BATRA COACHING CENTRE
1326/14, Arjun Nagar, Rohtak, : 9416124333
TRIANGLES
AREAS OF SIMILAR TRIANGLES
We have already studied that in two similar
triangles, the ratio of their corresponding
sides is the same. Let us now study that the
ratio of the areas of two similar triangles is
equal to the square of the ratio of their
corresponding sides.
Theorem The ratio of the areas of two
similar triangles is equal to the ratio of the
squares of their corresponding sides.
Given: ∆ ABC is similar to ∆DEF
AB AC BC
i.e.,
=
=
DE DF EF
BC2 AC2 AB2
= 2=
=
EF DF2 DE2
Corollary 1 The ratio of the areas of two
similar triangles is equal to the ratio of the
squares of their corresponding altitudes.
Given: Let ∆ ABC and ∆ DEF be the given
triangles such that ∆ ABC ∼ ∆ DEF
and AL ⊥ BC, DM ⊥ EF (Fig.)
ar(∆ABC) AL2
=
ar(∆DEF) DM 2
Proof: BY Theorem, we have
ar(∆ABC) AB2
=
ar(∆DEF) DE2
Now, consider ∆s ABL and DEM
Here,
∠ALB = ∠DME = 90°
[Q ∆ ABC ∼ ∆ DEF]
and ∠B = ∠E
By AA similarity criterion,
∆ ABL ∼ ∆ DEM
To Prove:
ar(∆ABC) AB2 BC2 AC2
=
=
=
ar(∆DEF) DE2 EF 2 DF 2
Construction: Draw AX ⊥ BC and DY ⊥ EF.
Proof: In ∆s ABX and DEY,
∠ABX = ∠DEY
[Q ∆s ABC and DEF are similar.]
∠AXB = ∠DYE
Hence, by AA criterion of similar triangles, ∆
ABX ∼ ∆ DEY.
AB AX BX
⇒
=
=
DE DY EY
1
.BC. AX
ar(∆ABC) 2
BC AX
Now,
.
=
=
ar(∆DEF) 1 .EF.DY EF DY
2
BC AB
=
.
EF DE
BC BC
=
.
EF EF
To Prove:
AB2 AL2
=
⇒
[By Theorem]…(2)
DE2 DM2
From Eq. (1) and Eq. (2), we have
ar(∆ABC) AL2
=
ar(∆DEF) DM 2
Corollary 2 The ratio of the areas of two
similar triangles is equal to the ratio of the
squares of their corresponding angles
bisectors.
Given: Let ∆ ABC and ∆ DEF be the given
triangles such that ∆ ABC ∼ ∆ DEF and AP,
DQ are bisectors of ∠A and ∠D respectively
ar(∆ABC) AX 2
=
ar(∆DEF) DY 2
Proof: we know that
ar(∆ABC) AB2
…..(1)
=
ar(∆DEF) DE2
Since ∆ ABC is similar to ∆ DEF.
AB BC CA
=
=
DE EF FD
AB 2BX BX
⇒
=
=
DE 2EY EX
AB BX BX
⇒
=
=
DE EY EX
By SAS similarity criterion, we have
∆ ABX ∼ ∆ DEY
AB AX
⇒
=
DE DY
To Prove:
ar(∆ABC) AP2
To Prove:
=
ar(∆DEF) DQ2
Proof: We know that
ar(∆ABC) AB2
=
ar(∆DEF) DE2
Now, ∆ ABC ∼ ∆ DEF [given]
⇒
∠A = ∠D
∠A ∠D
⇒
=
2
2
⇒
∠BAP = ∠DEQ
Now, in ∆s ABP and DEQ, we have
∠BAP = ∠EDQ
and ∠B = ∠E
So, by AA similarity criterion, we have
∆ ABP ∼ ∆ DEQ
AB2 AP2
⇒
…..(2)
=
DE2 DQ2
From Eq. (1) and Eq. (2), we have
ar(∆ABC) AP2
=
ar(∆DEF) DQ2
Corollary 3. The ratio of the areas of two
similar triangles is equal to the ratio of the
squares of their corresponding medians.
Given: Let ∆ ABC and ∆ DEF be the given
triangles such that ∆ ABC ∼ ∆ DEF and AX,
DY be the medians of ∆ ABC and ∆ DEF
respectively
AB2 AX 2
=
DE2 DY2
From Eq. (1) and Eq. (2), we have
⇒
….(2)
ar(∆ABC) AX 2
=
ar(∆DEF) DY 2
Example
Q.1
The areas of two similar triangles
ABC and PQR are 25 sq cm and 49 sq
cm. If QR = 9.8 cm, find the length of
BC.
Sol.
ar(∆ABC) BC2  BC 

=
=
ar(∆PQR) QR2  QR 
2
⇒
⇒
Q.2
2
25  BC 
=

49  9.8 
BC 5
5 × 9.8
=
⇒ BC =
= 7cm
9.8 7
7
In Fig., DE is parallel to BC and
AD: DB = 2 : 3. Determine
Sol.
ar(∆ ADE) : ar (∆ ABC).
Here, AD : DB = 2 : 3
DB 3
DB
3
= ⇒1+
=1+
AD 2
AD
2
AD + DB 5
AD 2
⇒
= ⇒
=
AD
2
AB 5
Since DE  BC, we have ∆ ADE ∼ ∆ ABC
⇒
ar(∆ADE) AD2  AD   2 
4
=
=
∴
 =  = .
2
ar(∆ABC) AB  AB   5  25
2
Q.3
In a ∆ ABC, DE  BC. If DE =
BC
3
and area of ∆ ABC = 81 cm2, find the
area of ∆ ADE.
Sol.
Since DE  BC, ∆ ADE ∼ ∆ ABC
ar(∆ABC) BC2
BC2
⇒
=
=
2
ar(∆ADE) DE2  2

 BC 
3

81
9
=
⇒
ar(∆ADE) 4
⇒
ar(∆ ADE) = 36 cm2.
Q.4
In Fig. DE  BC and AD : DB = 4 : 5.
ar(∆DEF)
Find
ar(∆BCF)
AD 4
Sol.
Since
=
DB 5
Let AD = 4x
∴
DB = 5x
∴
AB = AD + DB
= 4x + 5x
= 9x
∴
AD : AB = 4x : 9x = 4 : 9
AD 4
=
AB 9
2
2
Since DE  BC, we have ∆ ADE ∼ ∆ ABC
DE AD 4
=
=
BC AB 9
By AA similarity criterion, ∆ DEF ∼ ∆ BCF
ar(∆DEF)  DE   4  16
=
 =  = .
ar(∆BCF)  BC   9  81
2
∴
2
Q.5
In trapezium ABCD, O is the point of
intersection of AC and BD, AB  CD
and AB = 2CD. If the area of ∆AOB = 84
sq cm, find the area of ∆ COD.
Sol.
As AB  CD, ∠DCO = ∠BAO
∠CDO = ∠ABO
By AA similarity criterion, ∆ ABO ∼ ∆ COD
ar(∆COD) CD2
=
⇒
ar(∆AOB) AB2
 CD 
=

 AB 
 CD 
=

 2CD 
1
=
4
⇒
Q.6
2
2
1
× 84 sq cm
4
= 21 sq cm.
In Fig. ABC is a triangle. PQ is a line
segment intersecting AB in P and AC
in Q such that PQ 
 BC and divides
∆ABC into two parts of equal area.
BP
Find
.
BA
Area of ∆ COD =
Sol.
We are given that PQ  BC
∴
⇒
∠APQ = ∠B, ∠AQP = ∠C
∆ BAC ∼ ∆PAQ
ar(∆BAC) 2
=
[Given]
ar(∆PAQ) 1
⇒
2
∴
Let
∴
∴
∴
BA
 BA 2
= 2
  = ⇒
PA
 PA 1
BA = 2x
PA = x
BP = BA – PA
= 2 −1 x
(
BP
=
BA
)
( 2 −1)x
2x
2 −1
=
2
Prove that the area of the
Q.7
equilateral triangle described on a
side of a square is one--half of the
area of a similar triangle described
on any diagonal of the square.
Sol.
Let ABCD be the given square. Also
∆ BCE ∼ ∆ ACF
(
)
D
Now
⇒
∴
∴
AC2 = AB2 + BC2 = BC2 + BC2
AC = 2BC
1
ar(∆BCE) BC 2
BC 2
=
=
= .
2
2
2
ar(∆ACF) AC
2BC
(
ar ∆ BCE =
1
ar∆ACE
2
)
EXERCISE
1.
The areas of two similar triangles
ABC and PQR are in the ratio 9 : 16.
If BC = 4.5 cm, find the length of QR.
2.
ABC is an isosceles triangle, rightangled at B. Two equilateral
triangles are constructed with sides
BC and AC as shown in Fig. Prove
1
that ar (∆ BCD) = ar(∆ACE)
2
[Hint: Since ∆ ACE and ∆ BCD are
equilateral, they are equiangular and hence
similar.
ar(∆ACE) AC2 2BC2
∴
=
=
= 2.
ar(∆BCD) BC2 BC2
∆ ABC and ∆ DBC are two triangles
3.
on the same base as shown in the
Fig. If AD intersects BC at O, prove
ar(∆ABC) AO
=
that
ar(∆DBC) DO
4.
(i)
(ii)
ABCD is a trapezium in which
AB  DC and AB = 2 CD.
ar(∆AOB)
, where O is
Determine
ar(∆COD)
the point of intersection of the
diagonals of the trapezium.
If area of ∆ AOB = 84 sq cm, find the
area of ∆ COD.
In ∆ ABC ∼ ∆ PQR such that the
internal bisectors of the angles B
and Q are of respective length 5 cm
and 7 cm. If area of ∆ PQR is 98 sq
cm, then find the area of ∆ ABC.
Theorem If a perpendicular is drawn from
the vertex of the right angle of a right
triangle to the hypotenuse, then the
triangles on each side of the perpendicular
are similar to the whole triangles and to
each other.
Given: A right-angled ∆ ABC, right-angled at
B. BD ⊥ AC .
5.
To Prove: (i) ∆ ADB is similar to ∆ ABC.
∆ BDC is similar to ∆ ABC.
(ii)
(ii)
∆ ADB is similar to ∆ BDC.
Proof: (i)
In ∆s ADB and ABC,
∠A = ∠A
∠ADB = ∠ABC
∴ By AA criterion of similar triangles, ∆ ADB
is similar to ∆ ABC.
(ii)
In ∆s BDC and ABC,
∠C = ∠C,
∠BDC = ∠ABC
∴ By AA criterion of similar triangles, ∆ BDC
is similar to ∆ ABC.
(iii)
We have ∠ABD + ∠DBC = 90°
Also, ∠C + ∠DBC = 90°
∠ABD + ∠DBC = ∠C + ∠DBC
∠C = ∠ABD
…..(1)
So,
Now, in ∆s ADB and BDC
∠ABD = ∠C
(Just proved)
∠ADB = ∠BDC
(Each = 90°)
∴By AA criterion of similar triangles, ∆ADB
is similar to ∆ BDC.
Corollary: The square of the perpendicular
on the hypotenuse of a right triangle is
equal to the product of the lengths of the
two parts of the hypotenuse,
i.e., BD2 = AD x DC.
Proof: Since ∆ ADB ∼ ∆ BDC
AD DB
=
BD DC
⇒
AD x DC = BD x DB
⇒
BD2 = AD x DC
From Theorem (i), ∆ ADB ∼ ∆ ABC
AD AB
∴
=
AB AC
⇒
AB2 = AD x AC
Similarly ∆ BDC ∼ ∆ABC.
DC BC
∴
=
BC AC
⇒
BC2 = AC X DC.
Note: If ∆1 ∼ ∆2 then ∆2 ∼ ∆3. then ∆1 ∼ ∆3.
Theorem (Pythagoras Theorem)
In a right triangle, the square of the
hypotenuse is equal to the sum of the
squares of the other two sides.
Given: A right triangle ABC, right-angled at
B
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC.
Proof: Now In ∆ ADB and ∆ ABC
∠A = ∠A
[Common]
∠ADB = ∠ABC
[Each = 90°]
∠ABD = ∠ACB
[3rd angle]
∴
∆ADB ~ ∆ABC
AD AB
∴
=
AB AC
or
AB2=AD x AC
….(1)
Similarly ∆ BDC ∼ ∆ ABC
DC BC
∴
=
BC AC
or
BC2 = DC x AC
Adding (1) and (2) we get
AB2 + BC2=AD x AC + DC x AC= (AD+DC) x AC
= AC x AC = AC2.
Theorem
(Converse
of
Pythagoras
Theorem)
In a triangle, if the square of one side is
equal to the sum of the squares of the other
two sides, then the angle opposite to the
first side is a right angle.
Given: A triangle ABC such that
AB2 + BC2 = AC2
To Prove: ∆ ABC is a right triangle, rightangled at B.
Construction: Construct a right triangle
PQR, right-angled at Q such that PQ = AB
and QR = BC.
Proof: In ∆ PQR, since ∠Q = 90°
PQ2 + QR2 = PR2
[Pythagoras Theorem]
⇒ AB2 + BC2 = PR2 [By Construction] …(1)
[Given] ….(2)
But, AB2 + BC2 = AC2
From Eq. (1) and Eq. (2), we get PR2 = AC2
⇒
PR = AC
Therefore, by SSS congruence criterion, we
get ∆ ABC ≅ ∆ PQR
Which gives ∠B = ∠Q
But,
∠Q = 90° [By Construction]
∴
∠B = 90°
Hence, ∆ ABC is a right triangle, right-angled
at B.
Q.1
In Fig. PQR is an obtuse triangle,
obtuse-angled at Q. If PM ⊥ RQ,
prove that
PR2 = PQ2 + QR2 + 2QR x QM
Sol.
In right triangle PMQ, right-angled at
M,
We have
PQ2 = PM2 + MQ2
….(1)
Again, in right triangle PMR, right-angled at
M,
……(2)
We have
PR2 = PM2 + MR2
2
2
PR = PM + (MQ + QR)2
2
= PM + MQ2 + QR2 + 2MQ x QR
= PQ2 + QR2 + 2MQ x QR.
Q.2
In Fig. ∠B of ∆ ABC is an acute angle
and AD ⊥ BC. Prove that
(i)
AC2 = AB2 + BC2 – 2BC x BD
(ii)
AB2 + CD2 = AC2 + BD2.
Sol.(i) ∆ ADC is right triangle, right-angled
at D.
∴ AC2 = AD2 + DC2
= AD2 + (BC – BD)2
= AD2 + BC2 + BD2 – 2BC x BD
= (AD2 + BD2) + BC2 – 2BC x BD
= AB2 + BC2 – 2BC x BD.
(ii)
From right ∆ ABD,
AB2 = AD2 + BD2
…..(1)
From right ∆ ACD,
…..(2)
CD2 = AC2 – AD2
Adding (1) and (2)
AB2 + CD2 = AC2 + BD2.
Q.3
The perpendicular AD on the base
BC of a ∆ ABC intersects BC at D, so
1
that BD = CD. Prove that
3
2
2AC = 2AB2 + BC2.
Sol.
In right-angled ∆ ADB, we have
…..(1)
AB2 = AD2 + DB2
And in right-angled ∆ ADC, we have
AC2 = AD2 + DC2
……(2)
From Eq. (1) and Eq. (2), we have
2AC2 – 2AB2 = 2DC2 – 2DB2
2
2
3 
1 
= 2 BC − 2 BC
4 
4 
[Q DB = 3CD]
BC2
9
= BC2 −
= BC2.
8
8
Q.4(i) In ∆ PQR, ∠QPR = 90° and QR = 26
cm. If PS ⊥ SR, PS = 6 cm and SR = 8
cm, find the area of ∆ PQR.
(ii)
If in Fig. PQ = 24 cm, QR = 26 cm,
∠QPR = 90°, PS = 8 cm and SR = 6
cm, find ∠PSR.
Sol.(i) From right triangle PSR, we have
PR2 = PS2 + SR2 = 62 + 82 = 36 + 64 = 100
⇒
PR = 10 cm
Further, QPR is a right triangle.
So,
QR2 = PQ2 + PR2
⇒
262 = PQ2 + 102
⇒
PQ2 = 676 – 100 = 576
⇒
PQ = 24 cm
1
1
Now, ar(PQR) = × PQ × PR = × 24 ×10
2
2
Sq cm = 120 sq cm.
(ii)
∆ QPR is a right triangle
2
∴PR =QR2 – PQ2=262 – 242=100
⇒
PR= 10 cm
In ∆ PSR,
PS2 + SR2 = 82 + 62 = 100 = PR2
∠PSR = 90°.
In Fig. ABC is a right triangle, rightangled at B. AD and CE are the two
medians drawn from A and C
respectively. If AC = 5 cm and
3 5
cm, find the length of CE.
AD =
2
In right ∆ EBC,
Q.5
Sol.
AB2
CE = BE + BC = BC +
4
In right ∆ ABD,
BC2
+ AB2
AD2 = AB2 + BD2 =
4
2
∴
2
CE2 + AD2 =
2
(
2
)
5
5
BC 2 + AB2 = AC 2
4
4
5
5 2 3 5 

⇒ CE = AC2 − AD2 = (5) − 

4
4
 2 
2
(
)
= AD2 + BD2 + 2PD x BD
….(1)
In ∆ ABP
AB2 = AP2 + BP2
= AD2 – PD2 + (BD – PD)2
=AD2 – PD2 + BD2 + PD2 – 2BD x PD
= AD2 + BD2 – 2BD x PD
…(2)
Adding (1) and (2)
AC2 + AB2 = 2AD2 + 2BD2
Q.7
ABC is a right triangle, right-angled
at C. If p is the length of the
perpendicular from C to AB and
AB = c, BC = a and CA = b, then prove that
1 1 1
= +
(i)
pc = ab
(ii)
p2 a2 b2
1
Sol.(i) ar(∆ ABC) = ab
2
2
125 45 80
− = = 20
4
4
4
Hence, CE = 2 5cm.
Q.6
In ∆ ABC, if AD is the median, show
that
AB2 + AC2 = 2AD2 + 2BD2.
Sol.
=
Also, ar(∆ABC) =
∴
⇒
(ii)
⇒
Q.8
In right-angle ∆ ABC
AC2 = AP2 + PC2
= AD2 – PD2 + (PD + DC)2
(QAP2 = AD2 – PD2)
=AD2 – PD2 + PD2 + DC2 + 2PD x DC
= AD2 + DC2 + 2PD x DC
Sol.
1
pc
2
1
1
ab = pc
2
2
pc = ab.
pc = ab further gives
1 c
1
c2
=
⇒ 2= 2 2
p ab
p
ab
1 a 2 + b2 1 1
= 2 2 = 2 + 2.
p2
ab
b a
[Qc
]
= a 2 + b2
∆ ABC is right-angled at B. On the
side AC, a point D is taken such that
AD = DC and AB = BD. Find the
measure of ∠CAB.
Given AD = DC
2
⇒
⇒
⇒
⇒
⇒
⇒
Q.9
Sol.
BD is median to the side AC in
∆ ABC.
AB2 + BC2 = 2(BD2 + AD2)
AC2 = 2BD2 + 2AD2
[By Pythagoras Theorem]
2
(2AD) = 2BD2 + 2AD2
[Q D is the mid-point of AC]
2
2AD = 2BD2 ⇒ AD = BD = AB [Given]
∆ ABD is equilateral.
∠CAB = ∠DAB = 60°.
D and E are points on the sides CA
and CB respectively of ∆ ABC, rightangled at C. Prove that
AE2 + BD2 = AB2 + DE2.
From right ∆ ACE,
……(1)
AE2 = AC2 + CE2
[∴ By Pythagoras Theorem]
From right ∆ BCD,
BD2 = CD2 + BC2
…..(2)
Now, Eqs. (1) + (2)
⇒
AE2 + BD2 = (AC2 + CE2) + (CD2 + BC2)
= (AC2 + BC2) + (CE2 + CD2)
= AB2 + DE2.
[∴ In right ∆ ACB, AB2 = AC2 + BC2, and in
right ∆ DCE, DE2 = CD2 + CE2.]
Q.10 In ∆ ABC, right-angled at B, points D
and E trisects the side BC. Prove that
8AE2 = 3AC2 + 5AD2.
Sol.
Here, BD = DE = EC = x (say)
⇒
BE = BD + DE = 2x
And BC = BE + EC = 2x + x = 3x
From right ∆s ABD, ABE and ABC, we have
AD2 = AB2 + BD2 = AB2 + x2
AE2 = AB2 + BE2 = AB2 + 4x2
AC2 = AB2 + BC2 = AB2 + 9x2
Now, 3AC2+5AD2 = 3(AB2 + 9x2) + 5(AB2 + x2)
= 8AB2 + 32x2 = 8(AB2 + 4x2)
= 8AE2.
Q.11 If ABC is a right triangle with
∠C = 90° and AC = 3 BC, prove
that ∠ABC = 60°.
Sol.
Take D as the mid-point of the
hypotenuse AB. Then,
AD = DB = DC
From right ∆ ABC,
AB2 = BC2 + AC2
= BC2 + ( 3 BC)2
[Q AC = 3BC ]
AB = BC + 3 BC = 4BC2 = (2BC)2
AB = 2BC
1
BC = AB = DB = DC
2
∆ BCD is an equilateral triangle.
∠DBC = 60° ⇒ ∠ABC = 60°.
2
⇒
∴
⇒
⇒
…..(1)
2
2
Q.12
Sol.
acute.
In a ∆ ABC, both the angles, B and C
are acute. If BE ⊥ AC and CF ⊥ AB,
prove that BC2 = AB x BF + AC x CE.
Here angles B and C are given to be
⇒
Perpendiculars BE and CF lie in the
interior of ∆ ABC.
From right ∆ AFC,
AC2 = FC2 + AF2
= FC2 + (AB – BF)2
= (FC2 + BF2) + AB2 – 2AB x BF …(1)
[Q ∆ BFC is a right triangle]
= BC2 + AB2 – 2AB x BF …(1)
Again from right ∆ AEB,
AB2 = BE2 + AE2
= BE2 + (AC – CE)2
= (BE2 + CE2) + AC2 – 2AC x CE
[Q ∆ BEC is a right triangle]
= BC2 + AC2 – 2AC x CE
Now, Eqs. (1) + (2)
⇒
AC2 + AB2 = (BC2 + AB2 – 2AB x BF) +
(BC2 + AC2 – 2AC x CE)
⇒
0 = 2BC2 – 2AB x BF – 2AC x CE
⇒
BC2 = AB x BF + AC x CE.
Q.13 A ladder 10 m long reaches a
window 8 m above the ground. Find
the distance of the foot of the
ladder from the base of the wall.
Sol.
AP = 10 m and BP = 8 m
∴
By Pythagoras Theorem,
AB2 = AP2 – BP2
= 100 – 64
= 36
⇒
AB = 6 m
Hence, the distance of the foot of the
ladder from the base of the wall is 6 m.
EXERCISE
1.
The sides of certain triangles are
given below. Determine which of
them are right triangles:
(i)
5 cm, 8 cm, 11 cm
(ii)
8 cm, 15 cm, 17 cm
(iii)
7 cm, 24 cm, 25 cm
2.
A man goes 70 m due east and then
240 m due north. How far is the
from the start point.
3.
A ladder 26 m long reaches a
window of a building 24 m above
the ground. Determine the distance
of the foot of the ladder from the
building.
4.
A ladder reaches a window which is
12 m above the ground on one side
of the street. Keeping the foot at the
same point, the ladder is turned to
the other side of the street to reach
a window 9 m high. Find the width
of the street if the length of the
ladder is 15 m.
5.(i) ABC is a triangle in which AB = AC
and D is any point in BC. Prove that
AB2 – AD2 = BD x CD
(ii)
ABC is an isosceles triangle, with
AB = AC = 2a and BC = a. Now if
6.
(i)
(ii)
AD ⊥ BC, find AD.
From a point O in the interior of a ∆
ABC, perpendiculars OD, OE and OF
are drawn to the sides BC, CA and
AB respectively (fig.). Prove that
AF2 + BD2 + CE2 = OA2 + OB2 + OC2 –
OD2 – OE2 – OF2.
AF2 + BD2 + CE2 = AE2 + CE2 + BF2.
[Hints: (i) In right triangles OFA, ODB and
OEC, we have
OA2 = AF2 + FO2;
OB2 = BD2 + DO2;
and OC2 = EC2 + EO2.
Add all these to get the result.
(ii)
In right triangles ODB and ODC, we
have
OB2 = BD2 + DO2 and OC2 = CD2 + DO2
∴
OB2 – OC2 = BD2 – CD2
….(1)
Similarly, we have
OC2 – OA2 = CE2 – AE2
…..(2)
2
2
2
2
and OA – OB = AF – BF
…..(3)
Add (1), (2) and (3), to get the result.]
7.(i) In ∆ ABC, ∠B = 90° and D is the midpoint of BC. Prove that
AC2 = AD2 + 3CD2.
(ii)
In ∆ ABC, ∠C = 90°. If D is the midpoint of BC, prove that
AB2 = 4AD2 – 3AC2.
8.
In ∆ ABC, AB = BC = CA = 2a. AD ⊥
BC. Show that
(i)
AD = a 3
ar(∆ABC) = a2 3
(ii)
(iii)
3AB2 = 4AD2.
(iv)
AD2 = 3BD2
[Hints: (i) From Fig. BD = DC = a and AD⊥BC.
∴
AD =
AB2 − Bd 2
= 4a 2 − a 2 = 3a = a 3
1
(ii)
ar (∆ ABC) = × BC × AD
2
1
= × 2a × a 3 = a 2 3.]
2
9.
P and Q are the mid-points of the
sides CA and CB respectively of a ∆
ABC right-angled at C. Prove that
(i)
4AQ2 = 4AC2 + BC2
(ii)
4BP2 = 4BC2 + AC2
(iii)
4(AQ2 + BP2) = 5AB2
10.
In an isosceles triangle ABC, with AB
= AC, BD is perpendicular from B to
the side AC. Prove that
BD2 – CD2 = 2CD = AD
[Hint: In right triangle ABD (Fig.)
AB2 = AD2 + BD2
⇒
⇒
⇒
⇒
11.
AC2 = AD2 + BD2
[QAB = AC]
2
2
2
(AD + DC) = AD + BD
AD2 + DC2 + 2AD x DC = AD2 + BD2
BD2 – CD2 = 2CD x AD.]
Prove that three times the sum of
the squares of the sides of a triangle
is equal to four times the sum of the
squares of the medians of the
triangle.
12.
In an equilateral triangle ABC, the
side BC is trisected at D. Prove that
9AD2 = 7AB2.
13.
Prove that the sum of the squares of
the diagonals of a parallelogram is
equal to the sum of the squares of
its sides.
In Fig. ∠ABD = ∠CDB ∠PQB = 90°. If
AB = x units, CD = y units and PQ = z
1 1 1
units, prove that + = .
x y z
14.
In a right ∆ ABC, with ∠C = 90°, P
and Q are the points on the sides CA
and CB respectively which divide
these sides in the ratio 2 : 1. Prove
that
(i)
9AQ2 = 9AC2 + 4BC2
(ii)
9BP2 = 9BC2 + 4AC2
(iii)
9(AQ2 + BP2) = 13AB2
16.
In a right ∆ ABC, with ∠C = 90°, D is
the mid-point of BC. Prove that
(i)
AB2 = 4AD2 – 3AC2
(ii)
BC2 = 4(AD2 – AC2)
ANSWER EXERCISE
1.
(i), (ii)
2.
250 m
3.
10 m
4.
21 m
1
5.(ii)
15a
2
15.