Triangles
1) Let ABC DEF and their areas be, respectively, 64cm2 and 121 cm2. If EF=15.4 cm, find
BC.
Solution:
Given: ABC
DEF and A( ABC)=64cm2 & A( DEF)= 121cm2 and EF = 15.4cm.
To find: BC = ?
We know that, the ratio of the areas of two similar triangles is equal to the square of the ratio of
their sides.
i.e.
BC2 =
=
= 125.44
BC = 11.2 cm
2) In an equilateral triangle ABC, AD BC. Prove that 3AB2 = 4AD2.
Solution:
Given: ABC is an equilateral triangle and AD BC.
To prove that: 3AB2 = 4AD2
Proof:
Since, AD BC therefore, BD = DC.
Now, in right triangle ABD, we have
AB2 = AD2 + BD2 --- [Pythagoras theorem]
A
Since BD = DC therefore, BD = BC
AB2 = AD2 +
2
B
2
AB = AD +
D
AB2 = AD2 +
AB2 = AD2 +
AB2 -
------ [Since, AB = BC = AC – sides of an equilateral triangle]
= AD2
= AD2
3AB2 = 4AD2
C
Triangles
3) In the given figure, PQR is a right triangle right angled at Q & QS
= 4cm, find the QS, RS & QR.
Solution:
P
Given: In PQR, PQR= 90 .
PQ = 6cm, PS = 4cm.
4
To find: QS , RS & QR?
Since,QS PR,
PS = SR
And PS =4cm ----- (given )
SR = 4cm.
PR = PS + SR = 4 + 4 =8
PQR is right angled triangle.
by Pythagoras theorem,
PR2 = PQ2 + QR2
QR2 = PR2 - PQ2
QR2 = 82 - 62
QR2 = 64 -36 = 28
QR = 2√
PR. If PQ = 6cm and PS
S
6
Q
QS2 = PS SR -------- [ Since QS is perpendicular to PR]
QS2 = 4 4 = 16
QS = 4cm.
R
Triangles
4) In the given figure,
If BX 
and
1
BC , find the ratio of areas of ∆ABC
4
and ∆YCX.
Solution:
Since
Similarly

ACX  CXY
∆ABC ∆YCX …….(AA – similarity criterion)
ar (ABC ) BC 2  BC 



ar (YCX ) XC 2  XC 
1
Since, BX  BC ….. (Given)
4
1
 BC  XC   BX  BC
4
4BC – 4XC = BC


3BC = 4XC
BC 4


XC 3
2
A
C
B
X
ar (ABC )  4 
16
  
 16 : 9
ar (YCX )  3 
9
2
Putting values in eq.( i ), we get
5) PQR is an isosceles triangle right angled at B. Two equilateral triangles are constructed on
1
side QR and PR as shown in the given figure. Prove that area of ∆QRS = area of ∆PRT.
2
Solution:
T
R
In ∆PQR, Q  90 and consider PQ = QR = x

PR2 = PQ2 + QR2 --- (Pythagoras theorem)

PR2 = 2 x 2
 PR = x 2
S
P
Q
Triangles
Area of equilateral triangle ∆QRS =
3 2
x
4
Area of equilateral triangle ∆PRT =
3
3
2
 PR  
4
4

2x
Therefore area of ∆PRT =

2
3 2
x
2
2
3 2 2 3x 2
= 
x 
2 2
4
= 2 area of ∆QRS
6) The areas of two similar triangles are144 cm2 and 81 cm2 respectively. If the median of the
first triangle is 14.4 cm find the corresponding median of the other.
Solution:
Let two triangles be LMN and XYZ, such that ∆LMN ∆XYZ
R
ar (LMN ) [ Median (LMN )]

ar (XYZ )
[ Median (XYZ )]2
2

144 
14.4


 
81  Median (XYZ ) 

L
2
14.4 14.4  81
144
Median = 10.8 cm.
(Median of ∆XYZ)2 =
M
K
N
S
P
T
Triangles
7) The areas of two similar triangles are 144 cm2 and 81 cm2 respectively. If the altitude of
bigger triangle is 4 cm, find the corresponding altitude of other triangle.
Solution:
Let two triangles be LMN and RST,
ar (LMN ) LK 2

ar (RST ) RP 2



81 LK 2
 2
144
4
8116
LK 2 
144
LK = 3 cm.
8) If the diagonals of a quadrilateral PQRS divide each other proportionally, then prove that it is
a trapezium.
Solution:
MP MQ
In quad. PQRS, diagonals PR and QS intersect at M such that

MR MS
Draw MT SR
Q
P
Proof: In ∆PSR, MT SR
PT PM
(i)
[ B.P. Theorem ]

T
M
TS
MR
PM QM
( ii )

MR MS
Comparing eq. ( i ) and ( ii ), we get
PT QM
 MT PQ

TS
MS
Also MT SR PQ SR
Hence PQRS is a trapezium.
But
S
R
Triangles
MP LN

.
OP MN
9) In the given figure, OM  MN , OP  LM and LN  MN , prove that
Solution:
Since OM  MN and LN  MN ; hence OM LN (two line-segments are perpendicular
to same line)
OMP  PLN

L
(Alternate angles)
O
Now in ∆OMP and ∆LMN
OPM  LNM
(Each 90 )
OMP  PLN  OMP  MLN
Hence,
∆OMP ∆LMN
MP OM OP
(AA - criterion)


LN LM MN
MP LN


OP MN
10) In figure
P
M
and AB = 5cm. Find the value of DC.
Solution:
In AOB & COD, we have
AOB = COD ------- [vertically opposite angles]
AO BO
-------- [ Given]

OC OD
So, by SAS criterion similarity, we have
AOB
COD
AO BO AB


OC OD DC
1
5
----- [since AB = 5]

2 DC
DC = 10 cm
N
A
B
O
C
D
Triangles
11) D and E are points on the sides AB and AC respectively of a ∆ABC. State whether DE BC
if AD = 5 cm, BD = 5.5 cm, AE = 5 cm, CE = 5.5 cm.
A
Solution:
In the given figure,
AD = 5 cm
BD = 5.5 cm
AE = 5 cm
CE = 5.5 cm
AD
5
5  2 10
D
E
Here,



BD 5.5 5.5  2 11
AE
5
5  2 10
And



CE 5.5 5.5  2 11
B
C
AD AE


BD CE
DE
BC
(Converse of basic proportionality theorem)

12) In the given figure, ST YZ find TZ = ?
Solution:
In the given figure, we are given: ST YZ, XS = 2.5 cm, SY = 5 cm, XT = 1.4 cm, TZ =?
Since, ST YZ
X
XS XT


SY TZ
2.5 1.4


5 TZ
1.4  5
TZ =

 2.8 cm.
T
S
2.5
 TZ = 2.8 cm.
13) In the given figure, PX and RY are two attitudes of ∆PQR.
Prove that: (i) ∆PYZ ∆RXZ
Y
(ii) ∆PQX ∆RQY
(iii) ∆PYZ ∆PXQ.
Solution:
In the given figure,
PX  QR and RY  PQ in ∆PQR. Altitudes PX and RY intersect each other at Z.
Z
Triangles
Proof:
(i)
In ∆PYZ and ∆RXZ; PYZ  RXZ  90
(Vertical opposite angles)
PZY  RZX
∆PYZ ∆RXZ
(AA criterion)

Now in ∆PQX and ∆RQY
(Common in both)
PQX  RQY
(ii)
and
(iii)
R
X
PXQ  RYQ  90
∆PQX ∆RQY

Now in ∆PYZ and ∆PXQ,
Z
(AA criterion)
PYZ  PXQ  90
ZPY  XPQ

∆PYZ ∆PXQ
P
(Common in both)
Y
Q
(AA criterion)
14) In the given figure, MR  LN and NS  LM.
Prove that: (i) ∆LSN ∆LRM
NL
NS
(ii)

LM RM
Solution:
In the given figure,
MR  LN and NS  LM
Proof: In ∆LNS and ∆LMR,
LSN  LRM  90
and
NLS = MLR
(Common in both)
∆LNS ∆LMR
(AA criterion)

NL NS LS



LM MR LR
NL
NS


LM RM
N
R
L
S
M
Triangles
S
15) In the given figure, ∆RTS ∆RXY.
If
ST = 16 cm,
XY = 8 cm, SR = 13 cm
RX = 5.6 cm, find TR and RY.
Solution:
T
In the given figure, ∆RTS ∆RXY
ST = 16 cm, XY = 8 cm, SR = 13 cm, and RX = 5.6 cm.
Since
∆RTS ∆RXY
ST
SR RT



XY RY RX
16 13 RT



8 RY 5.6
RY = 6.5 cm and RT = 11.2 cm.

R
X
Y
16) Two poles of height ‘x’ meters and ‘y’ meters are z meters apart. Prove that the height of
intersection of the lines joining the top of each pole to the foot of the opposite pole is given by
xy
metres.
x y
Solution:
Let PQ and RS be two poles of height ‘x’ meters and ‘y’
meters respectively, such that the poles are z meters apart,
i.e. PQ = x, SR = y and PR = z Suppose the lines PS and Q
QR meet at ‘E’ such that DE = h meters.
Let PD =a , DR = b. Then, a + b = z
In ∆PSR and ∆PDE, we have;
SRP = EDP --[each 90 ]
P = P ----- [ common angles]
∆PSR ∆PED ----[ by A-A criterion of similarity]
x
E
P
D
z
S
y
R
Triangles
PS
SR PR


PE ED PD
SR PR
i.e.

ED PD
y z
i.e. 
h a
hz
………… (1)
a
y
In ∆RDE and ∆RPQ we have;
RDE = RPQ --[each 90 ]
R = R ----- [ common angles]
∆RDE ∆RPQ----[ by A-A criterion of similarity]
RD DE

RP PQ
b h

z x
zh
……………(2)
x
From (1) & (2) we have,
hz zh
a+b=
+
x
y
b
z = hz (
1 1
 )
x y
1=h(
1 1
 )
x y
1=h(
x y
)
xy
h= (
xy
) metres.
x y
Triangles
17) The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB =
3CD. Prove that 2 AB2 = 2AC2 + BC2.
Solution:
A
Given: DB = 3CD
2
2
2
We have to prove that: 2AB = 2AC + BC
Now, BC = DB + CD = 3CD + CD= 4CD
BC = 4CD
1
CD = BC
4
1
3
DB = 3CD = 3( BC) = BC
4
4
Now, in right-angled triangle ABD, we have
AB 2  AD 2  DB 2 -------------------- (1)
Again, in right-angled triangle ADC, we have
B
D
AC 2  AD 2  CD 2 -------------------- (2)
On subtracting (2) from (1), we have
AB 2  AC 2  DB 2  CD 2
2
2
8
3
 1

9 1
AB 2  AC 2 =  BC    BC      BC 2  BC 2
16
4
 4

 16 16 
1
AB 2  AC 2 = BC 2
2
 2 AB 2  2 AC 2  BC 2
2 AB 2  2 AC 2  BC 2
C
Triangles
18) Find the area of an equilateral triangle formed on the hypotenuse of a right triangle whose
sides are 3cm and 4 cm respectively.
Solution:
Given: PQR is a right angled triangle such that Q  90 , PQ = 4 cm, QR = 3 cm.
PR2 = PQ2 + QR2 ---[Pythagoras theorem]
= 42 + 32
S
P
= 25
PR = 5 cm.

Now ∆PRS is an equilateral triangle
 PR = PS = SR = 5 CM …. [Sides of the equilateral∆PRS]

Area of ∆PRS =
3
( side) 2
4

Q
R
3 2
=
5
4
Area of ∆PRS = 6.25 3 cm2.
19) In the given figure, LM ZY and
area(LNM )
XL 3
.
 . Find
LY 2
area(ZNY )
Solution:
Given : In ∆XYZ, LM ZY and
X
XL 3

LY 2
Since LM ZY
XLM  XYZ and XML  XZY

∆XLM ∆XYZ
[AA- criterion of similarity]

XL LM


XY YZ
XL 3
LY 2

Now


LY 2
XL 3
LY
2
LY  XL 5


1  1

XL
3
XL
3
Y
L
M
N
Z
Triangles
XY 5
XL 3



XL 3
XY 5
LM 3


YZ 5
In ∆LNM and ∆ZNY, we have LMN  NYZ and NLM  NZY .
∆LNM ∆ZNY

ar (LNM ) LM 2  3 
9
.

  
2
ar (ZNY ) YZ
25
5
2

20.ABC is an equilateral triangle of side 2a. find each of its altitude.
Solution:
Given: ABC is an equilateral triangle of side 2a units.
To find: each altitude
We draw AD
BC. Then D is the mid-point of BC.
BC 2a
BD 

a
2a
2
2
Now, ABD is right triangle right-angled at D.
 AB 2  AD 2  BD 2 ------ [by Pythagoras theorem]
(2a)2 = AD2 + a2
B
AD2=4 a2 - a2 =3a2
a
AD=√ a
And we know that in an equilateral triangle all the three altitudes are equal.
Hence, each of altitude = √ a units
A
2a
D
2a
a
C