Triangles Class 10 Notes+PYQs Shobhit Nirwan

TRIANGLES
HANDWRITTEN NOTES
(Previous Year Questions included)
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Designed with tar
.

Shobhit Nisman

①

Basic Proportionality Theorem (BPT)

THEOREM - I

If a line is drawn parallel to one side of a triangle to intersect the other two
sides in distinct points , then the other two sides are divided in the same ratio
.

(Also called Thales Theorem)
Given : A A ABC in which a line DE parallel to BC intersects AB at D and AC at Eg
- A

i. e.DE/lBC .

Tote:- ADIB = AI F ?
D f
Construction:- Join BE and CD and draw C-FLAB, DG LAC
. -

- -
-
Proof :- we know Area of D= Ix base x height - --

- -

B- - c
-
-
-

-

ara.Y.a.is#=E:::IIII=Efs-zo:aa:Y.a.zets- ¥::%÷=a± ⑦-

Since g DBDE and D DEC stands on same base and between same parallel Deandre
.

:o art DBDE) = ar (DDE C) -
Now, from ② g & t.,FwIe#gIeIut emebroove.d

K'B → Adding I both sides AEFIB + I = +L
,

ADITI = Affect I,Aff=A£Tf
-
similarly ,

THEOREM - 2 -

[ converse of BPD

If a line di vides a nytotwtohisriddessidoef a tria ngle in the same ratio , then the
.
G i line must be parallel cting AB atD and Al at t A
line Ef interse
ven : A DABO and a

such that ftp.B-AI ②- D Ff-
- - -→
To#or DE Il BC
construction :- Draw DF HBC . intersecting
AC at F
.

or Let us assume that in BABC DE is not parallel to BC . B C

b' ut we constructed of 11 , .

Bc

applying BP7g Ajay = Api ⑦-
from ② A Afdc = Api -

②

adding I both sides in g AE.jo#=Af-FFCC-

FI + d = Affect 1 EI = Aff ffC=fI

Thus we can say point E and f coincides i. e - Df coincides with DE .
,
Hence , DE ABCH. enc-eBoomed -
* HI theorems # generally # Type # Et maid a# For

TYPE #L : Unknown side length Hh1# I the aft 9¥; like : A

figuregivenLPI In the

I

soj Given:-
DEH DEHBC . If AD --3cm , DB - Yang AE = 6cm gfindfc . E
- 2016J
BC A [CBSE D
3¥y B
c

%by BM, FIB = Aff Cas De

3- = # EC = -8cm p

←

LIE In DPQRGSTHQR , f-= and PR- 28cm , find PT s y
-

Eggsfol : Given , ST HOR g =3 g PR -28cm a R.

PT = ?

Passby BPI (as Sellar) g = Pty

It = pP¥, of = zP÷p, 3128-177=57

84=5 PT -13107

BPHTYPE # 2 : Thet Results Proore AHH IT HTT PT = 20.50mg

converse til use ; like :

LPL : In a trapezium , show that any line drawn parallel to the parallel sides of the
sides proportionally.
trapezium , divides the non- parallel [CBSE 20113

SOI : Given : A trapezium ABCD g AB HDC and EFHAB A B
-

.

topee : Affj Eff- s ?' F

- E*

-

.

-

.

construction : Join AC to intersect C-fat G. D -c

Proofs : since AB Il DC and ABHEF
To EFHDC
AgtA÷=Now, in DADC . by BPT ②- f : EFHDC)

similarly for DABO , Caffe -_ LIB or A f f - B f f ⑤-
-

③

from ② k⑧ we get gamepad
.

LII given figure: In the , BABO and DDBC have same base BC and lie on the same side
PQHBA PRHBD , then QRHAD
of Bb . If and prove that .

A- - - - - - - D- [CBSE 20153

- -

- - -

Q - - - - - - - - - -- R-
-

Bp a

Sdl 's given: DABC and DDBC have same base BC and lie on same side of BC
.

Totoro re: OR HAD
construction : Join QR and AD
.

Ipg CadaProof:- by BPT g = t ' POHAB) -20
.

¥3 CARD& = t' PRI IBD)
.

from ② 4④ , we get Aaa - LTS
10RHAIfeemePwYdbyso
,
converse of BPI

Similar Triangles [symbol → n) Ip

Two triangles are said to be similar if :- -

gifting.gg:9/EnnaY:ngYsfdYskasoeaYrop9ttafonae B" " ca " "r
.

i.e ②LA = LP

if.
In DA BC and DPQR LB - Ld g LC=LR -
-
then }MmIptconvexlyGtf AABc rasper , , g

and gtfo = Bff = FIR ⑤-

'ABC ~ D POR .

Lc - LR and
Apba Bday ApfaLA =LP g LB - La , = =
-

,

# 314T TUT entail ② HTT condition check that ETTY - tret ! Staten
CRITERIA FOR SIMILARITY OF TRIANGLES 46¥ lifted THT 311419 IT TIPI I

4) AAA similarity criterion :- A D
In two triangles, it corresponding angles are equal , then their

corresponding sides are In the same ratio ; i.e. they are proportional
and hence the two triangles .
Tf LA = LD g LB = Lf
ft Afp PEE }i.e. and LC - LF Hence, BABU DDEF
-

then = = B cf f

⑨

B zH¥FhItfK' af corresponding angle of equal # # sit similar triangle et.at# called
A-A similarity criterion . (: Angle som Property) (More explanation in )lecture on our Yt channel '

(2) SSS similarity criterion :- AD
It in two rtaritaion)gletos t,hseidessideosf proportional lie
one triangle are . In
the same
of other triangle , then their
corresponding angles are equal and hence the two triangles are
similar
. B cE E

P e- . if Apte = Eff = Effy } Hence, DABCNDDEF

then , LA = LD g LB= LE g k - LF

Hitif It&AT
91¥ GHAK'B AAA andSss converse # , :. et AIT ② *④

check that at Eid HIEI

(3) SAS Similarity criterion : A D

If one angle of saidetrsiainndguledinisgtehqeusaleatongoknse_anagrele of other

triangle and the

proportional , then the two triangles are similar B
.

ALI AffFfFe. cE F
= and LA - LD

then , DA BCN DDE F

LII CD and GH are respectively the bisector of LA CB and LEG f such that
D and f lies on sides AB and FE of DABC and DEFG respectively. If
.

DABC N Df EG then show that :

cis DDCBNDHGE dis Eff =A£g Eius DDCAN DHGF F

A

set :- Given AABC ~ DFE G D H
CE
Io LA =Lf g LB = LE and LACB = Lf GE B g

since CD and GH are angle bisectors
ooo LD CB = LHGE and LACD =L f GH

'So DAC D - DFGH (AA rule)
,
similarly , DDCBNLHGE CAA rule) - Bst proved
jig proved
Now, for D ACD and DFGH
c¥=A and LACD =L FGH

Hence , DD CAN DHGF (SAS rule) - Eiiisrdprooved .

K3B THcysHprieimTtsepTorlTyifoT anqanugtleeftsTiodtrioeitana#snfg:-lIevasluestiemetihlairdtghaa tf HH EIHT Et DF Ignis IIT stem
angles eqsuimal ilparrovtehtti
o f til att at 4T sides ko proportional
o
prove tht tf Holt ate It if simply Triangles ll ¥ Hot find that Htt htt

prove UT

⑤

¥2: A veotide pole of length Gm cast a shadow am long on the ground and at
the same time a tower casts a shadow 28M
long HeightLol: ⑧ . of tower ? !
Let AB be tower and CD be pole. → tf Lqt A
c
shadow of AB B BE and shadow of is
( ,Ef*¥ )As we side valve
Eso 't similar
DF
. PEE tt

know the light rays from sun will fall on
,
and pole and at same time
tower at same angle .

So , LDCF =LBAE

LDFC =LBEA D7 F 7 E

LCDF = LABE (tower and pole are vertical to ground) B

ooo BABE N DCDF (AAA rule)

Sog AI = BI

CD DF

Alfa = 2yd A- 13=42 ooo height 42mgof tower =

.

Areas of similar Triangles

theorem: The ratio of theareas of two similar triangles is equal to the square of the ratio of
their corresponding sides Ap
.

⑧ aa.ria.az#=HBaHErt--KaaT

B C9 R
E: In the given figure, line segment XY is parallel to side AC of DABC A
and ft divides the triangles into two parts of equal areas
. ×

find AI
°

AB B

Lol:- Since XYHAC , LBXY =LBAC (correspondingangles) c
L BY #Bca
Y

: A ABC NDXBY CAA rule)

o

Asthweeirknocworr,etshpeornadtiniog of two similar triangles is equal to the square of the ratio of

sides
.
i. aar.Y.EE#=HxFs5 ②-
a¥fgA_×B¥ )Mtg
2g (④ @= - °o° ATOgas ABC) = EarlDXB'D

from ② a ④ g faffs) ' = I 9,73 - Ey
-

ABAB Iorg
=

-AX

AB = IAB - EAX
(J2- DAB = FAX

TIZIAI = or IE

AB fZ

.

⑥

¥2 : prove that the area of the equilateral triangle described on the side of an
isosceles right angled triangle. is half the area of equilateral triangle described
on its hypotenuse . E

be a square of side
diagonalsat:Let To - Ea
ABCD a aa
.

Aed, two desired triangles are formed as BABE and DDBF a a B
.

As we know all angles of equilateral triangles are 60
.

)(' Fa
- . BABE N BBFD AAA Rule a a%
. a
c
Asthweeirknocworr,ethspeornadtiniog of two similar triangles is equal to the square of the ratio
sides
. D

(Ffa)Io Tar (EBABHE) = " Ea

a

=I
%
Pythagoras Theorem and its converse

AWE. It a perpendicular Is drawn from the vertex of right
aonngleboothf asidreigshot f angled triangle to the hypotenuse , then triangles
the perpendicular B
are similar to the whole D y

and to each other .

ie. it in right angled DABC, right angled at BGBDLAC AB
then g Is AADBNDABC
E
dB DBDC N DABC

of theorem Ciii, DADBN DBDC that A tin 3111*1

pytha theorem tf prove

# Pythagoras Theorem : In a right angled triangle , sthideessquare of the hypotenuse

is equal to the sum of the squares of the other two . B

Gheen: A right angled AABC, right angled at B Yl

To prod :- ACK AB'tBE 1

Construction:- Draw BDLAC A 17 c

Proof :- By above theorem , ②DADBNDABC - D

⑤and DBDCNDABC -
from ② AAIB = AIN

from ④ ①ABE AD- AC -

Dfg =B€

BCI DC - AC -130
Now adding ① and⑨ AB't BC - = ACCADTDC)

AB' + BE = AC - AC

IABZ-BCKAC.TT Roared

-

Hence A

⑦

Phythagoras# Converse of Theorem:- In a triangle , if square of one side is equal to sum
of the squares of the other two sides , then the angle opposite to first side is right angle.

IPI 's In the figure, DABC is drawn such that ADLBC , then show that A

AT = AB't BE-213C . BD ( CBSE - 20167
.

70¥19Self Given : In DABC , ADLBC B 7 C
ACE AB2tBC- 2BC - BD D

Pntof 's ;E=AD2tDC2IIn DADC g (by pytha)
A- DE AE- Do ②-

In D ABD g AB' = BD't AD' ( by pytha)

⑤AD! AB'-13132 -

from ② 9 ' '
←
-
AC AB BDDC' '
-

A C'= ABIDE - BDZ

feogxcpafnulcahrvntahiansentriietolno-u,yrouYT At = AB't CBC - )BD ' BD'
-

ACK ABZTBCZTBDZ - 2. BC - BD - B D2

ProvedAct ABZTBCZ-213C - BD

Hence -

LII : In an isosceles right angled D.gif hypotenuse is 552 ,then find the length of
the sides of triangle. CCBSE 2015T
AB - BC C : ' Isosceles D) ②-
Sdf:- A

552 Now by pythag AE = ABZ + BC'

=

B7 " A-a = ZABZ (from

C (5542=2 AB'

at # 3¥ # tis et A-BE 25 IA→B=5cmT
angled D # tacket side Itta Isf ' in right angled triangle , HT tht et Right
k3B equation prove a# FIAT att Irf IIHF
square # terms of
above) . ; suet af Hink Fadil Pythagoras Theorem
( like in LPL 541917T MT use

etat fi

AB THODE SAWAI KARLO

t

2020 PREVIOUS YEAR QUESTIONS

aagfg¥BI④
Given DABO - DPQR , if ABIPQ - 43 then ?=
-
,

-

¥the④ In a triangle, it square of one side is equal to the
squares of the other two sides then prove that the angle opposite
,

to first side is right angle

aretz: we know that area of similar triangles are equal to the
proportion of the squares of proportional sides
.

So , aayfft.ae = Apj. = I

←

an Given : In DABC, DEHAC

To peeve : BI = Bpf

solution : BIA BffHACIn A ABC DE g so ②-
,
f. HenceforthRydz BcgIn DABP g DCHAP g
= from ② e

= Bff = Beep

ans or Cliven: LD = LE , which means AE- ED
.

7573 - AEC

-

as A D= A E : DB
.

Now g ADT DB = AETEC isosceles d
.
AB- AC - Io DA BC Ps an

ans - Given : ABC is a triangle

o

ACE AB't Be

To pave : 43=900

construction: construct a DPQR right angled at 9 such that , Pa - AB and QR -43C.

Poyet :- In DPQR , PRI Pce' +ORL (by pytha)

as AB - Pa and QR = BC .

②I
.
PR2= ABZTBCZ -

given⑤A- E- AB'
t BC' ( )
-

from ② & ④ we have ACE PR '
,
→ ACE PR -

Now , in DABC and DPQR
AB=pQ

Bc = OR
AC - PR (from )
by:O DABC I DPQR ( SSS
byis CPCT g LB=LQ congruency)

pqnuethordLd - 900 .

43=900

2019 goings AB - AD + BD = H2 - 3cm
-
Pythagoras theorem (Pg - 6)
→ Also DEHBCCgiven)
,

LADE = LABC - ② (corresponding angles)
In DA BC and DADE

LA - LA (common)

LABC = LADE ( by③)
AA rule)
by:
-
4ABc~ DADE (

Now ,aAH¥DeAB5CF) Tn55

f) I'

-

I

2018

④

Is NCERT Question .

Is Pythalpg -6)

aagffttpBY-g-f.BGAfg Ians⑥ Given ' ABC - Dear & = s so
,

= ¥2 f

t

amf⑦ AP
B Cd R

Given DABCNDPQR , ArloABC) - ar IDPQR)
To prover DABC DPQR
ask.at:7, Bata .ae#l:s:ual: eotna: aotot: : Y:n: asgno'9
- Ha t i .
-

But , as given , arlAAB# =L

AND peer)

=P¥j- AB' =P A

→ Bl ' = OR' = OR
By

Ofi→ ACE = PR
Hewett'oYd:

o
By SSS congruence axiom ABC EPQR A

2016

value of X. D E

8- In DABC , right angled at B , prove ACZZAB B C
.

FIB Afc② In DABC , Dell BC so by thales the08mg =

(AD) ( EC) = CAE) ( DB)

n ca - L) = @-2) Cut2)

⑧ Given: DABCg 43=90 R2- K = 22-4

To Pave : ACE ZABZ 2=91
To

Now , by Pythagoras theorem g AB't BC 'EAC2
AB= BC
by isosceles triangle property g A# + AB AC ' ( r: AB-43C)

Sog
Provedooo 2ABE AC 2
Flume←

⑤ As BABE is right angled triangle ,

CAEJE CAB)'t @E)2 ③-

similarly , DDBC is right angled B.

On So EDIE ④D) ' t (BC) ' ⑤-
9. ② d ④ ' fagot
add, ing Z ftB)2+4342-143135+4302

CAE) ' t (CDT - (AB'tBa) t (BEZ + BDD -

In D ABCg ACK AB't BC '

EgIn DDB DEZ = BE't B D2 -

putting values of in CAEJ-ICCDY-f-ACDTCDeyunwpmfl.tl

④ Join A to midpoint of BC at D
.

#So ED= BE = ⑤BC -
,

In DA ED g ④AE -= AD't EDL -
In DABD ,
AD ' A 132 - BD2 -

Putting value of A-D2 from ( ) FBI)2into ⑤
BIAE' - ABZ- BD' + fD2=AB2 - 't

@ E) )BD -
#BC and C-D= BC from 20

so we get ftbAEZ.BA#
-