn-1 e.g. S = n + 2n T = 2n + 1 n n 2 \ If a, b, c are 3 terms of an A.P. then : a + c = 2b. Common difference
m term from endth
Find A.P. whose n term is given ?
th
Find T when S is givenn n
Condition of an A.P. a , a , a ,... a , a . form an A.P. 1 2 3 n-1 n e.g. 2, 5, 8, 11, ... up to 25 terms. 2 n 2 n 2 25 2 ) 1 n ( n + 28 2 8 7 , !
Airthmetic Progression
Similar Triangles
Similar
Triangles
Criteria
If s are similar
D
Pythagoras
theorem
Area
theorem
Means 'Same shape' e.g. All circles are similar.
All squares are similar.
NOTE :
ABC ~ PQR doesn't mean ABC ~ QPR. D D D D Then : P P P Q Q Q R R R A A A B B B C C B A D C C AA criterion : If A = P and B = Q then ABC ~ PQR. Ð Ð Ð Ð D D SSS criterion : If then ABC ~ PQR D D If ABC ~ PQR :D D ( ) : D D Ð D D D ´ ´ ´
ABC is righ angledt B = 90°
ABC ~ BDC ~ ADB BD = AD BC. BC = CD AC. AB = CA AD. AB + BC = AC ® P.Th. 2 2 2 2 2 2 SAS criterion : If and B = Q then ABC ~ PQR Ð Ð D D QR BC PR AC PQ AB , , Q R B C P Q A B , 2 2 2 QR BC PR AC PQ AB ) PQR ( Area ) ABC ( Area 0 1 2 3 4 5 , 0 1 2 3 4 5 , 0 1 2 3 4 5 , 6 6 QR BC PR AC PQ AB , , and A = P B = Q C = R Ð Ð Ð Ð Ð Ð A B C D E If DE | | BC then, EC AE DB AD ,
Converse of B.P.T. :
EC AE DB AD If , then DE | | BC. Proved by Basic Proportionality theorem (THALES THEOREM)Height & Distance
Heights
& Distance
Elevation
Applications
Depression
A C C A B B q a Height of tower BC = AB tan (given AB & ) ´ q q Height of tower AB = (given CD,a, b) Height of tower PD = (given AB,a, b) Height of tower BD = (given AB,a, b) Height of tower AB = tan (given & BC) q a ´BC q =Angle of elevation Horizontal Horizontal a= Angle of Depression Definition : Definition :
Angle formed by the line of sight with the horizontal when the point is above the horizontal.
e.g.
e.g.
e.g.
@ Navigation @ Land surveys @ Buildings @ Optics @ Statics @ CrystallographyAngle formed by the line of sight with the horizontal when the point is below the horizontal. a a a a b b b b ) tan (tan tan tan CD . " / / ! . ! AB tan ) tan (tan ! . . " / ) tan (tan tan AB / " . / !
NO TANGENT ONE TANGENT AND
TANGENT
QT= 1/2 QR In PQR PQ=PR, hence PQR = PRQ = 1/2 (180– ) Since OQR = 90º, OQR = OQP – PQROQR = 90º – (90 – 1/2 ) =1/2 2 OQR = QPR. D q q q Þ Ð Ð Ð Ð Ð Ð Ð Ð Ð Tangent Length of tangents from an
external point to a circle are equal .
In OQP & ORP = -OQ = OR (radii), OP = OP OQP ORP PQ = PR D D \ D Þ ÐQPR + QOR = 180º QPO = RPO QOP = ROP Ð Ð Ð Ð Ð 7
P
P
P
P
Q
O
R
B
&*
Tengent
Distance
formula
y
y
y
x
x
x
Section
formula
Area of
Triangle
Coordinate Geometry
(x ,y )1 1 (x ,y )1 1 (x ,y )2 2 (x,y) (x,y) (x ,y )2 2 (x ,y )2 2 (x ,y )3 3 (x ,y )2 1 (x ,O)1 (x ,O)2 P A B A B R S R S T Q P R Q T C C Q O O O RS = (x –x ) = PT SQ = y ; ST = y ; QT = (y – y ) Applying Pythagoras in PTQ PQ = PT + QT = (x – x ) + (y – y ) PQ = 2 1 2 1 2 1 2 1 2 1 D 2 2 2 2 2 (Distance formula) Coordinates of P(x–xy) = A(x y ) B(x ,y ) C(x ,y ) Area = 1/2 base AltitudeArea = 1/2 [x (y –y ) + x (y –y ) +x (y -y )] If Area = 0,
then points are
1 1 2 2 3 3 1 2 3 2 3 1 3 1 2 ´ ´ collinear 2 1 2 2 1 2 x) (y y) x ( " + " 2 1 1 2 2 1 m m x m x m + + 2 1 1 2 2 1 m m y m y m + + , , 2 1 1 2 2 1 m m x m x m " " 2 1 1 2 2 1 m m y m y m " " (m : n externally) Verifying collinearity A(x ,y ), B(x ,y ), C(x ,y )
Find AB,BC,CA using distance formula If AB + BC = CA
or BC + CA = AB or AB + CA = BC
Then 3 pointare collinear
1 1 2 2 3 3
® ®
®
Application of distance formula
Verifying collinearity
A(x ,y ), B(x ,y ), C(x ,y ), D(x ,y ) Fin AB,BC,CD,DA using distance formula
Find diagonals A & BD
Check with different properties of quadrilaterals 1 1 2 2 3 3 4 4 ® ® ® d C
Verifying triangle formations A(x ,y ), B(x ,y ), C(x ,y )
Fin 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 using pythagoras therom
If no, it is not a 1 1 2 2 3 3 ® ® ® d D D (m : n Internally)
{
{
{
{
1 AC BC AC AB 2 2 2 2 , + =sin A + cos A = 12 2
Divide both sides by Interrelationship between T-ratios
T-ratios
Complimentry Angles
Graphs
Identities
Simplified trigonometric values
sinA = cosA = tanA =
cosecA = = sinA = = sin ÐA of ABC =D ÐA of AMPD sin A = = Note ;
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.
0° 30° 45° 60° 90° cos tan cosec sec cot =cot A+1 = cosec A2 2
cosA = = secA = = = 1+tan A = sec A2 2 tanA = = cotA = = 2 2 2 2 BC AC 1 BC AB , + 2 2 2 2 AB AC AB BC 1+ , AC BC H P BC AC P H AC AB H B AB AC B H AB BC B P BC AB P B AC BC 0 0 4 1 4 1 4 2 4 2 4 3 4 3 4 4 4 4 4 0 4 0 3 1 2 2 1 3 1 4 1 4 2 4 2 4 3 4 3 4 4 4 4 4 1 3 2 2 3 1 4 0 AP MP A A T R I G O O M E TR Y N C C x' 0 0 B B 90° M P 2 38 " "8 2 8 " 2 8 2 8 y y y' y' -1 -1 1 1 0.5 0.5 0.5 0.5 ecA cos 1 A sec 1 A cot 1 AB + BC = AC2 2 2 AB2 x x x 2 38 2 38 8 8 8 2 8 2 38 8 2 8 2 0 y y' tan(90-A)= cosec(90-A)= sin(90-A) =
cosA = secA = secA =
cos(90-A) =
sinA = tanA = cosecA =
cot(90-A) = sec(90-A) = AC AB BC AB AB AC AC AB BC AB AB AC AC BC AB BC BC AC AC BC AB BC BC AC
9
9
9
9
AC2 BC2TAN
SIN
COS
Triangle
Scaling
CALINGS
Constructing
Tangents
Division of a
line segment
in m:n(3:2)ratio
Alternate-Method
Construction
1. Scaling up 5/3 2. Scaling down 3/4 B C'||B C C'A'||CA 3 4 Join A B3 2 B C'||B C C'A'||CA 5 3 Join A C such that A C is parallel to A B 3 3 51. Join PO and bisect it (at O )1 2
2. O as centre & O O as radius draw a circle
2 1 2
3. 2 circles intersect at Q & R 4. Join PQ & PR (Tangents) DABC ~ A'BC' DAA C ~ AA BD hence 3 5 = = DABC ~ A'BC' DAA C ~3 DBB C2 = = = = = = = = = , , A A B A B A B C B C x x C x B1 B1 B1 B1 B2 B2 B2 B2 B3 B3 B3 B3 B4 B4 B4 B4 B5 B5 B B A A' A' A C' O2 O1 Q R P C C C x B ' A AB ' C ' A AC ' BC BC BC ' BC 3 5 BB BB 3 5 AB B ' A AC ' C ' A BC ' BC 4 3 A A B B X X 3 2 5 3 3 A A AA CB AC 2 3 A B A B X X Y Y B2 B2 B1 B1 A1 A1 A2 A2 A3 A3 3 2 2 3 BB AA BC AC 2 3
Sector of a circle
Area = R
p
2Area of combination
of plane figures
Area related to circle
Major sector r q r P B Q e.g. = 60° r = 3m q Generally sector implies minor sector
Area of segment (APB) =
Area of sector (AOB) – Area of AOBD
Note:
2 Rn = speed Time = Distance covered n = number of rotations For rotating wheel
p ´
r + r + r + r = length of square 4r = ; r = 1/2
Area of shaded reg. = Area of square – Area of 4 circles = (2 2) – 4 = 4 – Þ ´ ´ p
Area of shaded reg. = Area of
square ABCD –Area of (1+2+3+4) = 4 – 2(4 – ) = 2 – 4 p p e.g. e.g. Length of sector AB = ´2 Rp AOBP = ´ pR 2 AB = ´2p ´3 AOBP = ´ p ´3 2 m Þ p Þ e.g. e.g. Area of sector e.g. Sol.
Area I + III = A(ABCD) – Area of semicircle with AD & BC as diameter = 4 – p A 360 : 360 : 360 : 360 60 2 38 m
r
I I II II III III IV IV A A A B B B 2 units 2 units 2 units D D D C C C 4 ) 1 ( 2 8 Sol. Sol.Cuboid
Cube
Frustum
Hemisphere
Sphere
Cylinder
Cone
Surface area & volume
1. T.S.A®2[lb +bh + hl]
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 ® ® ® ® ® ® ® ® 1. T.S.A®6a2 1. T.S.A®2 r(r+h)p
Volume = Volume of cylinder
– Volume of hemisphere
Total surface area = C.S.A. of 2 hemispheres + C.S.A of cylinder
1. C.S.A = T.S.A. = 4 rp 2 1. C.S.A = 2 rp
2
1. C.S.A =p1r1 1l – r ( – )p 2l1l 1. T.S.A® pr( + r)l
2. C.S.A®2[bh +hl]
2. C.S.A®4a2
2. C.S.A®2 rhp 2. Volume = 4/3 rp 3 2. T.S.A. = 3 rp
3 2. T.S.A. =pl(r +r ) + (r + r )1 2 p 1 2 2 2 3. Volume = 2/3 rp 3 3. Volume = 1/3ph(r + r + r r ) 4. Slant height = 1 2 1 2 2 2 2. C.S.A® p lr 3. Volume®l´b´h 3. Volume®a3 3. Volume® pr h2 3. Volume ® 4. Diagonal ® 2 2 2 4. Diagonal ® h b + + 3a r h 3 1 2 8 b r r r r r1 r2 r r a a h h h h h1 l a l l1-l 2 2 1 2 (r r ) h + "
Probability
ROBABILITY
P
Elementary event (E) : Only one outcome
sum of the probabilities of all elementary events is 1.
0 £ P(E) £ 1
Applications
Gambling, insurance & statistics control theory.
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) =
P(E) > 4 Probability of getting 5 & 6.
® Examples
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.
Complimentary Events(E) Not E = E, P(E) = 1 – P (E) ln case 1 : P(1) = ; P(1) = 1 – = In case 3 In case 2 In case 1 EVENT(E) P(1) = not getting 1 ; = getting 2,3,4, 5 and 6 6 1 6 1 6 5
Certain event P(E) = 1 Impossible event P(E) = 0
P(sum 13) is impossible. In case 2 : Sum = 8 for (2,6) (3,5) (4,4) (5,3) (6,2) P(E) > 4 =P(5) + P(6) = = P(E) = 4 = P(1) + P(2) + P(3) + P(4) = = Probability of picking up black P(B) = P(E) = hence, P (sum 13) = 0
Red Red Red
Red Green Green Black Black Green P(sum 8) =36 5 outcomes possible all of . No E to favourable outcomes of . No 6 2 3 1 6 4 3 2 9 2 1 2 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 36 posibilities 6 1 6 , 5 , 4 , 3 , 2 , 1 1 ,
Grouped
Ungrouped
Statistics
Class of Intervals 10-25 25-40 40-55 55-70 70-85 85-100 NO. of Students 2 3 7 6 6 6 n = 30 ; n/2 = 15 55-70 is median class lower limit of the median class ( ) = 55c.f. = cum frequency of median preceeding class
l
Class size (h) = 15 Max. frequency f = 7 Modal class = 40-55
Lower limit of modal class = 40 f = 3 (Previous class f value) f = 6 (next class f value)
Mode = 40 + 4/5 15 = 52 1 0 2 Mode = l+ ´ x = x , x , ... x1 2 n®observations n is even MEAN MEAN MODE MODE MEDIAN MEDIAN n is odd f , f , ... f1 2 n®frequencies
Average of & observation.
The value of the observation having the max. frequency
3 Methods Direct Method x = = 62 x = a + hu x = a + d x = a + = 62 x = a + h =62
Step deviation Assumed mean method
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 Median = l + h = 62.5 f . f . c 2 / n ! " h f f f 2 f f 2 0 1 0 1 ! 0 0 1 2 3 3 4 5 " " "
