Triangles Class 10

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Ex 6.1
Fill in the blanks using the correct word given in brackets:
(i) All circles are (congruent, similar)

Both circles have same shape
Similar but do not have same size.
because all circles are of same shape(round) but,
size may not be same.
Hence, these are similar but not congruent.

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Ex 6.1, 2
Give two different examples of pair of
(i) Similar figures
Two different examples of pair of similar figures are:
(aJAny two rectangles
(b) any two squares.

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Ex 6.1, 3
State whether the following quadrilaterals are similar or not :
D 3cm c
S_15em_R 3em 3em
1.5 of js cm

PiTsem & A 3cm B
Two polygons of same number of sides are similar if;
{a) Their corresponding angles are equal and
(b) Their corresponding sides are in the same ratio.
In case of PQRS and ABCD,
The corresponding angles do not appear equal.
And, Z DAB = 90°
But, 2 SPQ # 90°
Hence, first condition not satisfied. So, they are not similar.

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Ex 6.2, 1
In figure(i) ,DE || BC. Find EC in (i)
A
Sem yj om
D E
3cm ?
It is given that
B Cc
DE || BC
Hence ,
AD AE {Line drawn parallel to one side of triangle,
DB EG intersects the other two sides in distinct points,
Then it divides the other 2 side in same ratio)
Putting the values
18 4
3° EC
15xXEC=1x3
pc = 23
15
EC=2cm

Ex 6.2, 2 teachoo.com
E and F are points on the sides PQ and PR respectively of a APQR.
For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ=3 cm, PF = 3.6 cm and FR=2.4cm
Note- If a line divides any two sides of the triangle in the same
ratio, then the line is parallel to the third side.
P
, 3.9 cm, 3.6cm
So, in order to prove EF II QR
E F
we have to prove that = = "= 3m 2.4m
EQ FR
Q R
L.H.S R.H.S
PE 39 PF _36
FO. 3. FR 24
39 _ 36
= 30 ~ 24
_13 =3
~ 10 2
Since L.H.S # R.H.S
Hence, EF and QR are not parallel

E and F are points on the sides PQ and PR respectively of a PQR. For each of the following cases, state whether EF || QR :
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and FR = 9 cm

E and F are points on the sides PQ and PR respectively of a PQR. For each of the following cases, state whether EF || QR :
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

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Ex 6.2,3
7 + AM AN
In figure, if LM || CB and LN || CD, prove that — =—
AB AD
Given: LM II CB B
And LN II CD M
A Cc
to prove: AM = 4" x
oO prove: AB AD A
Proof:
In A ACB InA ACD
LM ILCB N D
(Line drawn parallel to one side of (Line drawn parallel to one side of
triangle, intersects the other two triangle, intersects the other two
sides in distinct points, Then it divides | sides in distinct points, Then it divides
the other 2 side in same ratio) the other 2 side in same ratio}
AL AM
LC ND

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Ex 6.2, 4
. BF BE
In figure, DE || AC and DF || AE. Prove that, prove that FE EC
Given: DE II AC and A
DF Il AE
D
To prove: SF = 8 L\\
O prove: 7 =a Bé 7 . Cc
Proof:
In A ABC In AAEB
DE IAC DF II AE
(Line drawn parallel to one side of (Line drawn parallel to one side of
triangle, intersects the other two triangle, intersects the other two
sides in distinct points, Then it divides | sides in distinct points, Then it divides
the other 2 side in same ratio) the other 2 side in same ratio)
BE _ BD BF _ BD
EC DA (1) FE DA (2)

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Ex 6.2, 5
In figure, DE | | OQ and DF || OR. Show that EF || QR.
. P
Given: DE Il OQ and L
DF II OR EA“ PAF
To prove: EF Il QR BW
Q R
Proof:
In APQO, In A PRO
DE 110Q DF ILOR
(Line drawn parallel to one side of (Line drawn parallel to one side of
triangle, intersects the other two triangle, intersects the other two
sides in distinct points, Then it divides | sides in distinct points, Then it divides
the other 2 side in same ratio) the other 2 side in same ratio}
PE _ PD PF _ PD
EQ DO (1) SO, oe bo (2)

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Ex 6.2, 6

In figure, A, Band C are points on OP, OQ and OR respectively

such that AB || PQ and AC || PR. Show that BC [| QR.

Given: P

AB Il PQ and AC II PR Le iN

To prove: BC I|QR EE oN

Proof:

InAOPQ in A OPR
(Line drawn parallel to one side of (Line drawn parallel to one side of
triangle, intersects the other two triangle, intersects the other two
sides in distinct points, Then it divides | sides in distinct points, Then it divide.
the other 2 side in same ratio) the other 2 side in same ratio)

OA _ OB (1) OC _ OA 2)

AP BQ CR AP 7

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Ex 6.2, 7
Using Theorem 6.1, prove that a line drawn through the mid-
point of one side of a triangle parallel to another side bisects the
third side. (Recall that you have proved it in Class IX).
A
Given: Let us assume A ABC
Where DE is parallel to BC D E
& Dis the mid point of AB
B c

To prove: E is the mid point of AC
Proof:
In AABC, DE II BC
We know that if a line drawn parallel to one side of triangle,
intersects the other two sides in distinct points,
then it divides the other 2 side in same ratio

AD _ AE

DB EC

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Ex 6.2, 8
Using Theorem 6.2, prove that the line joining the mid-points of
any two sides of a triangle is parallel to the third side. (Recall
that you have done it in Class IX). A
Given: Let us assume A ABC
D E
Where D is the mid point of AB
& Eis the mid point of AC 3 c
To prove: DE II BC
Proof:
In AABC,
D is the mid-point of AB E is the mid-point of AC
=> AD = DB => AE=EC
AD AE
“os (1 “=
=o =1 (1) = a=l (2)

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Ex 6.2 ,9
ABCD is a trapezium in which AB || DC and its diagonals
intersect each other at the point O. Show that 0-8
BO DO
Given: ABCD is a trapezium A B
& diagonals AC & BD intersect at O
To prove: 42 = £2 D c
© prove: 75 = 55
Construction:
Let us draw a line EF II AB II DC passing through point O.
Proof: Now, in AADC
EO IIDC (Because EF II DC)
AE AO (Line drawn parallel to one side of triangle,
So, DE CO intersects the other two sides in distinct points, |--{1)
Then it divides the other 2 side in same ratio}

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Ex 6.2, 10
The diagonals of a quadrilateral ABCD intersect each other at the
point O such that = = . Show that ABCD is a trapezium
. : . A B
Given: ABCD is a quadrilateral
where diagonals AC & BD intersect at O E [— 7 \ F
® 0 o L™
BO DO
D c
To prove: ABCD is a trapezium
Construction: Let us draw a line EF Il AB passing through point O.
Proof: Given “2 = 2
— BO DO
= 40 _ BO
co” DO (1)

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Ex 6.3,1 {i)
State which pairs of triangles in figure are similar. Write the
similarity criterion used by you for answering the question and
also write the pairs of similar triangles in the symbolic form
P
p>
A
In A ABC and A PQR oY
Q ghee Wc ght? 47.
ZA = ZP :
(0)
ZB =2Q
ZC =2ZR
AABC ~APQR (AAA similarity}

State which pairs of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form

State which pairs of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form

State which pairs of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form

State which pairs of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form

State which pairs of triangles in figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form

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Ex 6.3, 2
In figure, AODC ~ AOBA, ZBOC = 125° and ZCDO = 70°. Find
ZDOC, ZDCO and ZOAB.

ao
Given: Shot
AODC ~ AOBA /\
Z BOC = 125° a 8
ZCDO = 70°
To find: ZDOC, ZDCO and ZOAB
Solution:
Here, BD isa line,
So, we can apply linear pair on it.
ZBOC + ZDOC = 180° (Linear Pair }
125° + ZDOC = 180°
ZDOC = 180° — 125°

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Ex 6.3, 3
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect
each other at the point O. Using a similarity criterion for two
: OA _ OB
triangles, show that — =—
oc” OD
A B
a
Given: ABCD is a trapezium with >
<o
and diagonals AB & CD intersecting atO LA —_
D Cc
To prove: OA _ OB
“OPIOVE: oc ~ op
Proof: In A OAB and AOCD
ZAOB = ZDOC (Vertically opposite angles)
ZABO = ZCDO (since AB Ii CD with BD as traversal,
alternate angle are equal
ZBAO = ZOCD (since AB If CD with AC as traversal,
alternate angle are equal
AOAB ~ AOCD (AAA Similarity)

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Ex 6.3, 4
. OR _ OT
In figure, 05 <BR and 21 = 22. Show that APQS ~ ATQR.
: _QR _ QT P x
Given: — = —
== as” PR
a LN,
oa LN R

To prove: APQS ~ ATQR s
Proof:
Given 21=22
PR=QP (Sides opposite to equal angles are equal) ...(1)
Given 22 = 27

QS PR
Putting (1)
QR _ QT wu(2)
QS OP

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Ex 6.3, 5
S and T are points on sides PR and QR of APQR such that ZP =
ZRTS. Show that ARPQ ~ ARTS.
P

Given: A PQR 5
and the points S and T on sides PR and QR.
Such that 2P = ZRTS

Q T R
To Prove: ARPQ ~ ARTS.
Proof:
In ARPQ and ARTS.
ZP = ZRTS (Given)
And ZPRQ= ZTRS=ZR (Common)
So, ARPQ ~ ARTS. (AA similarity)
Hence proved

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Ex 6.3, 6
In figure, if AABE = AACD, show that AADE ~ AABC.
A
Given: AABE = AACD
D E
To Prove: AADE ~ AABC.
B c
Proof:
Given A ABE = AACD
Hence , AB = AC (CPCT) (1)
And AE = AD (CPCT)
i.e. AD = AE «.(2)
Dividing (2) by (1)
AD _ AE

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Ex 6.3, 7
In figure, altitudes AD and CE of AABC intersect each other at the
point P. Show that:
(i) AAEP ~ ACDP C
ps
Given: A ABC
and, altitude AD and CE of triangle
A al B
intersects each other at the point P. E
To Prove : AAEP ~ ACDP
Proof:
In AAEP and ACDP
ZAEP = ZCDP (Since AD & CE are altitudes}
ZAPE = ZCPD (Vertically opposite angles)
Hence, AAEP ~ ACDP (AA Similarity)
Hence proved

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Ex 6.3, 8
E is a point on the side AD produced of a parallelogram ABCD
and BE intersects CD at F. Show that AABE ~ ACFB

A D E
Given: A parallelogram ABCD f
where E is point on side AD produced
& BE intersects CD at F B c
To Prove: AABE ~ ACFB.
Proof:
In parallelogram ABCD
, opposite angles are equal,
Hence, ZA = ZC {1}

.

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Ex 6.3, 10
CD and GH are respectively the bisectors of ZACB and ZEGF
such that D and H lie on sides AB and FE of AABC and AEFG
respectively. If AABC ~ AFEG, show that:

») CD _ AC
O Gizte

G
Given: A ACB and A EGF respectively
And, CD is the bisectors of 2 ACB
Z ACD = 2 BCD => 4 ACB cl OF H
& GH is the bisectors of 2 EGF /~
B

ZFGH = 4 EGH =+ 2 EGF AD
& AABC ~ AFEG
To Prove: ele
=<" GH FG

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Ex 6.3, 11
In figure, E is a point on side CB produced of an isosceles triangle
ABC with AB = AC. If AD L BC and EF L AC, prove that AABD ~ AECF.

A
Given: lsosceles triangle ABC
Where , AB = AC F
And, AD 1 BC, EF L AC
E: BD Cc

E is a point on side CB
To Prove: AABD ~ AECF
Proof :-
A ABC is isosceles triangle
Where , AB = AC
So, 2C = ZB {Angles opposite to equal sides are equal) ...(1)

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Ex 6.3, 12
Sides AB and BC and median AD of a triangle ABC are respectively
proportional to sides PQ and QR and median PM of APQR (see
figure). Show that AABC ~ APOR.
P

Given: a
AABC where AD is the median J
APQR where PM isthe median 8 D CQ M R
g AB Be _ AD

PQ. QR PM
To Prove: AABC ~ APQR.
Proof:-
Since AD is the median,
BD = CD =>.BC
Similarly, PM is the median,
QM = RM =5QR

Ex 6.3, 13 teachoo.com
Dis a point on the side BC of a triangle ABC such that ZADC =
ZBAC. Show that CA? = CB.CD
. A
Given: AABC where
ZADC= 2 BAC
To Prove: CA? = CB.CD
._ CA_CB B D c
i.e, — =—
CD CA

Proof:-
In ABAC and A ADC

ZACB = ZACD (Common angles }

ZBAC = ZADC (Given)
Hence by AA similarity criterion

ABAC ~ AADC

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Ex 6.3, 14
Sides AB and AC and median AD of a triangle ABC are
respectively proportional to sides PQ and PR and median PM of
another triangle PQR. Show that AABC ~ APQR.
P
A
Given: AABC and APQR
AD is the median of A ABC Q R
B D Cc M

,PM is the median of A PQR

AB _ AC _ AD (1)

PQ. PR PM
To Prove:- AABC ~ APQR.
Proof:

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Ex 6.3, 15
A vertical pole of length 6 m casts a shadow 4 m long on the
ground and at the same time a tower casts a shadow 28 m long.
Find the height of the tower. A
Given: 6m
Height of pole = AB = 6m D
Length of pole of shadow = BC=4m Bo4m. ©
Length of shadow of tower = EF = 28
To Find : Height of tower i.e ED
Solution:-
solution E :
In A ABC and A DEF
ZB=Z2E=90° (Both 90° as both are vertical to ground}
LC =2F (Same elevation in both the cases as both
shadows are cast at the same time}
“ AABC ~ ADEF {AA similarity criterion)

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Ex 6.3, 16
If AD and PM are medians of triangles ABC and PQR, respectively
where AABC ~ APQR, prove that AB _ AD

PQ. PM

P

Given: AABC and APQR A
AD is the median of A ABC /~
»PM is the median of A PQR B D c@ M
& AABC ~ APQR.
To Prove: ="
——=" PQPM
Proof:
Since AD is the median
BD = CD => BC
Similarly, PM is the median
QM = RM==QR

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Example 1
If a line intersects sides AB and AC of a AABC at DandE
: : AD _ AE
respectively and is parallel to BC, prove that ap ac
A
Given: AABC ,
where line intersects sides AB
«__D E
and AC at D and E.
And DE II BC
B
AN
To Prove 42 = 48
"AB OAC
Proof:
We know that if a line drawn parallel to one side of triangle,
intersects the other two sides in distinct points,
then it divides the other 2 side in same ratio
Therefore , AD _ AE
DB EC

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Example 2
ABCD is a trapezium with AB | | DC. E and F are points on non-
parallel sides AD and BC respectively such that EF is parallel to
AB (see Fig. 6.14). Show that e_P
ED FC A B

Given: ABCD is a trapezium
where AB II DC

D c
E and F are points non parallel sides AD and BC
such that EF Il AB
To Prove: AF BF
——="ED FC
Proof:
Given AB II DC & EF I|AD
So, EF II DC (Lines which are parallel to same line are

parallel to each other)

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Example 3
. PS _ PT :
in figure , 30 "TR and Z PST = Z PRQ, Prove that PQR is an
isosceles triangle.
P
Given:
Ps _ Pr
sq.TR s T
ZPST=2Z PRQ
Q R
To prove: PQR is an isosceles triangles
Proof:
Given Ps Pt
SQ TR

“ STILQR (if a line divides any two sides of a

triangles in the same ratio,

then the line is parallel to the third side)

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Example 4
In figure, if PQ || RS, prove that APOQ ~ ASOR.
R
Given: PQ II RS P
(¢)

To Prove :- APOQ~ ASOR

Q s
Proof:
In A POQ & ASOR
ZPOQ=2Z SOR (Vertically opposite angles)
ZP=24S (As PQ HRS, Alternate angles)
ZQ=Z2R (As PQ II RS, Alternate angles)
Using AAA similarity criterion
Therefore, APOQ~ASOR
Hence proved

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Example 5
Observe figure and then find 2 P.
R
A 63 16
fe
ot

pO; cP 7 Q
Given: AB = 3.8, PQ= 6V3, BC=6, QR=12, AC=3V3, PR=7.6
To find:- angle Pi.e. 2 P
Solution:-
in AABC andARQP
Let us find the ratio of sides

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Example 6
In figure, OA. OB = OC. OD. Show that 2A=2Cand2B=2D.
Cc
Given: OA x OB=O0OCx OD A
6
To Prove: 2A=2Cand 2B=2D D
B
Proof:-
OA.OB=O0C.O0D
OA _ OD (1)
oC” OB
In AAOD & ACOB
oa oD (From (1))
OC” OB
ZAOD=2Z COB (Vertically opposite angles }
Using SAS similarity criterion
So, AAOD ~ ACOB

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Example 7
A girl of height 90 cm is walking away from the base of a lamp-
post at a speed of 1.2 m/s. If the lamp is 3.6 m above the
ground, find the length of her shadow after 4 seconds.
A
Given: Lamp post (AB) = 3.6m 7
Height of girl (CD)=90 cm = =~ m=0.9m
Speed = 1.2 m/sec.
. Cc
To Find : Length of her shadow i.e. DE - -
D
t
. Shadow
Solution: -
The girl walks BD distance in 4 seconds
We know that,
Speed = Distance
Time

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Example 8
in Fig. 6.33, CM and RN are respectively the medians of A ABC
and A POR. If A ABC ~ A PQR, prove that :
(i) AAMC ~ APNR

Q N P
Given: A ABC and APQR

A
CM is the median of A ABC
and RN is the median of A POR M
Also , AABC ~ A PQR Cc
B R

To Prove: AAMC ~ A PNR
Proof:
CM is median of A ABC
So, AM = MB =3AB (1)
Similarly, RN is the median of A PQR
So, PN = QN==PQ (2)

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Example 9
In figure, the line segment XY is parallel to side AC of A ABC and
it divides the triangle into two parts of equal areas. Find the
tio
ratio =F A
espera A»
XY is parallel to AC i.e. XY ILAC ad “N NN
ar(A BXY ) = ar(AXYC)
To find :“%
———=" AB
Proof:
In A ABC & AXBY,
ZABC=ZXBY (Common)
ZACB=ZXYB_ (Since XY |l AC, corresponding angles are equal )
A ABC ~ AXBY (AA similarity)

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Example 10
In figure, 2 ACB = 90° and CD L AB. Prove that ae ==

AC? AD
Given:- ACAB, s
2 ACB = 90°
And CD 1 AB

A D B

To Prove :- ae = 2
——=<" Ac? ~ AD
Proof:
From theorem 6.7,
If a perpendicular is drawn from the vertex of the right angle to
the hypotenuse then triangles on both sides of the
Perpendicular are similar to the whole triangle and to each other
So, AACD ~ AABC (2)
& ABCD ~ABAC. ...(2)

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Example 11
A ladder is placed against a wall such that its foot is at a distance
of 2.5 m from the wall and its top reaches a window 6 m above
the ground. Find the length of the ladder. —,
ih
Given :- 4
Distance from wall = BC = 2.5m ZB
Height of window = AC = 6m
KZ

To Find : Length of ladder i.e. = AB ll |!

° 2.5m
Solution:
Since the wall will be perpendicular to ground
Z ACB = 90°
A ACBisa right angle triangle

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Example 12 eames
In figure, if AD L BC, prove that AB? + CD? = BD* + AC’.
c

Given: A ABC where

AD 1 BC ,.

To Prove:- AB? + CD? = BD? + AC2

Proof:

Since AC 1 BD

Z ADC = Z ADB =90°
So, A ADB is a right triangle So, A ADC is a right triangle
Using Pythagoras theorem Using Pythagoras theorem
(Hypotenuse)? = (Height)?+ (Base)’ | (Hypotenuse)? = (Height)? + (Base)?

(AB)? = (AD) + BD? (4) (AC)? = (AD)? + CD2.(2)

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Example 13
BL and CM are medians of a triangle ABC right angled at A. Prove
that 4 (BL? + CM2) = 5 BC2.
B
Given:
M
A ABC right angled at A ,i.e., 2 A = 90°
Where BL and CM are the medians
c iL A
To Prove: 4(BL? + CM?) = 5 BC?
Proof :-
Since BL is the median,
AL=CL=>AC (1)
Similarly, CM is the median
AM = MB = 5 AB (2)

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Example 14
O is any point inside a rectangle ABCD (see Fig. 6.52}. Prove
that OB? + OD? = OA? + OC?.
Given : Rectangle ABCD , and A D
a point O inside rectangle . Na
P Q
To prove :- OB? + OD? = OA? + OC? Loo
B c
Proof :-
Let us draw a line PQ, through O which is parallel to BC.
Hence, PQ II BC
=> PQIIAD (Opposite sides of rectangle are parallel)
All angles of a rectangle are 90° ,,
So, ZA=ZB=Z2C=2D=90°

Theorem 6.1: feachoo.com
If a line is drawn parallel to one side of a triangle to intersect the
other two side in distinct points, the other two sides are divided in
the same ratio. A
Given: A ABC where DE || BC lax
AD _ AE 8 c
To Prove: — =—
==" DB EC
A
Construction: Join BE and CD
D
Draw DM L AC and EN L AB.
B Cc
Proof:

Theorem 6.2: teachoo.com
If line a divides any two side of a triangle in the same ratio, then
the line is parallel to third side.
Given: A ABC and a line DE intersecting AB at D and AC at E,
AD _ AE A
such that De = EC
r=

To Prove: DE || BC D
Construction: Draw DE’ parallel to BC. J \

B Cc
Proof:
Since DE’ || BC,
By Theorem 6.1 :/f a line is drawn parallel to one side of a triangle
to intersecting other two sides not distinct points, the other two
sided are divided in the same ratio.

Theorem 6.3 (AAA Criteria} teackoo.com

If in two triangles, corresponding angles are equal, then their

corresponding sides are in the same ratio (or proportion) and

hence the two triangle are similar.

Given: Two triangles AABC and ADEF such that D

ZA=2D, ZB=ZE& ZC=ZF A

To Prove: AABC ~ ADEF /\ p a
B CE F

Construction: Draw P and Q on DE & DF

such that DP = AB and DQ = AC respectively

and join PQ.

AA Criteria teachoo.com
If two angles of one triangle are respectively equal to two angles
of another triangle, then the two triangles are similar.
Given: Two triangles AABC and ADEF such that D
2B=ZE& ZC= ZF A
To Prove: AABC ~ ADEF /\
B CE F

Proof:

In A ABC, In A DEF,

By angle sum property By angle sum property

ZA+ 2ZB+2ZC=180° ...(1) 2D+2ZE+2ZF=180° _ ...{2)

Theorem 6.4 (SSS Criteria) : teachoo.com
If in two triangles, sides of one triangle are proportional to (i.e., the
same ratio of} the sides of the other triangle, then their corresponding
angles are equal and hence the two triangle are similar. D
. . A
Given: Triangle AABC and ADEF
AB_ BC _CA
B CE F

To Prove: ZA = ZD, ZB = ZE, 2C = ZF

and AABC ~ ADEF

Construction: Draw P and Q on DE & DF

such that DP = AB and DQ = AC respectively

and join PQ.

Theorem 6.5 (SAS Criteria) teachoo.com
If one angle of a triangle is equal to one angle of the other triangle
and sides including these angles are proportional then the
triangles are similar.

D
Given: Two triangles AABC and ADEF such that /\
ZA =2D A
AB _ AC
DE DF P Q
To Prove: AABC~ ADEF 8 CE F
Construction: Draw P and Q on DE & DF
such that DP = AB and DQ = AC respectively
and join PQ.

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Theorem 6.6:
The ratio of the areas of two similar triangles is equal to the
square of ratio of their corresponding sides.
Given: AABC ~ APQR A P

Bom ca y R

ar(ABC) /fAB\2 /(BC\2_ ac\?

To Prove: ————~- = [—_} =[—_} =(—
————— ar (PQR) \PQ QR PR
Construction: Draw AM L BC and PN L OR.

Theorem 6.7: teachoo.com
If a perpendicular is drawn from the vertex of the right angle of a
right triangle to the hypotenuse then right triangle on both sides of
the perpendicular are similar to the whole triangle and to each other
Given: AABC right angled at B
& perpendicular from B intersecting AC at D. (he. BD 1 AC}
sy,
To Prove: AADB ~ AABC
ABDC ~ AABC
& AADB~ ABDC
A D Cc

Theorem 6.8 (Pythagoras Theorem) : teachoo.com
If a right triangle, the square of the hypotenuse is equal to the sum
of the squares of other two sides. B

YY?
Given: AABC right angle at B (~
To Prove: AC? = AB* + BC?

A D Cc

Construction: Draw BD 1 AC
Proof: Since BD L AC
Using Theorem 6.7: /f a perpendicular is drawn from the vertex of
the right angle of the a right triangle to the hypotenuse then
triangle on both side of the perpendicular are similar to whole
triangle and to each other.

Theorem 6.9: teachoo.com
In a triangle, if square of one side is equal to the sum of the square!
of the other two sides, then the angle opposite to the first side is a
right angle. A
Given: A triangle ABC in which
AC? = AB? + BC?
B Cc

To Prove: ZB = 90°
Construction: Draw A POR right angled at Q, such that

A P
PQ = AB and QR = BC.

B ca R

Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles.The height of Vijay’s house if 20m when Vijay’s house casts a shadow 10m long on the ground. At the same time, the tower casts a shadow 50m long on the ground and the house of Ajay casts 20m shadow on the
ground.
Question 1
What is the height of the tower?
(a) 20 m
(b) 50 m
(c) 100 m
(d) 200 m
Question 2
What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12m?
(a) 75 m
(b) 50 m
(c) 45 m
(d) 60 m
Question 3
What is the height of Ajay’s house?
(a) 30 m
(b) 40 m
(c) 50 m
(d) 20m
Question 4
When the tower casts a shadow of 40m, same time what will be the length of the shadow of Ajay’s house?
(a) 16 m
(b) 32 m
(c) 20 m
(d) 8 m
Question 5
When the tower casts a shadow of 40m, same time what will be the length of the shadow of Vijay’s house?
(a) 15 m
(b) 32 m
(c) 16 m
(d) 8 m

Rohan wants to measure the distance of a pond during the visit to his native. He marks points A and B on the opposite edges of a pond as shown in the figure below. To find the distance between the points, he makes a right-angled triangle using rope connecting B with another point C are a distance of 12m, connecting C to point D at a distance of 40m from point C and the connecting D to the point A which is are a distance of 30m from D such the ADC = 900 .
Question 1
Which property of geometry will be used to find the distance AC?
(a) Similarity of triangles
(b) Thales Theorem
(c) Pythagoras Theorem
(d) Area of similar triangles
Question 2
What is the distance AC?
(a) 50 m
(b) 12 m
(c) 100 m
(d) 70 m
Question 3
Which is the following does not form a Pythagoras triplet?
(a) (7, 24, 25)
(b) (15, 8, 17)
(c) (5, 12, 13)
(d) (21, 20, 28)
Question 4
Find the length AB?
(a) 12 m
(b) 38 m
(c) 50 m
(d) 100 m
Question 5
Find the length of the rope used.
(a) 120 m
(b) 70 m
(c) 82 m
(d) 22 m

A scale drawing of an object is the same shape at the object but a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents. The scale is written as a ratio. The ratio of two corresponding sides in similar figures is called the scale factor
Scale factor = length in image / corresponding length in object
If one shape can become another using revising, then the shapes are similar. Hence, two shapes are similar when one can become the other after a resize, flip, slide or turn. In the photograph below showing the side view of a train engine. Scale factor is 1:200
This means that a length of 1 cm on the photograph above corresponds to a length of 200cm or 2 m, of the actual engine.
The scale can also be written as the ratio of two lengths.
Question 1
If the length of the model is 11cm, then the overall length of the engine in the photograph above, including the couplings(mechanism used to connect) is:
(a) 22 cm
(b) 220 cm
(c) 220 m
(d) 22 m
Question 2
What will affect the similarity of any two polygons?
(a) They are flipped horizontally
(b) They are dilated by a scale factor
(c) They are translated down
(d) They are not the mirror image of one another.
Question 3
What is the actual width of the door if the width of the door in photograph is 0.35cm?
(a) 0.7 m
(b) 0.7 cm
(c) 0.07 cm
(d) 0.07 m
Question 4
If two similar triangles have a scale factor 5 : 3 which statement regarding the two triangles is true?
(a) The ratio of their perimeters is 15 : 1
(b) Their altitudes have a ratio 25 : 15
(c) Their medians have a ratio 10 : 4
(d) Their angle bisectors have a ratio 11 : 5
Question 5
The length of AB in the given figure:
(a) 8 cm
(b) 6 cm
(c) 4 cm
(d) 10 cm

Question Seema placed a light bulb at point O on the ceiling and directly below it placed a table. Now, she put a cardboard of shape ABCD between table and lighted bulb. Then a shadow of ABCD is casted on the table as A'B'C'D' (see figure). Quadrilateral A'B'C'D' in an enlargement of ABCD with scale factor 1: 2, Also, AB = 1.5 cm, BC = 25 cm, CD = 2.4 cm and AD = 2.1 cm; ∠A = 105°, ∠B = 100°, ∠C = 70° and ∠D = 85°.
Since the two quadrilaterals are similar
Angles will be the same
Sides have the ratio as that of scale factor, i.e. 1 : 2

Question An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour.
We know that
Speed = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 )/𝑇𝑖𝑚𝑒
Distance = Speed × Time
Thus, OA = 1500 km & OB = 1800 km

If in Fig 6.1, O is the point of intersection of two chords AB
and CD such that OB = OD, then triangles OAC and ODB are
equilateral but not similar
(B) isosceles but not similar
(C) equilateral and similar
(D) isosceles and similar
Here,
∠ AOC = ∠ DOB

D and E are respectively the points on the sides AB and AC of
a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE ∥ BC. Then, length of DE (in cm) is
(A) 2.5 (B) 3 (C) 5 (D) 6
Since DE ∥ BC
∠ ADB = ∠ ABD
∠ AED = ∠ ACB

In Fig. 6.2, ∠BAC = 90° and AD ⊥ BC. Then,
BD . CD = BC2 (B) AB . AC = BC2
(C) BD . CD = AD2 (D) AB . AC = AD2
From Theorem 6.7,
If a perpendicular is drawn from the vertex of the right angle to
the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other
So, ∆𝐵𝐴𝐷 ~ ∆ 𝐵𝐶𝐴
& ∆ 𝐶𝐴𝐷 ~ ∆ 𝐶𝐵𝐴
& ∆𝑩𝑨𝑫 ~ 𝜟 𝑨𝑪𝑫
Since ∆𝐵𝐴𝐷 ~ Δ 𝐴𝐶𝐷
And sides are proportional in similar triangles
𝐵𝐷/𝐴𝐷=𝐴𝐷/𝐶𝐷
BD × CD = AD2

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is
9 cm (B) 10 cm
(C) 8 cm (D) 20 cm

If ∆ ABC ~ ∆ EDF and ∆ ABC is not similar to ∆ DEF, then which of the following is not true?
(A) BC . EF = AC. FD (B) AB . EF = AC . DE
(C) BC . DE = AB . EF (D) BC . DE = AB . FD
Since

If in two triangles ABC and PQR, AB/QR = BC/PR = CA/PQ, then
∆ PQR ~ ∆ CAB (B) ∆ PQR ~ ∆ ABC
(C) ∆ CBA ~ ∆ PQR (D) ∆ BCA ~ ∆ PQR
Here,
A ⟷ Q
B ⟷ R
C ⟷ P

In Fig.6.3, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠ APB = 50° and ∠ CDP = 30°. Then, ∠ PBA is equal to

If in two triangles DEF and PQR, ∠ D = ∠ Q and ∠ R = ∠ E, then which of the following is not true?
(A) EF/PR = DF/PQ (B) DE/P𝑄 = EF/RP
(C) DE/QR = DF/PQ (D) EF/RP = D𝐸/QR
Our triangles look like

In triangles ABC and DEF, ∠ B = ∠ E, ∠ F = ∠ C and AB = 3 DE. Then, the two triangles are
congruent but not similar
(B) similar but not congruent
(C) neither congruent nor similar
(D) congruent as well as similar
Since two angles are equal
∴ Both triangles are similar

It is given that ∆ ABC ~ ∆ PQR, with BC/QR = 1/3. Then, (𝑎𝑟 (𝑃𝑅𝑄))/(𝑎𝑟 (𝐵𝐶𝐴)) is equal to
(A) 9 (B) 3 (C) 1/3 (D) 1/9
We know that
For similar triangles, ratio of Area of triangle is equal to the ratio of square of corresponding sides
(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝑃𝑄𝑅)=(𝐵𝐶)^2/(𝑄𝑅)2
(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝑃𝑄𝑅)=(𝐵𝐶/𝑄𝑅)^2
(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝑃𝑄𝑅)=(1/3)^2
(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝐴𝐵𝐶)/(𝐴𝑟𝑒𝑎 𝑜𝑓 ∆ 𝑃𝑄𝑅)=1/9

It is given that ∆ ABC ~ ∆ DFE, ∠ A = 30°, ∠ C = 50°, AB = 5 cm, AC = 8 cm and DF = 7.5 cm. Then, the following is true:
(A) DE = 12 cm, ∠ F = 50° (B) DE = 12 cm, ∠ F = 100°
(C) EF = 12 cm, ∠ D = 100° (D) EF = 12 cm, ∠ D = 30°
So, our triangles look like

If in triangles ABC and DEF, AB/DE = BC/FD , then they will be similar, when
(A) ∠ B = ∠ E (B) ∠ A = ∠ D
(C) ∠ B = ∠ D (D) ∠ A = ∠ F
Our triangles look like

If ∆ ABC ~ ∆ QRP, (ar (ABC))/(ar (PQR))=9/4 , AB = 18 cm and BC = 15 cm, then PR is equal to
(A) 10 cm (B) 12 cm (C) 20/3 cm (D) 8 cm
Our triangles look like

If S is a point on side PQ of a ∆ PQR such that PS = QS = RS, then
PR . QR = RS2 (B) QS2 + RS2 = QR2
(C) PR2 + QR2 = PQ2 (D) PS2 + RS2 = PR2

Ex 6.4, 1 teachoo.com
Let AABC ~ ADEF and their areas be, respectively, 64 cm? and 121
cm2. If EF = 15.4cm, find BC.
D
Given:- AABC ~ ADEF A
ar AABC = 64 cm?
ar ADEF = 121 cm?
EF=15.4cm
B ? Cc
E 15.4cm F

To find : BC ,
Solution :-
Since AABC ~ ADEF
We know that if two triangle are similar,
Ratio of areas is equal to square of ratio of its corresponding sides

ar AABC BC\2
Hence, ——— = (=)

ar A DEF EF
Putting the values

Ex 6.4, 2 teachoo.com
Diagonals of a trapezium ABCD with AB || DC intersect each
other at the point O. If AB = 2 CD, find the ratio of the areas of
triangles AOB and COD.
Given: ABCD is trapezium where AB II DC
d diagonals intersect at O y X
and diagonals intersect a 7
—
A B
. ar AAOB
To find: ar ACOD
Solution:
Since we need to find ratio of area of AAOB and ACOD.
Lets first prove A AOB and ACOD are similar

teachoo.com

Ex 6.4, 3
In figure, ABC and DBC are two triangles on the same base BC. If
AD intersects BC at O, show that ar(ABC) _ AO

ar(DBC) DO

; A c
Given: A ABC and A DBC NC
Having common base BC °
B D

To prove: 72486 _ 40
“OPTOVE: Gr ADBC DO
Proof:
Since AO and OD are not part of AABC and ADBC,
we cannot directly use the theorem
We know that
Area of triangle = ; xX Base x Altitude

teachoo.com
Ex 6.4, 4
if the areas of two similar triangles are equal, prove that they
are congruent. A R
/\ - /\
Given: B c
Let triangles be A ABC & A DEF
Both triangles are similar, i.e.,A ABC ~A DEF
and
Areas are equal, i.e., ar AABC = ar A DEF
To prove: Both triangles are congruent, i.e.A ABC = A DEF
Proof:

teachoo.com
Ex 6.4, 5
D, E and F are respectively the mid-points of sides AB, BC and CA
of AABC. Find the ratio of the areas of ADEF and AABC.

A

Given: A ABC
& D,E,F mid-points of AB,BC & CA respectively D F
To find; 7-7" 8 E c
———" ar AABC
Note: Since we need to find ratio of area of ADEF and AABC.
We first need to prove these triangles are similar
Solution:

teachoo.com
Ex 6.4, 6
Prove that the ratio of the areas of two similar triangles is equal
to the square of the ratio of their corresponding medians.

P
Given: Let A ABC ~A PQR A
Here AD is median /\\
Hence BD = CD ==BC B C

2 D
Q 5 R

Similarly, PS is median
Hence QS = RS => QR
r . arene _( 4D)"
=O Prove: ar APOR \ Ps
Proof:

teachoo.com
Ex 6.4, 7 (Introduction)
Prove that the area of an equilateral triangle described on one
side of a square is equal to half the area of the equilateral
triangle described on one of its diagonals.
Concept 1 D
Two equilateral triangle are always similar

A 12cm 12 cm
In A ABC and A DEF
6 6cm
cm

DE_ 12 _4
AB 6 8 ¢ :
EF _12 _, 6 cm E 12cm
BC 6
DF _ 12 _5
AC 6
Hence by SSS similarity
A ABC ~ A DEF

teachoo.com
Ex 6.4, 8 (Introduction)
Tick the correct answer and justify :
ABC and BDE are two equilateral triangles such that D is the mid-
point of BC. Ratio of the areas of triangles ABC and BDE is
(A)2:1 (B)1:2 (C)4:1 (D)1:4
Two equilateral triangle are always similar D
In A ABC and A DEF A 12c

2cm

DE _ 12 _ 2 6cm
ABO 6cm
EF 12 _ B c
BO. 6 2 6cm E 12cm F
DF _12 _94
AC 6
Hence by SSS similarity
A ABC ~ ADEF

teachoo.com
Ex 6.4, 9
Tick the correct answer and justify :
Sides of two similar triangles are in the ratio 4 : 9. Areas of these
triangles are in the ratio
(A)2:3 (B)4:9 (C)81:16 (D) 16:81
Given, Ratio of sides of similar triangles =5
We know that if two triangle are similar ,
ratio of areas is equal to the ratio of squares of corresponding sides
s area of triangle1 __ (side of triangle 17
°, area of triangle 2 ~ (side of triangle 2)°
-()
~\o
16
“81
Hence, option (D) is correct .

teachoo.com
Ex 6.5,1
Sides of triangles are given below. Determine which of them are
right triangles. In case of a right triangle, write the length of its
hypotenuse.
(i) 7 em, 24cm, 25 cm
Given sides 7 cm, 24cm, 25cm
Using Pythagoras theorem ,
(Hypotenuse)? = (Height )* + (Base)?
Here , Hypotenuse Is largest side that is 25 cm
L.H.S R.H.S
(Hypotenuse)? (Height )? + (Base)?
= 25? = (24) + (7)
= 625 =576+49
=625

teachoo.com

Ex 6.5,2 (Method 1)
PQR is a triangle right angled at P and M is a point on QR such that
PM LQR. Show that PM?=QM.MR

R
Given:
A PQR where 2 RPQ = 90° M
& PM LOR

Pp Q
To prove: PM? = QM .MR
Proof: In A PAR,
ZRPQ=90°
So, A PQR is a right triangle
Using Pythagoras theorem in A PQR
Hypotenuse? = (Height)? + (Base)?
RQ? = PQ? + PR2 .(1)

teachoo.com
Ex 6.5,3
in figure, ABD is a triangle right angled at A and AC LBD. Show that
(i) AB2=BC.BD
D
ABD is a triangle right angled at A . c
&ACL BD
B A
To prove: AB? = BC. BD
. AB_ BC
iQ. on ap
Proof:
From theorem 6.7,
If a perpendicular is drawn from the vertex of the right angle to
the hypotenuse then triangles on both sides of the
Perpendicular are similar to the whole triangle and to each other
So, ABAD ~ AACB

teachoo.com
Ex 6.5, 4
ABC is an isosceles triangle right angled at C. Prove that AB? = 2AC?
Given: A
AABC is right triangle
Also A ABC is isosceles
To prove: AB? = 2AC? c
Proof:
Here,
Hypotenuse = AB
Also we know that AABC is isosceles
Hence AC = BC (Isosceles triangle has two sides equal)

teachoo.com
Ex 6.5,5
ABC is an isosceles triangle with AC = BC. If AB* = 2AC?, prove that
ABC is a right triangle.
Cc
Given: A ABC is an isosceles triangle.
where, AC = BC
& AB? = 2 AC?
To prove: ABC is a right angle triangle .
B A
Proof:
Given
AB? = 2AC?
AB? = AC? + AC?
AB2 = AC2+ BC2 (As AC = BC)
So, AB will be the largest sice, i.e. Hypotenuse = AB

teachoo.com
Ex 6.5,6
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
A
Given:
Equilateral triangle ABC with each side 2a 2a 2a
Altitude AD is drawn such that AD LBC
B D Cc
To find: AD 2a
Solution:
In A ADB and A ADC
AB=AC (Both are 2a as it is equilateral triangle )
AD = AD (Common)
2 ADB = 2 ADC (Both 90° as AD 1BC)
Hence AADB = AADC (By R.H.S congruency)
Hence , BD = DC (CPCT)

teachoo.com

Ex 6.5,7
Prove that the sum of the squares of the sides of a rhombus is
equal to the sum of the squares of its diagonals.

D C
Given:- Xo]
Rhombus ABCD
with diagonals AC & BD intersecting at O aN

A B

To prove:
Sum of square of all sides = Sum of the squares of it’s diagonals
= AB’ + BC? + CD? + AD? = AC? + BD?
Proof:
Since sides of a rhombus are equal
AB = BC =CD=AD

\

teachoo.com
Ex 6.5,9
A ladder 10 m long reaches a window 8 m above the ground.
Find the distance of the foot of the ladder from base of the wall.
a

LC]
Given :- ia
Height of window = AC = 8m fy
Length of ladder = AB = 10m 10%
To Find : Distance of foot of ladder from wall i.e. BC

ff J
Solution: ae: i
Since the wall will be perpendicular to ground B c
Z ACB = 90°
=> A ACBis a right angle triangle
So, using Pythagoras theorem
{Hypotenuse}? = (Height)? + (Base)?
(AB)? = (AC)? + (BC)?

teachoo.com
Ex 6.5, 10
A guy wire attached to a vertical pole of height 18 m is 24 m long
and has a stake attached to the other end. How far from the base
of the pole should the stake be driven so that the wire will be taut?
Given: A
Let Height of vertical pole = AB = 18 m
Let length of wire = AC = 24m
24m 18m

To Find : Distance from the base of the

pole to the another end of

the wire i.e. (BC) c 8
Solution
Since the pole will be perpendicular (vertical} to ground
2 ABC = 90°
=> A ABCisa right angle triangle

teachoo.com

Ex 6.5, 11
An aeroplane leaves an airport and flies due north at a speed of
1000 km per hour. At the same time, another aeroplane leaves
the same airport and flies due west at a speed of 1200 km per
hour. How far apart will be the two planes after 1 hours?

B
Given:
Speed of north flying aeroplane = 1000 km/hr.
Speed of west-flying aeroplane = 1200 km/hr.||
To find: Distance between the two planes

after 1.5 hours , i.e. , BC
Solution:
We know that
Speed = Distance
Time

Distance = speed X time

teachoo.com
Ex 6.5, 12
Two poles of heights 6 m and 11 m stand ona plane ground. If
the distance between the feet of the poles is 12 m, find the
distance between their tops. D
Given: Height of first pole = AB=6m B | E
; 12m 11m
Height of second pole = CD =11m 6m 6m
Distance b/w feet of poles = AC = 12 m d
A 12m c
To Find :- Distance between the tops of the pole ,i.e., BD
Solution :-
Let we draw a line BE perpendicular to DC i.e. BE L DC
Since AC is also perpendicular to DC as pole is vertical to ground,
So, BE=AC=12m
Similarly , AB=EC=6m

teachoo.com
Ex 6.5, 13
D and E are points on the sides CA and CB respectively of a triangle
ABC right angled at C. Prove that AE? + BD? = AB? + DE?
Given : triangle ABC, right angled at C
Two points D and E are on the sides CA and CB
B
To prove: AE* + BD? = AB? + DE?
E

Proof:

A D c
Let us join the points D with EandB.
And point E with A

teachoo.com
Ex 6.5, 14
The perpendicular from A on side BC of a AABC intersects BC at
D such that DB = 3 CD (see figure). Prove that 2AB? = 2AC? + BC?
Given: AABC with AD L BC A
Also DB = 3 CD
To prove: 2AB? = 2AC? + BC? c O B
D
x
Proof:
Let BC =x
CD +DB=x DB=3CD
(As DB = 3 CD given) =3 x2
CD +3CD=x
= 3*
4CD =x 4
cD ==
4

teachoo.com
Ex 6.5, 15
In an equilateral triangle ABC, D is a point on side BC such that
BD = 2BC. Prove that 9AD? = 7 AB? A
Given: Equilateral triangle ABC x x
Dis a point an BC
Such that BD == BC BODE c
x
To prove: 9 AD? = 7 AB?
Construction: Lets draw AE L BC
Proof:
All sides of equilateral triangle is equal,
AB = BC = AC
Let AB = BC = AC =x

teachoo.com

Ex 6.5, 16
In an equilateral triangle, prove that three times the square of one
side is equal to four times the square of one of its altitudes.

A
Given:-

a a

Equilateral triangle ABC with each side a
& AD as one of its altitudes B D c

a
To Prove :-
3 X Square of one side = 4 X square of one of it’s altitude
=> 3a* = 4AD?
Proof:-

teachoo.com
Ex 6.5, 17
Tick the correct answer and justify : In AABC, AB = 6 V¥3cm, AC =
12 cm and BC =6cm.
The angle B is :
(A) 120° (B) 60° (c) 90° (D) 45°
A
Let us check whether it is a right angle triangle
63 12
To prove any triangle to be the right triangle.
We use Pythagoras theorem B z Cc
(Hypotenuse)? = (Height }* + (Base)?
Here, Hypotenuse is longest side i.e. AC = 12 cm
L.H.S R.H.S
(Hypotenuse)? (Base)? + (Height)?
=12 = (6 V3)? + (6)
= 144 = (6x 6 x V3 x V3) +(6X 6)
= (36 x 3) + (36) = 144

Ex 6.6, 1 feachoo.com
In Fig. 6.56, PS is the bisector of 2 QPR of A PQR. Prove that S = =

AT

ta

c
é
Given : A PQR /
f
and PS is the bisector of ZQPR p’
i.e. ZQPS = ZRPS
To Prove: os Pe
———==" SR PR
Q Ss R
Construction : Draw RT || SP such that RT cuts QP Produced at T.
Proof:
In A QRT,
RT || SP (By construction)

Ex 6.6, 2 teachoo.com
In Fig. 6.57, Dis a point on hypotenuse AC of A ABC, DM 1 BC and
DN 1 AB. Prove that :
(i) DM2=DN.Mc
(ii) DN?=DM.NA
A
Given: ABC is a triangle
D

and D is a point on hypotenuse AC such that N

BD 1 AC

Cc B

DM L BC M
And DN LAB
To Prove:

(i) DM? = DN. MC

(ii) DN? = DM. NA

teachoo.com
Ex 6.6, 3
In Fig. 6.58, ABC is a triangle in which 2 ABC > 90° and AD | CB
produced. Prove that AC? = AB? + BC? + 2 BC. BD.
A

Given : ABC is a triangle in which
ZABC > 90°
and AD | CB extended D.

D B Cc
To Prove: AC? = AB? + BC? + 2BC.BD
Proof:

Ex 6.6, 4 teachoo.com
In Fig. 6.59, ABC is a triangle in which 2 ABC < 90° and AD L BC.
Prove that AC? = AB? + BC? — 2 BC. BD.
Given: ABC is a triangle where A
ZABC < 90° and AD 1 BC
To Prove: AC? = AB? + BC? - 2BC.BD
B Cc
D
Proof:
In right A APB In right A ADC
A A

Z NN
By Pythagoras Theorem, By Pythagoras Theorem,
AB?= AD? +BD? (1) ACS= AD? + CD? (2)

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Ex 6.6, 5
In Fig. 6.60, AD is a median of a triangle ABC and AM L BC. Prove
that :
2
(i) AC? =AB2+BC2—2BC.DM+ (=) .
A
Given: ABC is a triangle
AD is a Median of AABC
1
~BD=CD=>BC _ ...(1)
B MD Cc
Also, AM L BC
2
To Prove: (i) AC? = AB? + BC?-2 BC. DM + (FF)
Proof :

Ex 6.6, 6 teachoo.com
Prove that the sum of the squares of the diagonals of parallelogram
is equal to the sum of the squares of its sides. , B
Given: ABCD is a parallelogram Lx</
To Prove: D Cc
Sum of squares of diagonals = Sum of squares of its sides
AC? + BD? = AB? + BC + CD? + DA?

Construction: Draw AX L CD and BY 1 DC extended to Y.

A B
Proof: Dx

BD x Cc Y

Ex 6.6, 7 feachoo.com
In Fig. 6.61, two chords AB and CD intersect each other at the point
P. Prove that :
(i) AAPC ~ ADPB D
Given: A Circle with two chords AB and CD
Cc
intersecting at point P
To Prove: (i) AAPC ~ ADPB D
A
A
Proof: B
In AAPC and ADPB c
ZAPC = ZDPB {Vertically opposite angles}
(Angles in the same segment
ZCAP = ZBDP of a circle are equal }
» AAPC~ ADPB (AA Similarity)
Hence Proved.

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Ex 6.6, 8
In Fig. 6.62, two chords AB and CD of a circle intersect each other at
the point P (when produced) outside the circle. Prove that
(i) APAC~APDB B
Given: A Circle with two chords AB and CD

A
which meet at a Point P D
outside the circle. P ¢c
To Prove: (i) A PAC ~ A PDB
Proof: We know that,
ABDC is a cyclic quadrilateral,
-. Sum of its opposite angle is 180°

Ex 6.6, 9 feachoo.com
. . . . BD _ AB
In Fig. 6.63, Dis a point on side BC of A ABC such that wac’
Prove that AD is the bisector of 2 BAC. A
Given: A AABC
Where = =
cD AC
B D Cc
E
To Prove: AD is the bisector of ZBAC /
f
ie. ZBAD = ZDAC /
x
c
é
¢
/
Construction : A
Produce BA to E such that AE = AC.
Now join CE.
B D

Ex 6.6, 10 feachoo.com
Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m
above the surface of the water and the fly at the end of the string
rests on the water 3.6 m away and 2.4 m from a point directly
under the tip of the rod. Assuming that her string (from the tip of
her rod to the fly} is taut, how much string does she have out (see
Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second,
what will be the horizontal distance of the fly from her after 12
seconds? ;

'

‘ HS

‘ Wwe
Let A be the position . potas

a ot
of the tip of the fishing = = ina Pm [

wa ma J
— SL
rod above the surface ”
2.4m 1.2m
of the water. A
Thus, AB=1.8m
Nazima
c 2.4m 8 2m?
oo