B AAL OSW MIND MA...1 10–9 = x 8 = 1 y 10 and = x 8 1 1 = y 10 1 1 ⇔ i.e. xy = 8 and By Cross-Multiplication = 10 ax by c 11 1 ++ = 0 = 0 ax 22 2,,,,, –R eal numbers abc abc

MIN

D M

AP

AN INTER

ACTIV

E M

AG

ICA

L TO

OL

Wha

t?

Wh

?en

Wh

?y

How

?R

esu

lt

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

!!

any

time, as freque

ny as yo

u like

tltill 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

!

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

OSW

AA

LB

OO

KS

LEA

RN

ING

MA

DE

SIM

PLE

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

Rea

l Num

bers

Theorem

sEuclid

's

Ste

p 3

: C

on

tin

ue

the

pro

cess

till

th

e re

main

der

is

zer

oP

rim

e F

acto

riz

ation M

eth

od

Fund

am

enta

l Theorem

of A

rithm

etic

Ever

y c

om

po

site

nu

mb

erca

n b

e ex

pre

ssed

as

a p

rod

uct

of

pri

mes

, an

d t

his

fact

ori

sati

on

is

un

iqu

e,ap

art

fro

m t

he

ord

er i

n w

hic

h t

he

pri

me

fact

ors

occ

ur

HC

F (

,)

× L

CM

(,

) =

×a

ba

ba

b

Fo

r E

xam

ple

()

= 3

fx

xy

2

()

= 6

gx

xy

2

HC

F =

3xy

LC

M =

6x

y2

2

Co

mp

osi

te N

um

ber

= P

1×

P2×

P3..

.×P

,x

nw

her

e P

1P

2..

. P

are

np

rim

e n

um

ber

s

Ste

p 2

: If

= z

ero,

is t

he

rd

HC

F o

fan

dIf

0,

ap

ply

Eu

clid

' s

cd

r

Div

isio

n t

oan

dd

r

Ste

p 1

: A

pp

ly E

ucl

id’s

Div

isio

nL

emm

a,

to&

.=

+c

dc

dq

r

2.

2

, 3

are

irr

ati

on

al

1.

Let

be

a p

rim

e n

um

ber

.p

Ifd

ivid

es,

then

div

ides

,p

ap

a2

wh

ere

is a

po

siti

ve

inte

ger

a

3.

Let

be

a r

ati

on

al

nu

mb

er w

ho

sex

dec

imal

exp

an

sio

n t

erm

inate

s. T

hen

can

be

x

exp

ress

ed i

n t

he

form

,w

her

e&

are

pq

cop

rim

e, t

he

pri

me

fact

ori

sati

on

of

is o

fq

the

form

25

wh

ere

,are

no

n-

nm

nm

neg

ati

ve

inte

ger

s

p — q

4.

Let

= b

e a r

ati

on

al

nu

mb

er s

uch

x

that

the

pri

me

fact

ori

sati

on

of

is o

fth

eq

form

25

wh

ere

,are

no

n-n

egati

ve

nm

nm

inte

ger

s. T

hen

,h

as

dec

imal

exp

an

sio

nx

a

wh

ich

ter

min

ate

s.

5.

Let

= b

e a r

ati

on

al

nu

mb

er,

such

x

that

the

pri

me

fact

ori

sati

on

of

is n

ot

of

q

the

form

of

25

wh

ere

,are

no

n-n

egati

ve

nm

nm

inte

ger

s. T

hen

,h

as

dec

imal

exp

an

sio

nx

a

wh

ich

is

no

n-t

erm

inati

ng r

epea

tin

g

p — q

Giv

en p

osi

tive

inte

ger

sa

an

d,

ther

e ex

ist

un

iqu

e in

teger

sb

an

dsa

tisf

yin

gq

r

abq

rr

b=

+;

0<

<

Div

isio

n L

emm

a

Ste

ps

to o

bta

in t

he

HC

F o

ftw

op

osi

tive

inte

ger

s, s

ay

an

d,

cd

wit

h>

cd

Div

isio

n A

lgo

rith

m

p — q

Fo

r an

y t

wo

po

siti

ve

inte

ger

s,

an

da

b

CHAP

TER

: 1 R

eal

Numb

ers

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

If(

) an

d(

) are

px

gx

two

po

lyn

om

ials

wit

h

()

0,

then

–g

x≠

()=

()

×(

) +

()

px

gx

qx

rx

wh

ere,

()

= 0

or

rx

deg

ree

of

()

< d

egre

er

x

of

()

gx

Hig

hes

t p

ow

er o

fin

Po

lyn

om

ial,

()

xp

x

Div

isio

n A

lgorithm

Rela

tionship

-Zeroes a

nd

Coeffic

ient

of P

oly

nom

ials

Gra

phic

al R

ep

resenta

tion

Qua

dra

tic P

oly

nom

ial

Degree o

f P

oly

nom

ial

Par

abol

a

yx

x=

–3

–4

2yy

xx

Poly

nom

ials

Case

1-

Gra

ph

cu

ts-a

xis

at

2 p

oin

tsx N

um

ber

of

Zer

oes

2

Case

2-

Gra

ph

cu

ts-a

xis

at

exact

ly o

ne

po

int

x

Nu

mb

er o

fZ

ero

es 1

Case

3-

Gra

ph

do

es n

ot

cut

-ax

isx

Nu

mb

er o

fZ

ero

es 0

Po

lyn

om

ial

Deg

ree

Gen

eral

Fo

rm

Lin

ear

1ax

b+

Qu

ad

rati

c2

ax

bx

c2+

+

0a

≠

Cu

bic

3ax

bx

cxd

32

++

+

0a

≠

Typ

es

αβ

an

dare

zer

oes

of

Qu

ad

rati

cP

oly

no

mia

l

Th

en,

Su

m o

fzer

oes

,

ax

bx

c2

++

αβ

+=

–b a

Pro

du

ct o

fzer

oes

αβ=

c a

Qu

ad

rati

c

αβ

γ,

an

dare

zer

oes

of

Cu

bic

Po

lyn

om

ial

Su

m o

fzer

oes

,

ax

bx

cxd

32

++

+

αβ

γ+

+=

–b a

αββγ

γα+

+=

c a

αβγ

=–

d a

Su

m o

fp

rod

uct

s o

fth

ezer

oes

tak

en t

wo

at

a t

ime

Pro

du

ct o

fzer

oes

Cu

bic

Zeroes o

f P

oly

nom

ial

Gra

phic

ally

CHAP

TER

: 2 p

oly

nomial

s


Pa

ir o

f Li

nea

r Eq

uations

in T

wo V

ari

ab

les

Alg

eb

ra

ic M

eth

od

s

Gra

phic

al R

ep

resenta

tion

So

lve:

2+

3–46 =

0–(i

)x

y

3+

5–74 =

0–(i

i)x

y

So

luti

on

: B

y c

ross

-mu

ltip

lica

tio

n m

eth

od

xy

13 5

–46

–74

2 3

3 5

Then

,=

x3(–

74)–

5(–

46)

y

(–46)(

3)–

(–74

)(2)

=1

2(5

)–3(3

)

=x

–222+

230

=y

–13

8+

148

1

10–

9

=x 8

=1 1

y 10an

d=

x 8

1 1=

y 10

1 1⇔

i.e.

xy

= 8

an

d=

10

By C

ross

-Mu

ltip

lica

tio

n

ax

by

c1

11

++

= 0

++

= 0

ax

by

c2

22

,,

,,

,,

–R

eal

nu

mb

ers

ab

ca

bc

11

12

22

Genera

l Fo

rm

Each

so

luti

on

(,

), c

orr

esp

on

ds

xy

to a

po

int

on

th

e li

ne

rep

rese

nti

ng

the

equ

ati

on

an

d v

ice-

ver

sa

Solu

tion G

ra

phic

ally

Gra

ph

ical

Rep

rese

nta

tio

n

Inte

rsec

ting

Lines

Pair

of

Lin

es =

–2

= 0

xy

3+

4–20 =

0x

y

a1

a2=

1 3,

b1

b2

=–2 4

,c 1 c 2

=0

–20

Co

mp

are

th

e R

ati

os

=≠

a1

a2

b1

b2

Alg

ebra

ic I

nte

rpre

tati

on

: E

xact

lyo

ne

solu

tio

n–

con

sist

ent

(un

iqu

e)

Pair

of

Lin

es =

2+

3–9 =

0x

y

4+

6–18 =

0x

y

a1

a2=

2 4,

b1

b2

=3 6

,c 1 c 2

=–3

–18

Co

mp

are

th

e R

ati

os

==

a a2

b1

b2

Alg

ebra

ic I

nte

rpre

tati

on

= I

nfi

nit

ely

man

y s

olu

tio

ns

–D

epen

den

t=c 1 c 2

Gra

phic

al R

epre

senta

tion

Coi

nci

dent

Lines

Pair

of

Lin

es =

+2

–4 =

0x

y

2+

4–12 =

0x

y

a1

a2=

1 2,

b1

b2

=2 4

,c 1 c 2

=–4

–12

Co

mp

are

th

e R

ati

os

==

a1

a2

b1

b2

Alg

ebra

ic I

nte

rpre

tati

on

: N

o s

olu

tio

n–

Inco

nsi

sten

t

≠

c 1 c 2

Gra

phic

al R

epre

senta

tion

Par

alle

l Li

nes

So

luti

on

: F

rom

eq

uati

on

(ii

),=

3–2

xy

sub

stit

ute

valu

e o

fin

eq

. (i

)x

7(3

–2

)–15

= 2

yy

–29

=–19

=y

y⇔

= 3

–2

x=

19

29

19

29

49

29

So

lve:

7–15

= 2

–(i

)x

y

+2

= 3

–(i

i)x

yBy

Subs

titu

tion

So

lve:

+3

= 6

xy

2+

3=

12

xy

No

w,

Ad

din

g e

qu

ati

on

(i)

an

d (

ii)

3=

18 o

r=

6x

x

Again

, fr

om

(i)

×2–

(ii)

3=

0 o

r,=

0y

y

Hen

ce,

= 6

,=

0x

y

By

Elim

inat

ion

No

w,

fro

m=

3–

2x

y

–(i

)

–(i

i)

CHAP

TER

: 3 p

air

of

line

ar e

qua

tions

in t

wo v

ariabl

es


Ro

ots

are

2 3,

–1 2–1 2

2 3x

=o

r=

x

Th

e ro

ots

of

6–

–2=

0x

x2

(3–2)

= 0

or

(2+

1)

= 0

xx

By F

act

ori

zati

on

Fin

d r

oo

ts o

f6

––2=

0x

x2

So

luti

on

: 6

+3

–4

–2 =

0x

xx

2

3(2

+1)–

2(2

+1)

= 0

xx

x

(3–2)(

2+

1)

= 0

xx

Qua

dra

tic

Eq

ua

tions

Eq

uati

on

of

deg

ree

2,

in o

ne

vari

ab

le

Mea

nin

g

ax

bx

c2+

+=

0,

,–

real

nu

mb

ers

ab

c

0a

≠

Genera

l Fo

rm

Qua

dra

tic F

orm

ula R

oo

ts o

f+

+=

0 a

re g

iven

by

ax

bx

c2 –

±b

bac

2–4

2a

Fo

r q

uad

rati

c eq

uati

on

++

= 0

,a

xbx

c2

–4

is D

iscr

imin

an

t (D

)b

ac

2

3.

D <

0N

o r

eal

roo

ts (

imagin

ary

)

So

luti

on

: 2

–5

+3 =

0x

x2

x2–

x+

= 0

5 2

3 22

2

–+

= 0

⇔

⇔

x–

5 4

x–

5 4

5 4

3 2

=±

1 4

x=

or

=x

5 4

1 4+

5 4

1 4–

x=

or

= 1

x3 2

2

–=

0x–

5 4

1 16

2

=1 16

x–

5 4

So

lve:

2–5

+3 =

0x

x2

By

Com

plet

ing

the

Squ

are

Solu

tion o

f a

Qua

dra

tic E

qua

tion

Na

ture o

f R

oots

2.

D =

0T

wo

eq

ual

real

roo

ts

1.

D >

0T

wo

dis

tin

ct r

eal

roo

t

CHAP

TER

: 4 q

uadr

atic e

qua

tions


Arithm

etic

Prog

ressio

ns

Defin

itio

n

Genera

l fo

rm

Com

mon D

ifference

Exa

mp

les

Ho

w m

an

y 2

-dig

it n

um

ber

sare

div

isib

le b

y 3

?

2-d

igit

nu

mb

ers

div

isib

le b

yare

12,

15,

18,

...

99

= 1

2,

= 3

,=

99

ad

an

=+

(–1)

aa

nd

n

99 =

12 +

(–1)3

n

n=

30

n–1 =

i.e.

,=

29

87 3

Su

m o

ffi

rst

n p

osi

tive

inte

ger

sL

et s

= 1

+ 2

+ 3

+ .

..n

n

= 1

, la

st t

erm

=a

ln

s=

nn

al

(+

)

2=

n1

n(

+)

2

Nth

term

Wh

en f

irst

& l

ast

ter

ms

are

giv

en :

or

S=

(

+)

nn

aa

n 2S

= (

+)

na

ln 2

a–

firs

t te

rm–

tota

l te

rms

n

–te

rma

nn

th

–la

st t

erm

l

Wh

en f

irst

ter

m a

nd

co

mm

on

dif

fern

ce a

re g

iven

:

a–

firs

t te

rm–

com

mo

n d

iffe

ren

ced

–to

tal

term

snS

= (

2+

(–1)

)n

an

dn 2

Sum

(s)

If,

,,

are

in

AP,

ab

c

bis

ari

thm

etic

mea

n

b=

ac

+ 2

Arithm

etic m

ea

n

aa

da

da

da

nd

,+

,+

2,

+3

, ..

.+

(–1)

Lis

t o

fn

um

ber

s in

wh

ich

each

ter

m i

so

bta

ined

by a

dd

ing a

fix

ed n

um

ber

to

th

e p

rece

din

gte

rm e

xce

pt

the

firs

t te

rm.

Fix

ed n

um

ber

is

call

ed c

om

mo

n d

iffe

ren

ce.

•Fix

ed n

um

ber

in

ari

thm

etic

pro

gre

ssio

n w

hic

h p

rov

ides

th

eto

an

d f

ro t

erm

s by a

dd

ing/

sub

tract

ing

fro

m t

he

pre

sen

t n

um

ber

.•C

an

be

po

siti

ve

or

neg

ati

ve.

Fro

m b

egin

nin

g=

+(

–1)

aa

nd

n

Her

e–

firs

t te

rma

–co

mm

on

dif

fere

nce

d

–te

rma

nn

th

Fro

m t

he

end

=–

(–1)

al

nd

n

Her

e–

last

ter

ml

–co

mm

on

dif

fere

nce

d

–te

rma

nn

th

CHAP

TER

: 5 a

rith

metic

progre

ssions


Lines (

In T

wo-

Dim

ensio

ns)

Mea

nin

g

Centr

oid

Coord

ina

te a

xis

Ho

rizo

nta

l =

–ax

is (

Ab

scis

sa)

x

Ver

tica

l =

–ax

is (

Ord

inate

)y

ord

inate

Y

ab

scis

saXX

’

Y’

Dis

tance

form

ula

Ex

am

ple

Are

th

e fo

llo

win

g p

oin

ts v

erti

ces

of

asq

uare

: (

1,

7),

(4,

2),

(–1,

–1),

(–4,

4)?

A (

1,

7);

B =

(4,

2);

C =

(–1,

–1);

D =

(–4,

4)

Sin

ce,

AB

= B

C =

CD

= D

A a

nd

AC

= B

D.

All

fo

ur

sid

es a

nd

dia

go

nals

are

eq

ual

Hen

ce,

AB

CD

is

a s

qu

are

AB

=

BC

=

CD

=

DA

=

AC

=

BD

=√(4

+ 4

)+

(2

–4)

22

√(1 +

1)

+ (

7 +

1)

22

√(1 +

4)

+ (

7–

4)

22

√(–1 +

4)

+ (

–1

–4)

22

√(4 +

1)

+ (

2 +

1)

22

√(1–

4)

+ (

7–

2)

22

= = = = = =

Area

of T

ria

ngle

Mid

-poin

t Li

ne S

egm

ent

Section f

orm

ula

Ex

am

ple

Fin

d p

oin

t o

fT

rise

ctio

n o

fli

ne

segm

ent

AB

, A

(2,

–2)

an

d B

(–7,

4)

Co

ord

inate

of

P

i.e.

,(–

1,

0)

Co

ord

inate

of

Q

i.e.

,(–

4,

2)

=1(–

7)

+ 2

(2) ,

1 +

2

1(4

) +

2(–

2)

1 +

2

2(–

7)

+ 1

(2) ,

2 +

1

2(4

) +

1(–

2)

2 +

1

A (

,)

xy

11

(,

)x

y2

2(

,)

xy

33

BC

G

++

xx

x1

23

3

yy

y1

23

++

,3

G=

xx

12

+R

,

2

yy

12

+ 2

,m

xm

x1

22

1±

mm

12

±

my

my

12

21

±

mm

12

±

Inte

rnall

y (

+)

Ex

tern

all

y (

–)

PQ

=√(

–)

+ (

–)

xx

yy

21

22

12

Are

a =

|[

(–

) +

(–

)x

yy

xy

y1

23

23

1+

(–

)]|

xy

y3

12

1 2

Q=

Stu

dy

of

alg

eb

raic

eq

ua

tio

ns

on

gra

ph

s

IQ

uad

ran

t

IVQ

uad

ran

tII

IQ

uad

ran

t

IIQ

uad

ran

tX’

Y’Y

X

(–,+

)(+

,+)

(–,–

)(+

,–)

34 34 34 34 68 68

CHAP

TER

: 6 l

ines

(in t

wo d

imen

sions

)


Tria

ngle

s

1.

Ifa l

ine

is d

raw

n p

ara

llel

to

on

e si

de

of

a t

rian

gle

to

in

ters

ect

the

oth

er t

wo

sid

es i

n d

isti

nct

po

ints

,th

e o

ther

tw

o s

ides

are

div

ided

in t

he

sam

e ra

tio.

If,

DE

||

BC

AD

DB

=A

E

EC

then

2.

Ifa l

ine

div

ides

an

ytw

o s

ides

of

a t

rian

gle

in

th

esa

me

rati

o,

then

th

e li

ne

isp

ara

llel

to

th

e th

ird

sid

e.th

en,

DE

||

BC

AD

DB

=A

E

EC

If

3.

Ifin

tw

o t

rian

gle

s,co

rres

po

nd

ing a

ng

les

are

eq

ual,

then

th

eir

corr

esp

on

din

g s

ides

are

in

the

sam

e ra

tio

(o

r p

rop

ort

ion

) an

dh

ence

th

e tw

o t

rian

gle

s are

sim

ilar.

(AA

A c

rite

rio

n)

∆∆

AB

CD

EF

IfA

=D

,B

=E

,∠

∠∠

∠

C =

F∠

∠

then

,A

B

DE

=B

C

EF

=A

C

DF

4.

Ifin

tw

o t

rian

gle

s, s

ides

of

on

e tr

ian

gle

are

pro

po

rtio

nal

to (

in t

he

sam

e ra

tio

of

) th

ei.

e.,

sid

es o

fth

e o

ther

tri

an

gle

, th

enth

eir

corr

esp

on

din

g a

ng

les

are

equ

al

an

d h

ence

th

e tw

o t

rian

gle

sare

sim

ilia

r.(S

SS

cri

teri

on

)

then

,A

=D

,B

=E

,∠

∠∠

∠

C =

F∠

∠

AB

DE

=B

C

EF

=C

A

FD

If

AB

C∆

∆DE

F

Area

of S

imila

r T

ria

ngle

s

InA

BC

, le

t D

EB

C.

Th

en,

∆(i

)A

D

DB

=A

E

EC

(ii)

(iii

)

AB

DB

=A

C

EC

AD

AB

=A

E

EC

Theorem

s

Sum

mar

y

Sim

ilarity

Rig

ht

ang

led

tria

ngle

theorem

Py

tha

gora

s

(i)

Co

rres

po

nd

ing a

ng

les

are

eq

ual

(ii)

Co

rres

po

nd

ing s

ides

are

in

th

e sa

me

rati

o

A

BC

P

QR

∆∆

AB

CP

QR

~

1.

Ifa p

erp

end

icu

lar

is d

raw

n f

rom

th

e ver

tex

of

the

righ

t an

gle

of

a r

igh

t tr

ian

gle

to

th

e h

yp

ote

nu

seth

en t

rian

gle

s o

n b

oth

sid

es o

fth

e p

erp

end

icu

lar

are

sim

ilar

to t

he

wh

ole

tri

an

gle

an

d t

o e

ach

oth

er.

In r

igh

tA

BC

, B

DA

C,

∆⊥

then

,A

DB

AB

C∆

∆~ ~

∆∆

BD

CA

BC

∆∆

AD

BB

DC

~A

B

CD

3.

In a

tri

an

gle

, if

squ

are

of

on

esi

de

is e

qu

al

to t

he

sum

of

the

squ

are

so

fo

ther

tw

o s

ides

, th

en t

he

an

gle

op

po

site

the

firs

t si

de

is a

rig

ht

an

gle

.

IfA

C=

AB

+B

C2

22

then

,B

= 9

0°

A BC

2.

In a

rig

ht

tria

ng

le,

the

squ

are

of

the

hyp

ote

nu

se i

s eq

ual

to t

he

sum

of

the

squ

are

s o

fth

e o

ther

two

sid

es.

In r

igh

tA

BC

,∆

BC

= A

B+

AC

22

2

C

BA

5.

Ifo

ne

an

gle

of

atr

ian

gle

is

equ

al

to o

ne

an

gle

of

the

oth

er t

rian

gle

an

d t

he

sid

es i

ncl

ud

ing t

hes

ean

gle

s are

pro

po

rtio

nal,

then

th

e tw

o t

rian

gle

s are

sim

ilar.

(SA

S c

rite

rio

n)

then

,A

BC

DE

F∆

∆~

AB

DE=

AC

DF

&A

=D

∠∠

If

Th

e ra

tio

of

the

are

as

of

two

sim

ilar

tria

ng

les

is e

qu

al

to t

he

squ

are

of

the

rati

oo

fth

eir

corr

esp

on

din

g s

ides

Her

eA

BC

PQ

R∆

∆~

B

A

CM

Q

P

RN

ar(

AB

C)=

AB

2

ar(

PQ

R)

PQ

=B

C

QR

2 =C

A

RP

2

CHAP

TER

: 7 t

rian

gle

s


Cir

cle

s

Defin

itio

n

O

rra

diu

s cen

tre

pT

he

locu

s o

fa p

oin

teq

uid

ista

nt

fro

m a

fix

edp

oin

t. F

ixed

Po

int

is a

cen

tre

& s

epara

tio

n o

fp

oin

ts i

n t

he

rad

ius

of

circ

le.

Non-i

nte

rsecting lin

e

O

rP

P Q

No

co

mm

on

po

int

bet

wee

nli

ne

PQ

an

dci

rcle

.

Seca

nt

Tangent

and

ta

ngent

poin

t

O

rP

A

Q

On

ly o

ne

com

mo

np

oin

t b

etw

een

cir

cle

an

d P

Q l

ine.

Theorem

s

Facts

2.

Th

ere

is o

ne

an

d o

nly

on

e ta

ngen

t to

a c

ircl

ep

ass

ing t

hro

ugh

a p

oin

tly

ing o

n t

he

circ

le.

P

1.

Th

ere

is n

o t

an

gen

tto

a c

ircl

e p

ass

ing

thro

ugh

a p

oin

t ly

ing

insi

de

the

circ

le.

P

3.

Th

ere

are

ex

act

ly t

wo

tan

gen

ts t

o a

cir

cle

thro

ugh

a p

oin

t ly

ing o

uts

ide

the

circ

le.

T1

P

T2

Q

X

P

Y

O

OP

Q =

90°

Th

e ta

ngen

t at

an

yp

oin

t o

fa c

ircl

e is

per

pen

dic

ula

r to

the

rad

ius

thro

ugh

the

po

int

of

con

tact

PQ

= P

R

O

Q R

P

Th

e le

ngth

s o

fta

ngen

ts d

raw

nfr

om

an

ex

tern

al

po

int

to a

cir

cle

are

eq

ual

O

r

Q

B

AP

Tw

o c

om

mo

np

oin

ts b

etw

een

lin

e P

Q a

nd

circ

le.

CHAP

TER

: 8 c

ircl

es


Const

ructions

Tangent

to c

ircle

Ste

p 3

. Jo

in P

Qan

d P

R,

req

uir

edta

ngen

ts t

oth

e ci

rcle

Ste

p 1

. Jo

in P

Oan

d b

isec

t it

at

mid

-po

int

M

Defin

itio

n

To

dra

wgeo

met

rica

l sh

ap

es u

sin

gco

mp

ass

es,

rule

r et

c

Line S

egm

ent

Div

isio

nof lin

e in r

atio’s

Met

hod

1

Giv

en:

Lin

e se

gm

ent,

rati

o (

3 :

2)

Ste

p 3

. T

hro

ugh

A(

= 3

), d

raw

3m

lin

e p

ara

llel

to

BA

5

cutt

ing A

B a

t C

AC

:C

B =

3 :

2

Ste

p 2

. L

oca

te 5

po

ints

A,

A,

A,

A,

Aat

equ

al

12

34

5

dis

tan

ces

(A=

A=

A=

A=

A).

12

34

5

Join

BA

5

Ste

p 1

. D

raw

an

y r

ay A

X,

Mak

ing a

cute

an

gle

wit

h l

ine

segm

ent

AB

Met

hod

2G

iven

: L

ine

segm

ent,

rati

o (

3 :

2)

1.

Dra

w a

ny r

ay A

X m

ak

ing

an

acu

te a

ng

le w

ith

lin

ese

gm

ent

AB

2.

Dra

w r

ay B

Y |

| A

X

3.

Lo

cate

A,

A,

A1

23

(=

3)

on

AX

an

dm

B,

B(

= 2

) o

n B

Y1

2n

Join

AB

, in

ters

ecti

ng

32

AB

at

CA

C :

CB

= 3

: 2

Tria

ngle

sim

ilar t

o g

iven

tria

ngle

3.

Dra

w l

ine

para

llel

to

BC

4

fro

m B

inte

rsec

tin

g B

C a

t C

'.3

Dra

w l

ine

||

to

AC

fro

m C

'in

ters

ecti

ng A

B a

t A

'A

'BC

' is

th

e re

qu

ired

tri

an

gle

Ste

p 2

. M

as

cen

tre

an

dra

diu

s =

MO

dra

w a

circ

le,

inte

rsec

tin

ggiv

en c

ircl

e at

Q a

nd

R

Giv

en:

Cir

cle

wit

h c

entr

e O

an

d p

oin

t P

ou

tsid

e it

.

1.

Dra

w a

ny

ray B

X m

ak

ing

an

acu

te a

ng

lew

ith

BC

Co

nst

ruct

a t

rian

gle

sim

ilar

to a

giv

en∆

AB

C w

ith

sid

es o

fth

e co

rres

po

nd

ing s

ides

of

AB

C∆

3 4

2.

Lo

cate

4 p

oin

ts (

gre

ate

r o

f3

dis

tan

ce f

rom

each

oth

er

(BB

= B

B=

BB

= B

B)

11

22

33

4

Join

BC

4

an

d 4

in

)

on

BX

at

equ

al

3 4

CHAP

TER

: 9 c

ons

truc

tions


Intr

oduc

tion

toTr

igonom

etr

ya

nd

Trig

onom

etr

icId

entitie

s

Stu

dy o

fre

lati

on

ship

s b

etw

een

the

sid

es &

an

gle

s o

fa r

igh

t tr

ian

gle

Trig

on

om

etr

y

Trig

on

om

etr

y R

atio

Sin

e o

fA

∠B

C

AC

Co

sin

e o

fA

∠A

B

AC

Tan

gen

t o

fA

∠B

C

AB

Co

seca

nt

of

A∠

AC

BC

Sec

an

t o

fA

∠A

C

AB

Co

tan

gen

t o

fA

∠A

B

BC

=

=

=

===

Sid

eo

pp

osi

te

toA

∠

Sid

e o

pp

osi

te t

oC

∠

A

C B

Hyp

oten

use

No

t (∞

)d

efin

ed

2

∠A

30°

0°

45°

60°

90°

1 21

0si

n A

1

2

01

cos

A2

11 2

10

No

t (

∞)

def

ined

tan

A1 3

√ 21

cose

c A

21

No

t (

∞)

def

ined

sec

A2

10

No

t (∞

)d

efin

edco

t A

1

Va

lues

Com

ple

menta

ry

Angle

s

sin

(90°

–A

) =

co

s A

cos

(90°

–A

) =

sin

A

tan

(90°

–A

) =

co

t A

cot

(90°

–A

) =

tan

A

sec

(90°

–A

) =

co

sec

A

cose

c (9

0°

–A

) =

sec

A

No

te:

Ho

w t

o l

earn

th

e re

lati

on

“S

om

e p

eop

le h

ave”

“C

url

y B

row

n H

air

”

“th

rou

gh

pro

per

Bru

shin

g”

sin

=P H

cos

=B H

tan

=P B

Trig

on

om

etr

ic Id

entities

cos

A +

sin

A =

12

2

1 +

tan

A =

sec

A;0

A90

22

≤≤

°≤

≤°

cot

A +

1 =

co

sec

A;0

A90

22

Exa

mp

le

Ex

pre

ss t

an

A,

cos

A i

n t

erm

s o

fsi

n A

So

luti

on

: S

ince

, co

sA

+ s

inA

= 1

22

tan

A =

=si

n A

cos

A

sin

A

√1

–si

nA

2

cos

A =

1–

sin

Aco

s A

=2

2i.

e.√

1–

sin

A2

3

3 3 3

3

33

22 2 2

CHAP

TER

: 10 in

trodu

ction

to t

rigono

metr

y an

d tr

igono

metr

ic id

entities


Heig

hts

and

Dis

tance

sExa

mp

les

Ob

ject

Hei

gh

tD

eter

min

e h

eigh

t o

fo

bje

ct A

B

In�A

BC

B =

90°,

C =

60°

∠∠

Her

e, t

an

60°

=A

B

BC

i.e.

,A

B =

15√ 3

m

=A

B

15

√3

Mea

surin

g A

ngle

s

An

gle

of

Ele

vati

on

is

equ

al

to A

ng

le o

fD

epre

ssio

n

Angleof

Dep

ressio

n

Ap

plic

ation–T

rig

on

om

etr

icR

atios (

To d

ete

rm

ine) A

30°

60°

B20

0m

D CF

ind

an

dx

h

x

h

Dis

tan

ce b

etw

een

tw

o o

bje

cts

90°

ABCD

(I)

BD

is

a t

ree

AC

= D

Cif

CD

is

bro

ken

(ii)

AB

DFlag C

Find

flag

leng

thx

h

hei

ght

/ le

ngt

h o

f an

obj

ect

Det

erm

ine

wid

th A

B

Fro

m f

igu

re,

AB

= A

D +

DB

In r

igh

tA

PD

A30°

D90°

∆∠

∠=

=,

√3 m

tan

30°

=i.

e.,

AD

= 3

PD

AD

In r

igh

tB

PD

B45°

D90°

∆∠

∠=

=,

tan

45°

=i.

e.,

BD

= 3

PD

BD

∴A

B =

(3 +

3)m

= 3

( +

1)m

√ 3√ 3

Dis

tance

h 200

—ta

n 6

0º

= .

..(i

)

h

x+

200

—ta

n 6

0º

= .

..(i

i)

h x—

tan

= .

..(i

)

h+

DC

x—

tan

= .

..(i

i)

CHAP

TER

: 11 he

ight

s a

nd d

ista

nces


Area

s R

ela

ted

to

Cir

cle

s

Cir

cle

•C

ircu

mfe

ren

ce =

× d

iam

eter

= 2

��

r

•A

rea =

�r2

Secto

r

Mea

nin

g

Fo

rmu

la

Po

rtio

n o

fth

eci

rcu

lar

regio

n e

ncl

ose

dby t

wo

rad

ius

an

d t

he

corr

esp

on

din

g a

rc

Are

a

×are

a o

fci

rcle

�

360°

A =

A =

×�

r2�

360°

Len

gth

of

arc

× c

ircu

mfe

ren

ce�

360°

L =

L =

× 2

�r

�

360°

Segm

ent

Are

a =

Are

a o

fth

e co

rres

po

nd

ing s

ecto

r–A

rea o

fth

e co

rres

po

nd

ing t

rian

gle

×–

are

a o

fO

AB

�r2

∆�

360°

= =–

��

r2360°

1 2r2

sin�

Q

P

Are

a o

fS

ecto

r =

1 2×

L×

r

Fo

rmu

la

Area

- C

om

bin

ation o

f f

igures

Are

a o

fT

= A

rea o

fP

+ A

rea o

fQ

PQ

T

Mea

nin

g

Fin

d A

rea o

fsh

ad

ed r

egio

n

14

cm

Are

a o

fsq

uare

AB

CD

= 1

4×

14 c

m2

= 1

96 c

m2

Dia

met

er o

fea

ch c

ircl

e,D

= =

7 c

m14 2

Fo

r ea

ch c

ircl

e, r

ad

ius

()

=r

cm7 2

Are

a o

f1 c

ircl

e =

�r2

Are

a o

f4 c

ircl

es =

4 ×

2 cm2

22 7

7 2()

×

=1

54 4

× 4

cm2

= 1

54

cm

2

Are

a o

fsh

ad

ed r

egio

n =

Are

a o

fA

BC

D–

Are

a o

f4 c

ircl

es

= (

196

–154)

= 4

2 c

m2

Ex

am

ple

Po

rtio

n o

fth

e ci

rcu

lar

regio

nen

clo

sed

bet

wee

na c

ho

rd a

nd

th

eco

rres

po

nd

ing a

rc

Mea

nin

g Se

gm

en

t

Min

or

Seg

men

t

CHAP

TER

: 12

area

s r

elat

ed t

o c

ircl

es


Surfa

ce A

rea

s a

nd

Volu

mes

Su

m o

fall

of

the

surf

ace

are

as

of

the

face

s o

fso

lid

.

Qu

an

tity

of

3-D

sp

ace

encl

ose

d b

y a

ho

llo

w/

clo

sed

so

lid

.

Volu

me

Surfa

ce A

rea

Com

bin

ation o

f S

olid

s

Cu

rved

Su

rface

Are

a

r 2

r 1

lh

Frust

um

of

Cone

5 c

m

5 c

m

Giv

en–

Inn

er d

iam

eter

of

the

Cyli

nd

rica

l g

lass

= 5

cm

Hei

gh

t =

5 c

m

Fin

d–

Act

ual

cap

aci

ty o

fC

yli

nd

rica

l g

lass

So

luti

on

:–

Ap

pare

nt

cap

aci

ty

of

the

gla

ss =

�r

h2

=.

×.

×.

×3

14

25

25

5cm

3

=.

98

125

cm3

Act

ual

cap

aci

ty =

Ap

pare

nt

cap

aci

ty–

Vo

lum

e o

fh

emis

ph

ere

= 9

8.1

25

–32.7

1=

65.4

2 c

m3

Vo

lum

e o

fh

emis

ph

ere

=π

,if

=2.5

cmr

r3

= ×

3.1

4 ×

(2.5

)cm

= 3

2.7

1 c

m3

33

2 – 32 – 3

Conversio

n o

f S

olid

s

1 2

A c

op

per

ro

d–

Dia

met

er 1

cm,

len

gth

8cm

con

ver

ted

in

to a

wir

e o

fle

ngth

18m

Fin

d t

he

thic

kn

ess

of

the

wir

e.

So

luti

on

: V

olu

me

of

the

rod

=×

8cm

�3

=2�

cm3

Let

,is

th

e ra

diu

s o

fcr

oss

-sec

tio

n o

fth

er

wir

e, v

olu

me

=1800

cm�

××

r23

∴×

×=

��

r21800

2

=r2

=r

cm

=-

Th

ick

nes

sD

iam

eter

of

the

cro

ssse

ctio

n

==

.cm

007

cm

(

(2

1 — 900

1 — 30

1 — 15

Volu

me

CS

A =

π(

+)

lr

r1

2

wh

ere

=l

hr

r+

(–

)2

12

2

TS

A =

(+

)+�

�l

rr

rr

l2

2 l2 2

(+

)

To

tal

Su

rface

Are

a

l

r 2

r 1

r 2

r 1

V =

(+

+)

�h

rr

rr

2 12 2

12

1 — 3

CHAP

TER

: 13

sur

face

ar

eas a

nd v

olu

mes


Sta

tist

ics

A c

oll

ecti

on

,an

aly

sis,

in

terp

reta

tio

no

fq

uan

tita

tive

data

Defin

itio

n

Fre

qu

ency

ob

tain

ed b

y a

dd

ing

the

freq

uen

cies

of

all

th

e cl

ass

esp

rece

din

g t

he

giv

ing c

lass

Cum

ula

tive F

req

uency

Cla

ss M

ark

Group

ed

Da

ta

Mea

n

3 M

edia

n =

Mo

de

+ 2

Mea

n

Og

ive

Em

peric

al

Rela

tionship

Up

per

cla

ssli

mit

Lo

wer

cla

ssli

mit

+ 2[

[

Med

ian

=l

+

n – 2–

cf

(

(

f

×h

Mod

e=l

+

ff

10

–

(

(×h

2–

–f

ff

10

2

A C

um

ula

tive

Fre

qu

ency

Gra

ph

Rep

rese

nta

tio

n o

fcu

mu

lati

ve

freq

uen

cies

wit

hre

spec

t to

giv

en c

lass

inte

rvals

Mea

nin

g

Ass

um

ed M

ean

(S

ho

rt c

ut)

xa

=+

–

Her

e,=

(–

)d

xa

Dir

ect

Met

ho

d (

Lo

ng

cu

t)x

=–

∑f i

n i=1

i=1

n∑f

x ii

——

Ste

p D

evia

tio

nx

a=

+–

×h

Her

e,=

ux

a–

—— h

∑f i

n i=1

i=1

n∑f

u ii

——

∑f i

n i=1

i=1

n∑f

d ii

——

CHAP

TER

: 14

sta

tistics


Prob

ab

ility

Theoretica

l P

rob

ab

ility

Exp

erim

enta

l P

rob

ab

ility

Va

lue

Defin

itio

ns

Su

re o

r C

erta

in E

ven

t

Wh

en e

ven

t h

avin

gp

rob

ab

ilit

y t

o o

ccu

r as

1

Exa

mp

les

Dic

e

Tw

o d

ice

are

ro

lled

, w

hat

is p

rob

ab

ilit

yo

fget

tin

g 1

2 a

s a s

um

?

So

luti

on

: N

um

ber

of

po

ssib

le

ou

tco

me

= 6

=36

2

Nu

mb

er o

ffa

vo

rab

le o

utc

om

es =

1

Co

inW

hen

a c

oin

is

toss

ed,

wh

at

wo

uld

be

the

pro

bab

ilit

y o

fap

pea

rin

g h

ead

?

So

luti

on

: T

ota

l o

utc

om

es =

2F

avo

rab

le o

utc

om

es =

1

Req

uir

ed P

rob.

P(E

) =

1 — 2

Card

Wh

at

is t

he

pro

bab

ilit

y o

fget

tin

gan

ace

fro

m a

pack

of

52 c

ard

s?

So

luti

on

: N

um

ber

of

favo

rab

le

ou

tco

mes

= 4

Nu

mb

er o

fp

oss

ible

ou

tco

mes

= 5

2

4P

(E)

=— 52

1=

— 13

Even

t h

avin

go

nly

on

e o

utc

om

e o

fth

eex

per

imen

t

Ele

men

tary

Even

t

Su

m o

fp

rob

ab

ilit

ies

of

all

ele

men

tary

even

ts i

s 1.

Fo

r ev

ents

A,

B,

C;

P(A

) +

P(B

) +

P(C

) =

1

Nu

mb

er o

fo

utc

om

esfa

vo

rab

le t

o E

Nu

mb

er o

fall

po

ssib

leo

utc

om

es o

fth

eex

per

imen

t

P(E

) =W

hat

act

uall

y h

ap

pen

sin

an

ex

per

imen

t

Nu

mb

er o

ftr

ials

in

P(E

) =

wh

ich

th

e ev

ent

hap

pen

ed

To

tal

nu

mb

er o

ftr

ials

Wh

at

we

exp

ect

toh

ap

pen

in

an

ex

per

imen

t

0P

(E)

≤1

≤

Fo

r ev

ent

E,

com

ple

men

t ev

ent,

P(E–

) =

1–

P(E

)

Co

mp

lem

enta

ry E

ven

t

1 — 36

P(E

) =

CHAP

TER

: 15

proba

bility

Documents

B AAL OSW MIND MA...1 10–9 = x 8 = 1 y 10 and = x 8 1 1 = y 10 1 1 ⇔ i.e. xy = 8 and By Cross-Multiplication = 10 ax by c 11 1 ++ = 0 = 0 ax 22 2,,,,, –R eal numbers abc abc