p b aal osw mind ma

MIN

D M

AP

AN INTER

ACTIV

E M

AG

ICA

L TO

OL

Wha

t?

When?

Why

?

How

?R

esu

lt

any

time, as freque

ntly a

s yo

u like

till it b

ecom

es a habit!

prese

nting words and

con

cept

s as pictu

res!!

Learning made simple

‘a w

inning c

ombina

tion

’with a b

lank she

et of

pape

rcolou

red p

ens a

ndyo

ur c

reative imagina

tion

!

Min

d M

ap

s f

or e

ach c

ha

pte

r s

how

the b

rea

kthroug

h s

yst

em

of p

lannin

g a

nd

note

-ta

kin

gw

hic

h h

elp

ma

ke s

choolw

ork f

un a

nd

cut

hom

ew

ork t

ime in h

alf

!!

OSW

AA

LB

OO

KS

LEA

RN

ING

MA

DE

SIM

PLE

nlo

ck t

he im

agin

atio

nT

o u

and

co

me u

p w

ith

id

eas

em

em

ber

fac

ts

and

To

r

�gu

res

easi

ly

ake c

leare

r and

To

m

better

no

tes

onc

entra

te a

nd

save

To

c

tim

e

lan w

ith

ease

and

ac

eT

o p

exam

s

2 ] Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – 9

Num

ber

Syst

em

−3−

2−1

01

23

1,2,

3...∞

0,1,

2,3.

..∞

a,b

− re

al n

umbe

rs

n −

+ve

inte

ger

Prod

uct l

awam

an =

am

+n

am ÷

an

= a

m−

n

(am

)n =

am

n

a

Quo

tient

law

Pow

er la

w

Rec

ipro

cal l

aw

Num

ber l

ine

Succ

essi

vem

agni

ficat

ion

Ever

y po

int o

n th

e lin

ere

pres

ents

real

num

bers

Real

num

ber

Rat

iona

l num

ber (

Q)

nth root o

f rea

l num

ber

Irrati

onal number (Q)

Can

not

be

wri

tten

in p

/q fo

rm

a =

(b)1/

n

Form

q≠0,

(p,q

) ∈ z

Who

le n

um

bers

(W)

Inte

gers

(Z)

Natural n

umbe

rs (N

)

Square root

Cube

root

Exponents with

Integ

ral P

ower

s

3

bb a

Exam

ple

5,

7Ex

ampl

e 7

, 4

33

x =

x1/

2x

= (x

)1/3

Examples 2, 3

−m

m =R

atio

naliz

atio

n

Term 1 r

Rat

iona

lisi

ngfa

ctor

1 r − s

1 r + s

1r −

s

1r +

s

r + s

r − s

r + s

r − s

r

Tran

sfor

mde

nom

inat

or in

toa

ratio

nal n

umbe

r

−∞

... −

3,−

2,−

1,0,1

,2,

3 ...∞

p q

CHAP

TER

: 1 n

umbe

r sys

tem

Oswaal CBSE Chapterwise Mind Maps, MATHEMATICS, Class – 9 [ 3

Polynomial

Poly

nom

ials

Poly

nom

ial

Con

stan

t(o

r ind

epen

dent

)

Zer

o

Mon

omia

l

Deg

ree

not d

efin

ed(c

onst

ant p

olyn

omia

l 0)

4x 2x +

3Bi

nom

ial

Trin

omia

l

4, −

7/5

Exam

ple

Poly

nom

ial i

n on

eva

riab

le:

ax3 −

bx2

− cx

+ d

An

alge

brai

c ex

pres

sion

of t

hefo

rm:

f(x) =

anx

n +

an−

1 xn–

1 + ..

. a1x

1 + a

0x0 3x

2 + 7

x +

2

−1/

2 is

the

zero

s of

the

poly

nom

ial

Term

sTy

pes

Degree

Poly

nom

ial

Line

ar

Qua

drat

ic

Cub

ic

3x +

21 2 3

Exam

ple

Deg

ree

2x2 +

3x

+ 1

7y3 +

6y2 +

2y+

2

Alg

ebra

ic Id

entit

ies

(x+

y)2

(x−

y)2

(x +

y +

z)2

x3 +y3 +

z3 −3x

yz(x

+y+

z) (x

2 +y2 +

z2 −xy

−yz

−zx

)

x3 +

y3

x3 −

y3

(x–y

)(x2 +

xy+

y2 )

(x+

y)(x

2 −xy

+y2 )

(i) x

3 +y3 +

z3 =

3xy

zIf

x +

y +

z =

0

(x −

y)3

(x +

y)3

x3 −

y3

− 3

xy (x

− y

)

x3 +

y3 +

3xy

(x +

y)

x2 +y2 +

z2 +2x

y+2y

z+2z

x=

x2 +

y2 +z2 +

2 (x

y+yz

+zx

)

x2 +

(a +

b)x

+ a

b

x2 −

2xy

+y2

x2 +

2xy

+y2

(x −

y) (

x +

y)

(x+

a) (x

+b)

x2 −y2

(ii)

x2

y

2

z2

yzxy

xz

Theo

rem

s

Fact

or T

heor

em

Rem

aind

er T

heorem

(i) if

(x−

a) is

afa

ctor

of p

(x),

then

p(a

) = 0

if p(

x) p

olyn

omia

lof

deg

ree

n>1,

isdi

vide

d by

x−

a,p(

a) is

the

rem

aind

er

Div

iden

t = (D

ivis

or

× Q

uotie

nt)

+

Rem

aind

er

if p(

x) is

apo

lyno

mia

lof

deg

ree

n>1,

a: a

ny re

al n

umbe

r(ii

) if p

(a) =

0, t

hen

(x−

a) is

a fa

ctor

of p

(x).

Zeroes of p

olyno

mia

l

Example

Num

ber t

hat s

atis

fies

the

equa

tion

P(x)

= 2

x +

1fin

d ze

roes

of t

he p

olyn

omia

lP(

x) =

02x

+ 1

= 0

x =

−1/

2

++

= 3

CHAP

TER

: 2 p

oly

nomial

s


Qua

dran

tEx

ampl

eA

xis

(I) Q

uadr

ant

(+,+

)

(II)

Qua

dran

t(–

,+)

(III

) Qua

dran

t(–

,–)

(IV

) Qua

dran

t(+

,–)

x >

0,

y >

0

x <

0,

y >

0

x <

0,

y <

0

x >

0,

y <

0

(1, 2

),(3

, 4)

(–3,

4),

(–8,

9)

(–3,

–4)

, (–

5, –

6),

(3, –

6) (1

, –1)

Coor

dina

teG

eom

etry

To d

escr

ibe

the

posi

tion

of p

oint

in a

pla

ne

Meani

ng

O(0

,0)

axis

inte

rsec

t

Point w

here

x &

y

Fixed point

Absci

ssa

x-axis

Ord

inat

e

y-axis

Two

perp

endi

cula

rlin

es

Cart

esia

n sy

stem

x’

y’

x

y

x’

y’

x

y O(0

,0)

Coo

rdin

ate

syst

em

(I) Q

uadr

ant

(+,+

)(I

I) Q

uadr

ant

(–,+

)

(III

) Qua

dran

t(–

,–)

(IV

) Qua

dran

t(+

,–)

x’

y’

x

y

Plotting point in a

plane (5,4)

Plan

e di

vide

d in

to 4

par

tsby

co-

ordi

nate

axe

s

Quadrants

Step

1 :

Dra

w c

o-or

dina

teax

is a

nd s

elec

t uni

ts

Step

2 :

Star

ting

from

ori

gn,

coun

t uni

ts o

n x

and

y ax

is

Step

3 :

Mar

k th

eco

rres

pond

ing

poin

ts

Gra

ph-L

inea

r Equ

atio

n

R(2

,2)

Q(1

,0)

P(0,

–2)

y=2x

–2

x’

y’

x

y

4P(

5,4)

Poin

t, P(

5,4)

–4

4–4

5

3 –3

3–3

2 –22

–2

1 –101

–1

x’

y’

x

y

4 –4

4–4

5

3 –3

3–3

2 –22

–2

1 –101

–1O

rdin

ate

Abs

ciss

a

Step

1 :

Con

vert

giv

en e

quat

ion

in

the

form

y =

mx

+ c

Step

2 :

Sele

ct a

tleas

t 3 v

alue

s

of x

, suc

h th

at x

,y∈

I

Step

3 :

Dra

w ta

ble

for t

he

o

rder

ed p

air (

x,y)

Step

4 :

Plot

thes

e or

dere

d po

ints

on

the

grap

h pa

per

Step

5 :

Dra

w s

trai

ght l

ine

pas

sing

thro

ugh

plot

ted

poi

nts.

CHAP

TER

: 3 c

oord

inat

e geo

metr

y


Step

s to

find

solu

tion

Lin

ear E

quat

ion

in T

wo

Varia

bles Li

near

equ

atio

n: 2

x +

3y

+ 1

2

Step

1: F

rom

equ

atio

n, w

e ge

t

4x +

3y

+ 7

= 0

1

2 - 2

x

Step

2: P

ut a

rbitr

ary

valu

e of

x, y

.

x0

04

6y

1 -1-1 -2-2

-31

02

2

3

3

4A

(0,4

)

B(6,

0)4

56

Step

3: P

lot (

0,4)

and

(6,0

) on

the

grap

h an

d jo

in th

em.

Inte

rpre

tati

onEq

uati

onG

raph

ical

rep

rese

ntat

ion

Equa

tion

of y

-axi

s

Equa

tion

of x

-axi

s

Stra

ight

line

para

llel t

o y-

axis

Stra

ight

line

para

llel t

o x-

axis

Line

pas

sing

thro

ugh

orig

iny

= m

x

y =

K

x =

K

y =

0

x =

0

Linear equation

a, b, c - co

nsta

nts

(a,b) ≠ (0

,0)

x, y

- va

riab

les

Whe

re

Exam

ple

Equa

tion

of th

e fo

rmax

+ b

y +

c =

0

Step

1: W

rite

the

equa

tion

in tw

o va

riab

les,

if n

ot p

rese

nt

Step

2: R

educ

e it

to o

ne v

aria

ble

by p

uttin

g an

arb

itrar

y va

lue

ofan

y va

riab

le, t

o fin

d a

pair

of

solu

tion.

Step

3: R

epea

t ste

p 2

for a

noth

erar

bitr

ary

valu

e of

var

iabl

e to

find

anot

her p

air o

f sol

utio

n.

It c

an h

ave

one,

no

or in

finite

lym

any

solu

tions

.

Example- Graph of Linear Equation

3

Gra

phic

al R

epre

sent

atio

n

11

2

2

3

3

4x

y’y

4

x =

0

–4–3–2–1–1

–2–3

–4

x =

2.5

y =

1.5 y =

2x

y =

0

x’

11

2

2

3

3

4x

y’y

4 –4–3–2–1–1

–2–3

–4x’

11

2

2

3

3

4x

y’y

4 –4–3–2–1–1

–2–3

–4x’

11

2

2

3

3

4x

y’y

4 –4–3–2–1–1

–2–3

–4x’

11

2

2

3

3

4x

y’y

4 –4–3–2–1–1

–2–3

–4x’

x

y’y

x’

y =

CHA

PTER

: 4

line

ar e

qua

tion

in t

wo v

ariabl

es


Axi

omat

ic s

yste

m, i

n w

hich

all t

heor

ems

are

deri

ved

from

a sm

all n

umbe

r of a

xiom

s

Poin

tPo

int

Has

no

wid

th, n

o le

ngth

and

no d

epth

Col

lect

ion

of p

oint

s, c

anbe

ext

ende

d in

bot

h di

rect

ions

.

Two

- dim

ensi

onal

col

lect

ion

ofpo

ints

(has

leng

th &

bre

adth

only

)

Thin

gs w

hich

are

equ

al to

the

sam

e th

ing

are

equa

l to

one

anot

her.

1 2 3 4 5 6 7

If e

qual

s ar

e ad

ded

to e

qual

s, w

hole

s ar

e eq

ual.

If e

qual

s ar

e su

btra

cted

from

equ

als,

the

rem

aind

ers

are

equa

l.

Thin

gs w

hich

coi

ncid

e w

ith o

ne a

noth

er a

re e

qual

to o

ne a

noth

er.

The

who

le is

gre

ater

than

the

part

.

Thin

gs w

hich

are

dou

ble

of th

e sa

me

thin

gs a

re e

qual

to o

ne a

noth

er.

Thin

gs w

hich

are

hal

ves

of th

e sa

me

thin

gs a

re e

qual

to o

ne a

noth

er.

Line

Line

Surf

ace

Eucl

id’s

Geo

met

ry

Def

initi

ons

Surf

ace

Euclid’s Axioms

Eucl

id’s

Pos

tula

tes

Intr

oduc

tion

to E

uclid

'sG

eom

etry

1 32 4 5

A s

trai

ght l

ine

can

be

draw

n fr

om a

ny o

nepo

int t

o an

y ot

her p

oint

.

A te

rmin

ated

line

can

be

prod

uced

infin

itely

.

A c

ircl

e ca

n be

dra

wn

with

any

cen

tre

and

ofan

y ra

dius

.

All

righ

t ang

les

are

equa

l to

one

anot

her.

If a

str

aigh

t lin

e fa

lling

on

two

stra

ight

line

s m

akes

the

inte

rior

ang

les

on th

esa

me

side

of i

t, ta

ken

toge

ther

mak

es le

ss th

antw

o ri

ght a

ngle

s, th

en th

e tw

o st

raig

ht li

nes,

ifpr

oduc

ed in

defin

itely

,m

eet o

n th

at s

ide

on w

hich

the

sum

of t

he a

ngle

s is

less

than

two

righ

t ang

les.

PQ

AB

Rad

ius

Cir

cle

Cen

tre

A B ∠A

BC

= ∠

DEF

= 9

0°E

D CF

A C2

P

1Q

B

D

∠1

+ ∠

2 <

180°

→

CHAP

TER

: 5 in

trodu

ction

to e

uclid'

s g

eome

try


A p

art o

f a li

ne w

ithon

e en

d po

int

1. If

two

lines

inte

rsec

t eac

h ot

her,

then

the

vert

ical

ly o

ppos

ite a

ngle

s ar

e eq

ual.

2. If

a tr

ansv

ersa

l int

erse

cts

two

para

llel

lines

, the

n ea

ch p

air o

f alte

rnat

e in

teri

oran

gles

is e

qual

3. If

tran

sver

sal i

nter

sect

s tw

o lin

es s

uch

that

a p

air o

f alte

rnat

e in

teri

or a

ngle

s is

equa

l, th

en th

e tw

o lin

es a

re p

aral

lel

4. If

a tr

ansv

ersa

l int

erse

cts

two

para

llel

lines

, the

n ea

ch p

air o

f int

erio

r ang

les

on th

e sa

me

side

of t

he tr

ansv

ersa

l is

supp

lem

enta

ry.

5. If

a tr

ansv

ersa

l int

erse

cts

two

lines

such

that

a p

air o

f int

erio

r ang

les

on th

esa

me

side

of t

he tr

ansv

ersa

l is

supp

lem

enta

ry, t

hen

the

two

lines

are

para

llel.

6. L

ines

whi

ch a

re p

aral

lel t

o th

e sa

me

line

are

para

llel t

o ea

ch o

ther

.

7. T

he s

um o

f all

inte

rior

ang

les

of a

tria

ngle

is 1

80°

Lines

Poin

ts

Line segment

Ray

Definitions

AB

⎢⎢C

D th

en∠

BQR

= ∠

CR

Q

if ∠

BQR

= ∠

CR

Qth

en A

B ⎢⎢

CD

AB

⎢⎢C

D th

en,

∠A

QR

+∠

CR

Q=

180˚

if ∠

AQ

R +

∠C

RQ

= 1

80˚t

hen

AB

⎢⎢C

D

AB

P QR

S

CD

Ray

AB

colli

near

in s

ame

line

A,B

,C -

colli

near

poin

ts

AB

C

A,B

,C -

non

colli

near

poin

ts

AB

Cno

n-co

llin

ear

not i

n sa

me

line

thre

e or

mor

epo

ints

lyin

g

Theo

rem

s

∠A

OD

= ∠

CO

B∠

AO

C =

∠B

OD

A

BC

O

D

AB

P QR

S

CD

AB

P QR

S

CD

A A

A

BCB B

CD

P QR

S

CD

EF

AB

⎢⎢C

D a

ndC

D ⎢

⎢EF

then

AB

⎢⎢EF

∠A

+ ∠

B +

∠C

= 1

80˚

Axioms

Type

s of

Ang

les

Ang

les

Acu

te

Rig

ht

Obt

use

Stra

ight

line

Ref

lex

x

x

x

x

x =

90˚

90˚<

x <

180˚

0˚<

x <

90˚

Valu

e

x =

180˚

180˚

<x

<360˚

Angles

End

poi

nts

Ray

s mak

ing

an

angl

e

Vert

ex

The

incl

inat

ion

betw

een

two

stra

ight

line

sA

ngle

Vert

ex

Arm

sArm

x

1. If

a ra

y st

ands

on

a lin

e, th

enth

e su

m o

f tw

o ad

jace

nt a

ngle

s so

form

ed is

180

°

2. If

the

sum

of t

wo

adja

cent

angl

es is

180

°, th

en th

e no

n-co

mm

on a

rms

of th

e an

gles

form

a lin

e.

3. If

a tr

ansv

ersa

l int

erse

cts

two

para

llel l

ines

, the

n ea

ch p

air o

fco

rres

pond

ing

angl

es is

equ

al.

4. If

a tr

ansv

ersa

l int

erse

cts

two

lines

suc

h th

at a

pai

r of

corr

espo

ndin

g an

gles

is e

qual

,th

en th

e tw

o lin

es a

re p

aral

lel t

oea

ch o

ther

.

AO

C

AB

is a

line

then

, ∠A

OC

+ ∠

BOC

= 1

80˚

B

AO

C

BIf

∠A

OC

+ ∠

BO

C =

180˚

then

AB

is a

str

aigh

t lin

e

AB

QS

PR

C

Her

e PQ

⎢⎢

RS

then

∠A

BQ =

∠B

CS

D

AB

Q

S

P RC

If ∠

BQR

= ∠

QR

C a

nd∠

AQ

P =

∠Q

RD

then

, AB

⎢⎢ C

D

D

colle

ctio

n of

poi

nts

(can

be

exte

nded

in b

oth

dire

ctio

ns)

Line

AB

A li

ne w

ith tw

oen

d po

ints

Line

seg

men

tA

B

CHAP

TER

: 6 l

ines

and

ang

les


Tria

ngle

s

It h

as th

ree

- sid

es,

angl

es a

nd v

ertic

es e

ach

Congr

uent

Triangle

clos

ed fi

gure

form

edby

thre

e st

raig

ht li

nes

If an

y th

ree

para

met

ers

of g

iven

tria

ngle

s are

sam

e, th

e tr

iang

les

will

be

cong

ruen

t.

A

BA

BCC

∆ABC

=~ ∆

DEF

A

BC

D

EF

Con

grue

nce r

ule

Propertie

s

Inequalitie

s

Two

tria

ngle

s are

con

grue

ntif

two

side

s and

the

incl

uded

angl

e of

one

tria

ngle

are

equa

l to

the

two

side

s and

the

incl

uded

ang

le o

f the

othe

r tria

ngle

.

Two

tria

ngle

s ar

e co

ngru

ent

if tw

o an

gles

and

the

incl

uded

sid

e of

one

tria

ngle

are

equa

l to

two

angl

es a

ndth

e in

clud

ed s

ide

of o

ther

tr

iang

le.

1. S

AS

AB

CD

O

2. A

SA

3. A

AS

4. S

SS

If th

ree

side

s of

one

tria

ngle

are

equa

l to

the

thre

e si

des

of a

noth

er tr

iang

le, t

hen

two

tria

ngle

s ar

e co

ngru

ent.

Two

tria

ngle

s ar

e co

ngru

ent

if an

y tw

o pa

irs

of a

ngle

san

d on

e pa

ir of

corr

espo

ndin

gsi

des

are

equa

l.

If in

two

righ

t tri

angl

es th

ehy

pote

nuse

and

one

sid

eof

one

tria

ngle

are

equ

al to

the

hypo

tenu

se a

nd o

nesi

de o

f the

oth

er tr

iang

le,

then

the

two

tria

ngle

s ar

eco

ngru

ent.

5. R

HS

Rul

eSt

atem

ent

Figu

re

A

B

In ∆

ABC

and

∆DEF

C

D

EF

A

32

23

44

B

BA5

5 44

D EF

C

In ∆

ABC

and

∆D

EFA

C =

DF

AB

= D

E BC

= F

E ∴

A

BC =~

D

EFC

D

EF

A

In

AO

D a

nd

CO

B

B

O

C

D

C

O =

OD

∠C

OB

= ∠

AO

D

O

B =

OA

∠B

= ∠

EBC

= E

F ∠

C =

∠F

∆AO

D =~

∆C

OB

∴ ∴

∆A

BC =~

∆D

EF Giv

en A

B ||

CD

AO

B an

d

CO

D

∠A

BO =

∠D

CO

∠

AO

B =

∠D

OC

O

A =

OD

∴

A

OB

=~

CO

D

In ∆

ABC

and

∆D

EF

AC

= D

F =

5cm

B

C =

FE

= 4

cm

DE

=

DF2 –

EF2 =

52 –

42

= 3

AB

=

AC

2 – B

C2 =

52 –

42 =

3

∴ A

B =

DE

Hen

ce ∆

ABC

=~ ∆

DEF

In

Stat

emen

tFi

gure

In a

ny tr

iang

le, t

he a

ngle

opp

osite

to th

e lo

nger

sid

e is

larg

er.

In a

ny tr

iang

le, t

he s

ide

oppo

site

toth

e la

rger

(gre

ater

) ang

le is

long

er.

The

sum

of a

ny tw

o si

des

of a

tria

ngle

is g

reat

er th

an th

e th

ird

side

.

Diff

eren

ce o

f any

two

side

s of

atr

iang

le is

less

than

the

thir

d si

de.

CB

AIn

A

BCA

B –

BC <

CA

AB

– A

C <

BC

AC

– B

C <

AB

CB

AA

C is

the

long

est s

ide

∴ ∠

B is

larg

est

CB

A

CB

Aif

∠B

is th

ela

rges

t∴

AC

is lo

nges

t

In

ABC

AB

+ A

C >

BC

AB

+ B

C >

AC

AB

+ B

C >

AB

A

B

∠BA

C =

∠C

AD

∠A

DB

= ∠

AD

C∆A

BD ≅

∆A

CD

(ASA

rul

e)H

ence

, AB

= A

CD

C

Stat

emen

tFi

gure

Ang

les

oppo

site

to e

qual

sid

eof

an

isos

cele

s tr

iang

le a

re

equa

l

The

side

s op

posi

te to

equ

alan

gles

of a

tria

ngle

are

equ

al

A

B

AB

= A

D∠

B =

∠C

CD

CHAP

TER

: 7 t

rian

gle

s


Quad

rilat

eral

sQuadrilat

eral

Figu

re fo

rmed

by

join

ing

four

poi

nts

in a

n or

der

A ABC

D is

aQ

uadr

ilate

ral

DC B

It h

as fo

ur -

ver

tices

,an

gles

and

sid

es e

ach

Types

Prop

erty

Para

llel

ogra

mR

ecta

ngle

Rho

mbu

sSq

uare

Trap

eziu

m

Opp

osite

sid

esar

e pa

ralle

l and

co

ngru

ent

All

angl

es a

reco

ngru

ent

Opp

osite

ang

les

are

cong

ruen

t

Dia

gona

ls a

reco

ngru

ent

Dia

gona

ls a

repe

rpen

dicu

lar

Dia

gona

ls b

isec

tea

ch o

ther

Adj

acen

t ang

les

are

supp

lem

enta

ry

All

side

s ar

eco

ngru

ent

No

AB

B

DC

No

No

No

Yes

Yes

Yes

Yes

No

Yes

Yes

No

Yes

Yes

Yes

Yes

Yes

No

No

Yes

Yes

Yes

Yes

Yes

Yes

Yes

Yes

Yes

Yes

Yes

Yes

Yes

No

No

Yes

No

Para

llel b

utno

t con

grue

nt

Yes

No

Yes

AB

DC

DC

AB

A

A

B

DD

C

C

Stat

emen

tFi

gure

1. A

dia

gona

l of a

par

alle

logr

am

divi

des

it in

to tw

o co

ngru

ent

tr

iang

les.

2. In

a p

aral

lelo

gram

, opp

osite

s

ides

are

equ

al a

nd p

aral

lel.

3. If

eac

h pa

ir o

f opp

osite

sid

es

of a

qua

drila

tera

l are

equ

al a

nd

para

llel,

then

it is

a p

aral

lelo

gram

.

4. In

a p

aral

lelo

gram

, opp

osite

a

ngle

s ar

e eq

ual.

5. If

in a

qua

drila

tera

l, ea

ch p

air

o

f opp

osite

ang

le is

equ

al, t

hen

it

is a

par

alle

logr

am.

6. T

he d

iago

nals

of a

par

alle

logr

am

bis

ect e

ach

othe

r.

7. If

the

diag

onal

s of

a q

uadr

ilate

ral

b

isec

t eac

h ot

her,

then

it is

p

aral

lelo

gram

.

A

ABC

D is

apa

ralle

logr

amA

C d

iago

nal

then

∆A

BC ≅

∆A

DC

In p

aral

lelo

gram

ABC

D,

AB

|| C

D, A

D |

| BC

,an

d A

B =

CD

,

A

D =

BC

If A

B ||

CD

, AD

||

BC,

and

AB

= C

D,

AD

= B

C th

enA

BCD

is a

par

alle

logr

am

B

DC

AB

DC

AB

DC

AB

DC

AB

DC

AB

DC

O O

In p

aral

lelo

gram

ABC

D,

∠A

= ∠

C, ∠

B =

∠D

If ∠

A =

∠C

, ∠B

= ∠

Dth

en A

BCD

is a

para

llelo

gram

In p

aral

lelo

gram

ABC

D,

OA

= O

C th

enO

B =

OD

If O

A =

OC

O

B =

OD

then

ABC

D is

para

llelo

gram

AB

DC

Mid

-poi

nt th

eo

rem

Stat

emen

t

The

line-

segm

ent j

oini

ngth

e m

id-p

oint

s of

two

side

s of

a tr

iang

le is

para

llel t

o th

e th

ird

side

.

The

line

draw

n th

roug

hth

e m

id-p

oint

of o

ne s

ide

of a

tria

ngle

, par

alle

l to

anot

her s

ide

bise

cts

the

thir

d si

de

A

B

If E

and

F a

rem

id-p

oint

of A

B an

d A

C,

then

EF

|| B

C

If E

is th

em

id-p

oint

of A

BEF

||

BC, t

hen

AF

= F

C,

F is

the

mid

-poi

nt o

f AC

C

FE

Dl l

Figu

re

A

BC

FE

D

Prope

rtie

s

CHAP

TER

: 8 q

uadr

ilat

eral

s


A

A

D

Dar

A

DC

= a

r A

BC

CC

B

B

PQ

ar (A

BC

D) =

ar (

PQC

D)

A DCB

if ar

A

DC

= a

r A

BC

then

AB

|| C

D(i)

If A

and

B a

re c

ongr

uent

figu

res,

(ii) I

f a p

lana

r reg

ion

form

ed b

y a

f

igur

e T

is m

ade

up o

f tw

o no

n-

ove

rlap

ping

pla

nar r

egio

ns

for

med

by

figur

es P

and

Q

ar (A

) = a

r (B)

ar (T

) = a

r(P)

+ar

(Q)

if PQ

R ≅

DEF

then

ar (

A) =

ar (

B)

Are

a of

figu

re T

= A

rea

of fi

gure

P +

Are

a of

figu

re Q

Theo

rem

s

Stat

emen

t

1. P

aral

lelo

gram

s on

the

sam

eba

se a

nd b

etw

een

sam

epa

ralle

ls a

re e

qual

in a

rea.

2. T

wo

tria

ngle

s on

the

sam

eba

se (o

r equ

al b

ases

) and

betw

een

the

sam

e pa

ralle

lsar

e eq

ual i

n ar

ea

3. T

wo

tria

ngle

s ha

ving

the

sam

e ba

se (o

r equ

al b

ases

)an

d eq

ual a

reas

lie

betw

een

the

sam

e pa

ralle

ls

A

BC

h

h

AB

DC

AB

CD

is a

pa

ralle

logr

am

Figu

re

Properties

PD

EF

Q(A

)(B

)R T

PQ

Area

= B

ase

× H

eigh

t

Are

a of

tria

ngle

1

2×

Bas

e ×

Hei

ght

Med

ian

of a

tria

ngle

divi

des

it in

to tw

oeq

ual a

reas

Area of para

llelo

gram

Num

ber (

in s

ome

unit)

asso

ciat

ed w

ith p

art o

fth

e pl

ane

encl

osed

by

the

figur

e

BA

D -

med

ian

ar ∆

ABD

= a

r ∆A

CD

DA

C

Are

a of

tria

ngle

=

CHAP

TER

: 9 a

reas

of

para

llel

ogra

ms &

trian

gle

s


Circ

lesCirc

le

cent

reFi

xed

poin

t

rad

ius

(r)

Fixe

d di

stan

ce

A c

ircl

e is

the

locu

s of

a cu

rve

equi

dist

ant

from

a p

oint

Stat

emen

t

1. E

qual

cho

rds

of a

cir

cle

subt

end

e

qual

ang

les

at th

e ce

ntre

.

2. If

the

angl

es s

ubte

nded

by

the

c

hord

s of

a c

ircl

e at

the

cent

re a

re

equ

al, t

hen

the

chor

ds a

re e

qual

.

3. T

he p

erpe

ndic

ular

from

the

cent

re

of a

cir

cle

to a

cho

rd b

isec

ts th

e

cho

rd.

4. T

he li

ne d

raw

n th

roug

h th

e

cen

tre

of a

cir

cle

to b

isec

t a c

hord

is

perp

endi

cula

r to

the

chor

d.

5. T

here

is o

ne a

nd o

nly

one

circ

le

pas

sing

thro

ugh

thre

e gi

ven

n

on-c

ollin

ear

poin

ts.

6. E

qual

cho

rds

of a

cir

cle

are

e

quid

ista

nt fr

om th

e ce

ntre

.

A

Q

P

Ref

lex

∠PO

Q =

2 ∠

PAQ

O

7. C

hord

s eq

uidi

stan

t fro

m th

e ce

ntre

o

f a c

ircl

e ar

e eq

ual i

n le

ngth

.

PQ

OR

∠R

PS =

∠R

QS

S

8. T

he a

ngle

sub

tend

ed b

y an

arc

at

th

e ce

ntre

is d

oubl

e th

e an

gle

s

ubte

nded

by

it at

any

poi

nt o

n th

e

rem

aini

ng p

art o

f the

cir

cle.

9. A

ngle

s in

the

sam

e se

gmen

t of a

c

ircl

e ar

e eq

ual.

10. I

f a li

ne s

egm

ent j

oini

ng tw

o po

ints

s

ubte

nds

equa

l ang

les

at tw

o ot

her

p

oint

s ly

ing

on th

e sa

me

side

of t

he

line

con

tain

ing

the

line

segm

ent,

the

fo

ur p

oint

s lie

on

a ci

rcle

.

11. T

he s

um o

f eith

er p

air

of o

ppos

ite

an

gles

of a

cyc

lic q

uadr

ilate

ral

is 1

80°.

12. I

f the

sum

of a

pai

r of

opp

osite

angl

es o

f a q

uadr

ilate

ral i

s 18

0°,

t

hen

quad

rila

tera

l is

cycl

ic.

PS

QR

C A

∠A

CB

= ∠

AD

Bth

en A

,B,C

,D li

eon

the

circ

le

BD

Figu

re

All

the

four

ver

tices

of a

qua

drila

tera

lA

BC

D li

e on

cir

cle

Rad

ius

Cen

tre

Or

rA A

B =

Dia

met

erof

a c

ircl

e

B

r

D =

2r

chor

d w

hich

pas

ses

thro

ugh

the

cent

re o

f a

cir

cle

Dia

met

er (D

)

Inte

rior o

f the

cir

cle

Exterior o

f the c

ircle

M

ajor

Arc

Segment

Relate

d te

rms

Inte

rior

Exte

rior

Cir

cle

Minor

Maj

or a

rc

Min

orse

gmen

t

Min

or a

rcS

P

Minor

Maj

or

Cyclic quadrilateral

AD

DA

AB

= C

Dth

en∠

AO

B =

∠C

OD

BC

O

BA

BC

D -

cycl

icqu

adri

late

ral

C

Theo

rem

s

DA

BC

O∠

AO

B =

∠C

OD

then

AB

= C

D

AB

OM

⊥ A

Bth

en A

M =

MB

If A

M =

MB

then

OM

⊥ A

B

O M

AB

O O

RQ

P

OM

L

D

C

B

AB

= C

Dth

enO

L =

OM

A

If O

L =

OM

then

AB

= C

D

∠P

+ ∠

R =

180

°∠

Q +

∠S

= 1

80°

PS

QR

If ∠

P +

∠R

= 1

80°

∠Q

+ ∠

S =

180

°th

en P

QR

S is

cycl

icqu

adri

late

ral

M

OM

L

D

C

B

A

O

Maj

orse

gmen

t

CHAP

TER

: 10 c

ircl

es


Cons

truc

tions

BCA

AB

Proc

ess

of d

raw

ing

geom

etri

cal f

igur

e

Step

1 :

Take

A a

s ce

ntre

draw

arc

inte

rsec

ting

AB

Step

1 :

Take

B a

s ce

ntre

,dr

aw a

rcs

inte

rsec

ting

AB

&B

C (o

f any

radi

us)

Step

2 :

With

D a

s ce

ntre

and

sam

e ra

dius

, dra

w a

rcin

ters

ectin

g sa

me

arc.

Step

2 :

Join

PQ

Step

2 :

D &

E a

s ce

ntre

s an

dra

dius

> 1

DE,

dra

w tw

o ar

cSt

ep 3

: D

raw

AC

pas

sing

thr

ough

E,

∠C

AB

= 6

0˚

Step

3 :

PQ in

ters

ect A

B at

a p

oint

M to

fro

m p

erpe

ndic

ular

bis

ecto

r

Step

3 :

BF is

the

requ

ired

ang

le b

isec

tor

AB

D

AB

AB

E

D

A

E

D

C

B60˚

Step

1 :

With

A &

B a

s ce

ntre

& ra

dius

> 1

AB,

dra

w

inte

rsec

ting

arcs

on

both

side

s.

2

2A

B

P Q

× ×

AB

PM

Q

× ×

B

E

D

A C

B

E

D

A F Can

d jo

in B

to itGeo

met

ry b

oxRe

quir

emen

ts

Geometrical constru

ctio

n

Angl

e b

isecto

r

60

˚ Ang

le

Perpendicular bisector

Tr

iang

leGiven : Perimeter and tw

o base angles

diffe

renc

e of t

wo s

ides

Giv

en: B

ase,

Angles,

sum

of t

wo

sides

Giv

en :

Base

, Ang

les

Step

1 :

Dra

w B

C =

bas

e an

d m

ake

give

n an

gle.

Step

2 :

Cut

BD

= d

iffer

ence

of s

ides

join

DC

Step

3 :

Dra

w p

erpe

ndic

ular

bis

ecto

r DC

and

let

it in

ters

ect

BX. N

ame

it as

A.

Step

4 :

Join

AC

, AB

C is

the

requ

ired

tria

ngle

.

B

X

C

B

X

D

C

B

X

D

C

B

X

D

C

Step

1 :

Dra

w li

ne e

qual

to s

um o

f sid

es =

XY

Step

3 :

Bise

ct ∠

LXY

and

∠M

YX

, let

the

bise

ctor

sm

eet a

t A.

Step

4 :

Dra

w p

erpe

ndic

ular

bis

ecto

rs o

f AX

and

AY. E

xten

d th

em to

inte

rsec

t XY.

AB

C is

the

requ

ired

tria

ngle

.

Step

2 :

Dra

w g

iven

two

angl

es a

t X &

Y

XY

L

XYM

LA

XYM

LA

BC

XYM

1. S

cale

2. P

air o

f set

-

squ

ares

3. P

air o

f

divi

ders

4. C

ompa

ss

4. P

rotr

acto

r

Step

1 :

Dra

w B

C =

bas

e an

d m

ake

give

n an

gle.

Step

2 :

Cut

BD

equ

al to

sum

of s

ides

, joi

n D

C

Step

3 :

Dra

w p

erpe

ndic

ular

bis

ecto

r of D

C a

nd

l

et it

inte

rsec

t BD

. Nam

e it

as A

.

Step

4 :

Join

AC

, AB

C is

the

requ

ired

tria

ngle

.

B

X

C

B

DX C

B

D

A

X C

B

D

A

X

C

CHAP

TER

: 11 co

nstr

uctions


Her

on’s

Form

ula

× B

ase

× A

ltitu

de1 2

a2 ; a

= s

ide

of tr

iang

le3 4

Pa

a

a QR

Pb

c

a

QR

C ABa

se

Alti

tude

B

Are

a of

tria

ngle

=

s (s–

a) (s

–b) (

s–c)

whe

re, a

,b,c

=

side

s of

tria

ngle

s =

se

mi –

per

imet

er =

2

BC5

m

8 m

12 m

9 mA

D

90˚

Right-angled tr

iang

le

Equi

late

ral t

rian

gle

A

rea

Heron’s formula

App

licat

ions

Find

are

a of

tria

ngle

of s

ides

122c

m, 2

2cm

, 120

cm

Are

a of

qua

drila

tera

lA

BC

D w

ith g

iven

dim

ensi

ons

:-

By H

eron

’s fo

rmul

aA

rea

of tr

iang

le =

s (s

–a) (

s–b)

(s–c

)

a+

b+c

2

2

122

+22

+12

0 =

132

cm

∴ A

rea

=

132

(132

-122

) (13

2-22

) (13

2-12

0) c

m2

=

( 132

(10)

(110

) (12

))cm

2

=

1320

cm

2

Are

a of

AB

CD

= A

rea

of

ABD

+ a

rea

of

BC

D

A

rea

of A

BD =

s (s

-a) (

s-b)

(s-c

)

a

+b+

c

(9

+8+

13) m

Are

a =

15

(15-

9) (1

5-8)

(15-

13) m

2

∴ A

rea

of A

BC

D =

(30

+35

.496

) m2

= 6

5.49

6 m

2

= 1

5 (6

) (7)

(2) m

2

= 3

5.49

6 m

2

1

Her

e BD

, =

BC

2 +D

C2

BD =

12

2 +52

= 1

3 m

=

×

12

× 5

= 3

0 m

2

2

22

2

ABD

Her

e, a

rea

of

BC

D =

× B

C ×

CD

1

whe

re s

==

= 1

5 m

a+b+

c

whe

re, s

=

Her

e,

s =

CHAP

TER

: 12

Area

s :

hero

n's f

orm

ula


Are

a

πr2 h

cu u

nits

SA o

f Cub

oid

= 2

(lb+

bh+

hl)

4 πr

3 cu

units

3

Surfa

ce ar

ea

b

h

l

r

Volu

me

of c

ubio

d =

lbh

Volu

me

Sphere3-

D fi

gure

with

ever

y po

int o

n its

surf

ace

equi

dist

ant

from

its

cent

re

Sphe

re c

ut in

hal

f

2 πr

3 cu

units

3

1 πr

2 h c

u un

its

3

Hem

isph

ere

Right circular cone

3-D

obj

ect w

hich

tape

rs fr

oma

circ

ular

bas

e to

a p

oint

Basic concept

Sum

of t

he a

reas

of a

ll fa

ces

(or s

urfa

ces)

on

a 3-

D s

hape

Volu

me

=A

rea

of b

ase

× V

ertic

al h

eigh

t

Cub

e3-

D fi

gure

with

6 fa

ces

and

all

equa

l sid

es

Are

a

Are

a

l × b

× h

cu

units

a

a

a

b

h

l

Cu

boid

3-D

figu

re

with

6 fa

ces

A c

ylin

der w

ith

circ

ular

bas

esan

d ax

is jo

inin

g th

e tw

o ce

ntre

sof

the

base

spe

rpen

dicu

lar t

oth

e pl

anes

of t

hetw

o ba

ses

Cub

e

Cub

oid

Righ

t circ

ular

cylin

der

h

rR

ight

cir

cula

rcy

linde

r

Mea

sure

men

t of

ever

ythi

ng w

ithin

the

lines

of t

he s

hape

Qua

ntity

of t

hree

-dim

ensi

onal

spac

e en

clos

ed b

y a

clos

ed s

urfa

ce

Volum

e of a

ny o

bjec

t

of any object

Surface area

Total surface

Are

a =6a2 sq units

Lateral

surf

ace

Area = 4a

2 sq u

nits

Volu

me

= a

3 cu

units

Tota

l sur

face

Are

a =

2 (l

b+bh

+hl

)sq

uni

ts

Late

ral s

urfa

ceA

rea

= 2

(l +

b) h

sq u

nits

Volu

me

=Volu

me

=

Cur

ved

surf

ace

Are

a =

2πr

hsq

uni

ts

Tota

l sur

face

Are

a =

2πr

(r+

h)sq

uni

ts

Surf

ace

area

= 4πr

2 sq

uni

ts

Curved su

rface

Area = πrl sq

unit

To

tal su

rface

Area

=

πr (l+

r) sq unit

Curved surface

Total

surfa

ce

Area

=

3 πr

2 sq units

Area = 2 πr2 sq units

Area

Area

r

lh

r

Volu

me

=

Volu

me

=

Volu

me

=

CHAP

TER

: 13

sur

face

are

as a

nd v

olu

mes


Stat

istic

s

Data

Set o

f val

ues

of q

ualit

ativ

eor

qua

ntita

tive

info

rmat

ion

Mos

t fre

quen

tlyoc

cure

d ob

serv

atio

nD

ista

nce

(in k

m) o

f 20

stud

ents

from

thei

r res

iden

ce to

sch

ool i

sgi

ven

as -

Con

stru

ct a

gro

uped

freq

uenc

ydi

stri

butio

n ta

ble

6

7

5

7

7

8

7

6

9

74

10

6

8

8

9

5

6

4

8

Stat

istic

s

Tally

mar

ks

Dis

tanc

eTa

lly

Freq

uenc

y

4 5 6 7 8 9 10

ll ll llll

llll

llll ll l

2 2 4 5 4 2 1

Freq

uenc

y di

stri

butio

n ta

ble

Histog

ram

sGraphical representatio

n

grouped data

Freq

uenc

y po

lygo

nsBa

r gra

phs

Cent

ral T

ende

ncy

-Ung

roup

ed d

ata

Mod

e

M

ean

Med

ian

Mea

n, x

=_∑

xi

n i=1 n

Cal

cula

ted

by a

ddin

gal

l the

val

ues

and

divi

ding

it by

tota

l num

ber o

fob

serv

atio

ns.

Obs

erva

tion

odd

num

ber

even

nu

mbe

rO

bser

vatio

n

Valu

e of

the

mid

dle

mos

t obs

erva

tion.

Item

s

A20 30 50 60

B C D

Qua

ntit

ies

Med

ian

= v

alue

of th

obse

rvat

ion

n+1

2

Med

ian

=

obse

rvat

ion+

obse

rvat

ion

n 2th

n 2th

+ 1

2

Are

a of

stu

dy d

ealin

gw

ith th

e pr

esen

tatio

n,an

alys

is a

nd in

terp

reta

tion

of d

ata

valu

e of

CHAP

TER

: 14

sta

tistics


Pro

babi

lity

The

exte

nt to

whi

chso

met

hing

is li

kely

to h

appe

n

Act

ion

whi

ch re

sults

inon

e or

sev

eral

out

com

es

Impo

ssib

le

Less

like

lyM

ore

likel

y

Even

cha

nce

Cer

tain

Definition

Out

com

e

Tria

l

Even

tRe

late

d te

rms

Probability valu

e

Emperical probabilit

y

Exam

ples

Coi

nDice

A d

ice

is th

row

n2

times

. Fin

d th

epr

obab

ility

of g

ettin

g 11

Tota

l num

ber o

f tri

als

Out

com

es fo

r get

ting

11 =

(5,6

) (6,

5)

N

umbe

r of g

ettin

g 11

1 18Tota

l num

ber o

f tri

als

= 6

n , n is

the

num

ber o

f tim

esdi

ce is

thro

wn

= 6

2 = 3

6

=

2 36=

Num

ber o

f tim

es

t

hree

hea

ds o

ccur

ed

Tota

l num

ber o

f tri

als

Tota

l num

ber o

f tri

als

= 2

n , n is

num

ber o

f

tim

es c

oin

is to

ssed

i.e. i

f n=

3{H

HH

, HH

T, H

TT, H

TH,

TTH

, TH

H, T

TT, T

HT}

= 2

3

= 8

1 8=

A c

oin

is to

ssed

3 ti

mes

.Fi

nd p

roba

bilit

y of

thre

e he

ads?

Col

lect

ion

of s

ome

outc

omes

of a

n ex

peri

men

t

Som

ethi

ng th

at fo

llow

s as

are

sult

or c

onse

quen

ce

P (E

) =

P (E

) =

Nev

er g

reat

er th

an o

ne0<

P(E)

<1

Nev

er b

e ne

gativ

e N

umbe

r of t

rial

s in

whi

ch e

vent

hap

pene

d

Tota

l num

ber o

f tri

als

P (E

) =

· · ·

CHAP

TER

: 15

proba

bility

p b aal osw mind ma

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