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Revision Question Bank
Triangles
1.
2.
In the figure, DE  AC and DC  AP. Prove that
BE BC
.
EC CP
Solution:
In BPA, we have
DC  AP
[given]
Therefore, by basic proportionality theorem, we have
BC BD
..(i)
CP DA
In BCA , we have
DE  AC
[given]
Therefore, by basic proportionality theorem, we have
BE BD
..(ii)
EC DA
From Eqs. (i) and (ii)
BC BE
BE BC
or
Hence proved.
CP EC
EC CP
In figure, if EDC~ ESA BEC = 115° and EDC=70°. Find
AEB and EBA.
DEC,
DCE,
EAB,
Solution:
Since, BD is a line and EC is a ray on it.
DEC + BEC = 180° [linear pair axiom]
DEC +1150=1800 [ BEC =1150,given]
DEC = 180° – 115° = 65°
But
AEB = DEC [vertically opposite angles]
AEB=65°
In CDE, we have CDE + DEC+ DCE = 180°
[by property of sum of the angles of a triangle is 1800 ]
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70° + 65° + DCE =180°
DCE = 180° – 135° = 45°
It is given that, EDC  EBA
3.
EBA = EDC, EAB = ECD
EBA = 70° and EAB = 45°
[ ECD = DCE =45°]
Hence, DEC = 65°, DCE = 45°, EAB = 45°,
AEB = 65° and EBA = 70°
In the figure, ABCD is a quadrilateral with BA parallel to CD.
AC and BD meet at X, where CX=8 cm and XA = 10 cm.
(i) Given that, BD=27cm,find the length of BX.
(ii) Find the ratio
Area of a BXC: Area of AXD
Solution:
AXB  CXD
BX AX
XD XC
BX 10 5
XD 8 4
Since,
BD=27cm
5
BX
× 27 =15 cm
9
(ii) Consider BXA and
ar BXA
15 5
ar AXD 12 4
AXD, sharing common height
1
× b × h]
2
BXC 8 4
Consider BXC and AXD,
AXD 10 5
Ratio of area of BXC : Area of AXD
=4:5
In the given figure, ABC is a right angled triangle, right angled at C and DE
that ABC ADE and hence find the lengths of AE and DE.
[ area of triangle =
4.
AB. Prove
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5.
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Solution:
In ABC and ADE, we have
A= A
[common angle]
C= E
[each 90°]
ABC  ADE
[by AA similarity criterion]
In ABC, we have
AB2 =BC2 +AC2
[Pythagoras theorem]
AB2 =122 +52 =144+25=169
[ BC = 12 cm and AC = 5 cm given]
AB = 13 cm
Since,
ABC  ADE
AB BC AC
AD DE AE
[by basic proportionality theorem]
13 12
5
3 DE AE
36
DE
cm
13
15
AE
and
cm
13
In the figure, AB = 8 cm, BC=3 cm and BE is parallel to CD.
(i) Find the values of
Area of ABE
Area of ABE
BE
(a)
(b)
(c)
Area of Quadrilateral BCDE
Area of ACD
CD
(ii) What is the special name given to the quadrilateral BCDE?
Solution:
(i)(a) ABE and ACD ,
ABE = ACD
[corresponding angles, BE  CD]
AEB = ADC
[corresponding angles, BE  CD]
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and
BAE = CAD
[common angle]
ABE and ACD is similar.
BE AB
8
8
CD AC 8 3 11
Area of ABE
(b)
Area of ACD
8
=
11
6.
Length of side AB
Length of side AC
2
2
64
121
Area of ABE
64
64
(c)
Area of Quadrilateral CDE 121 64 57
(ii) It is a trapezium, which has 2 parallel sides BE and CD.
In the figure, ABC= AED.
(i) Explain why ABC and AED are similar.
(ii) Given, also that AD = 3cm, AE =5cm and EC=2cm. Calculate
(a) BD
Area of AED
(b)
Area of ABC
Solution:
(i) In ABC and AED ABC= AED
[given]
BAC= EAD
[common angle]
ACB  ADE
[by AAA similarity criterion]
(ii) (a) Since, ABC  AED
AB
AE
BC AC
ED AD
[by basic proportionality theorem]
AB 7
[  AE = 5 cm, AC = 7cm and AD = 3cm given)
5 3
3AB = 35
35
2
AB
11 cm
3
3
2
2
BD = AB – AD = 11 – 3 = 8 cm
3
3
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(b) Since,
AD
AC
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3
7
2
7.
Area of AED
3
9
Area of ABC
7
49
Two poles of heights p and q m are standing on a level ground, a metre apart. Prove that
the height of the point of intersection of the line segments joining the top of each pole to
pq
the foot of the opposite pole is given by
m.
p q
Solution:
In CQP and CAB ,
C= C
CQP = CAB
CQP  CAB
CQ
CA
In
PQ
AB
[common angle]
[each90°]
[by AA similarity criterion]
x
a
h
q
AQP and ACD,
A= A
AQP = ACD
AQP  ACD
AQ PQ
a x
AC CD
a
...(i)
[common angle]
[each 90°]
[by AA similarity criterion]
h
p
[by basic proportionality theorem
x h
1
..(ii)
a p
On adding Eqs. (i) and (ii), we get
h h
ph qh
1
1
q p
pq
1
h p q
pq
h
pq
m
p q
Hence proved.
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In the figure, PQ  BC and PR  CD. Prove that
QB DR
AR AQ
(ii)
AQ AR
AD AB
Solution:
In ABC, we have
PQ  BC
[given]
Therefore, by basic proportionality theorem, we have
AQ AP
..(i)
AB AC
In ACD,
PR  CD
Therefore, by basic proportionality theorem, we have
AP AR
...(ii)
AC AD
From Eqs.(i) and (ii),
AQ AR
AR AD
or
AB AD
AD AB
From Eqs. (i) and (ii),
AQ AR
AB AD
AB AD
AQ AR
[reciprocal the above equation]
AQ QB AR RD
QB
1
AQ
AR
AQ
RD
QB DR
=1
Hence proved.
AR
AQ AR
(i)
9.
In the figure. A, B and C are points on OP, OQ and Off respectively, such that AB  PQ and
BC  QR. Show that AC  PR.
Solution:
In OPQ, we have
AB  PQ
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OA
AP
OB
BQ
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...(i)
[by basic proportionality theorem]
In OQR, we have
BC  QR
OB OC
...(ii)
BQ CR
[by basic proportionality theorem]
From Eqs. (i) and (ii),
OA OC
AP CR
Thus, A and C are points on sides OP and OR, respectively of PQR, such that
OA OC
AC  PR
AP CR
[by converse of basic proportionality theorem]
10. D, E and F are the points on sides BC, CA and AB respectively, such that AD bisects A,
BE bisects B and F bisects C. If 48 = 5 cm, SC = 8 cm and A=4 cm, then determine
AF,CE and BD.
Solution:
Given in ABC, CF bisects C.
We know that, the internal bisector of an angle of a triangle divides the opposite side
internally in the ratio of the sides containing the angle.
AF AC
AF
4
FB BC
5 AF 8
[ FB = AB – AF = 5 – AF]
AF
1
5 AF 2
2AF = 5 – AF
2AF + 5 3AF = 5
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5
cm
3
Again, in ABC, BE bisects B.
AE AB
[by above theorem]
EC BC
4 CE 5
CE
8
[ AE =AC– CE =4 – CE, AB = 5cm and BC = 8 cm, given]
8(4 – CE) = 5CE
32 – 8CF =5CE
32=13CE
32
CE =
cm
13
Similarly, in ABC, AD bisects A.
BD AB
BD
5
DC AC
8 BD 4
[ DC = BC – BD = 8 – BD, AB = 5 cm, AC = 4 cm]
4BD = 5(8 – BD)
4BD = 40 – 5BD
9BD = 40
40
BD =
cm
9
5
32
40
Hence, AF = cm, CE =
cm and BD =
cm
3
13
9
AF =
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Chapter Test {Triangles}
M: Marks: 40
1.
2.
M: Time: 40 Min.
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.
[4]
Solution:
Given A ABC in which AB = BC = CA and AD BC.
To prove 3AB2 =4AD2
Proof In ADB and ADC,
AB = AC
[given]
B= C
[each 60°]
and
ADB = ADC
[each 90°]
ADB
ADC
[by SAA congruence rule]
1
BD DC
BC
..(i)
2
In right ADB, by Pythagoras theorem,
AB2 =AD2+ BD2
2
1
2
2
AB = AD +
BC [ from Eq. (i)]
2
1 2
AB2 AD2
BC
4
4AB2 = 4AD2 + BC2
4AB2 – ABD2 = 4AD2
[given, BC=AB]
3AB2 = 4AD2
Hence proved.
If the area of two similar triangles are equal, then prove that they are congruent.
Solution:
Given ABC  DEF such that
ar( ABC) = ar( DEF)
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To prove ABC
DEF
Proof Since, ABC - DEF and
ar( ABC) = ar( DEF)
B
E
..(i)
and
C
F
ar
Now,
ar
[ ar
3.
ABC
DEF
ABC
BC2
EF2
ar
ar
ar
ABC
ABC
BC2
EF2
DEF , given]
BC2
1
EF2 BC2
2
EF
EF = BC
[taking positive square root]
Now, in ABC and DEF, we get
B = E [from Eq. (i)]
C = F [from Eq. (i)]
and BC = EF
[proved above]
ABC
DEF
[by ASA congruence rule]
Hence proved.
If the sides AB and AC and median AD of a ABC are proportional to sides PQ and PR
and median PM of another PQR. Then, prove that ABC  PQR.
[4]
Solution:
Sol :
Given Two ABC and PQR in which AD and PM are medians such that
AB AC AD
PQ PR PM
To prove ABC  PQR
Construction Produce AD to E and PM to N such that AD = DE and PM = MN. Join CE and
NR.
Proof In ADB and EDC,
AD = ED
BD = CD
and
ADB = CDE
[by construction]
[since, D is mid-point of BC]
[vertically opposite angles]
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ADB
EDC
[by SAS congruence rule]
AB = EC
[by CPCT] ...(i)
Similarly, in PMQ and NMR,
PQ = NR
.
..(ii)
AB AC AD
Now,
[given]
PQ PR PM
EC AC 2AD
NR PR 2PM
[ from Eqs. (i) and (ii)]
EC AC AE
NR PR PN
[2AD = AE and 2Pm = PN]
ACE  PRN
[by SSS similarity criterion]
1= 2
...(iii)
[corresponding angles are equal]
Similarly,
3= 4
...(iv)
On adding Eqs. (iii) and (iv), we get
1 + 3 = 2 + 4 or A = P
Now, in ABC and PQR,
AB AC
A = P and
[given]
PQ PR
ABC  PQR
[by SAS criterion rule]
Hence proved.
Or
Given In the given figure,
AB  DC and AC and PQ intersect each other at the point O.
To prove OA . CQ = OC . AP
Proof Since, AB  DC and AC is a transversal.
OCQ = OAB
..(i)
[alternate angles]
Again, PQ is also a transversal.
OQC = APO
... (ii)
[alternate angles]
Now, in OAP and OCQ,
OAP = OCO,
[from Eq. (i)]
APO = OQC
[from Eq. (ii)]
and AOP= QOC
[vertically opposite angles]
OAP  OCQ
[by AAA similarity criterion]
OA OP AP
OC OQ CQ
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OA AP
OC CQ
OA = CQ = QC. AP
4.
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Hence proved.
In the given figure, in
that ar
ABC, XY  AC and XY divides the
AX
.
BXY = 2 ar (ACY X). Determine
AB
ABC into two regions such
[4]
Solution:
We have, ar( BXY) = 2ar( CYX)
ar( BXY)=2[ar( BAC) – ar( BXY)]
3ar( BAY)=2ar( BAC)
ar BXY
2
ar BAC 3
In BXY and BAC,
B= B
[common angle]
BXY = BAC
[corresponding angles]
and
BVX = BCA [corresponding angles]
BXY  BAC
[by AAA similarity criterion]
2
ar BXY
BX
ar BAC BA2
2 BX 2
from Eq. (i)]
3 BA 2
BX
2
[taking positive square root]
BA
3
BX
2
[multiplying both sides by –1]
BA 3
BX
2
[adding l on both sides]
1
1
BA
3
BA BX
3
2
BA
3
AX
3
2
BA
3
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5.
6.
In the given figure, DEFG is a square and
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BAC = 900 show that DE2 = BD × EC.
[4]
Solution:
Given In the given figure, DEFG is a square and BAC = 90°.
To prove DE2 = BD × EC
Proof In DBG and ECF
3 + 1 = 90° = 3 + 4
3+ 1= 3+ 4
1= 4
Also
D= E
[each 90°]
DBG  DEF
[by AA similarity criterion]
BD EF
[by basic proportionality theorem]
DG EC
BD × EC = EF × DG
But DG = EF = DE
[side of a square]
BD × EC = DE × DE
DE2 = BD × EC
Hence proved.
P is the mid-point of side BC of ABC. Q is the mid-point of AP, BQ when produced
1
AC .
meets AC at L. Prove that AL
[4]
3
Solution:
Given In ABC, P is the mid-point of BC and Q is the mid-point of AP. BQ when
produced meets AC at L.
1
AL= AC
3
Construction Draw PM  BL meeting AC at M.
Proof In ABLC, PM  SL
BP LM
[by basic proportionality theorem]
PC MC
But BP= PC
[ P is mid-point of BC]
To prove
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LM
1 LM MC
...(i)
MC
Now, in APM, Q is the mid-point of AP and QL  PM
AQ AL
[by basic proportionality theorem]
QP LM
But AQ=QP
[Q is mid-point of AP]
AL
1 AL LM
..(ii)
LM
From Eqs. (i) and (ii),we get
1
AL=LM=MC
AL = AC
3
Hence proved.
Gopal is walking away from the base of a lamp-post at a speed of 2 m/s and his height
(MN) is 0.9 m. Suppose the lamp-post 4.5 m above from the ground as shown in
adjoining figure.
(i) Find the distance of Gopal from the base of lamp-post in 5s.
(ii) Is the PQR and MNR similar?
(iii) Find the length of his shadow after 5 s.
[4]
Solution:
(i) Let N be the position of Gopal after 5 s, then distance of Gopal From the base of lamppost,
QN = 2 × 5=10m
(ii) In PQR and MNR, LQ = LN [since, each angle is 90° because lamp-post as well as
Gopal is standing vertically]
R= R
[common angle]
PQR  MNR
[by AA similarity criterion]
(iii) Let length of his shadow after 5 s in NR = x Since, PQR – MNR
QR PQ
QN NR PQ
Therefore,
NR MN
NR
MN
10 x 4.5
x
0.9
[QN = 10 m, PQ = 4.5 m, MN = 0.9 m]
9 + 0.9x = 4.5X
9 = 4.5x - 0.9x
9 = 3.6x
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9 90 10
2.5m
36 36 4
Hence, the shadow of Gopal after 5 s is 2.5 m.
In the given figure, FEC
GBD and 1 =
x
8.
Solution:
GBD and
Given FEC
To prove ADE  ABC
9.
1=
2. Prove that
ADE 
ABC.
[4]
2
FEC
GBD
Proof Since,
DG=EF
[by CPCT]
EFG = DGB
i.e., 5= 6
[angles opposite to equal sides are equal]
Also, BD=EC
[by CPCT]
DBC= ECB
i.e., 3 = 4
and
BC=FC
[by CPCT]
BDC= CEF
But also
1= 2
[given]
AE= AE
[sides opposite to equal angles are equal]
In ADE and ABC, AD=AE
Also, BD=EC
AD AE
BD EC
ADE  ABC
[by converse of basic proportionality theorem]
State and prove the basic proportionality theorem.
Solution.
Given A triangle ABC in which DE  BC, and intersects AB and D and AC in E.
AD AE
To prove
DB EC
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Construction Join BE, CD and draw EF BA and DG CA.
Proof Since EF is perpendicular to AB. Therefore, EF is the height of triangles ADE and
DBE.
1
1
Now, Area ADE
(base × height) = (AD. EF)
2
2
1
1
and, Area DBE = (base × height) = (DB. EF)
2
2
1
Area ADE 2 AD.EF AD
..(i)
Area ABE 1 DB.EF DB
2
Similarly, we have
1
AE. DG
Area ADE
AE
2
..(ii)
1
Area DEC
EC
EC. DG
2
But, DAC are on the same base DE and between the same parallels DE and BC.
Area DBE = Area DEC
1
Area
1
DBE
Area
DEC
[Taking reciprocals of both sides]
Area ADE
Area ADE
[Multiplying both sides by Area ( ADE)]
Area DBE
Area DEC
AD AE
[Using (i) and (ii)]
DB EC
10. In the given figure, ABCD is a quadrilateral P, Q, R and S are the points of trisection of
the sides AB, BC, CD and DA, respectively. Prove that PQRS is a parallelogram.
[4]
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Solution.
Given A quadrilateral ABCD in which P, Q, R and S are the points of trisection of sides
AB, BC, CD and DA respectively and are adjacent to A and C.
To prove PQRS is a parallelogram i.e., PQ  SR and QR  PS.
Construction Join AC.
Proof Since, P, Q, R and S are the points of trisection of AB, BC, CD and DA respectively.
BP = 2PA, BQ = 2 QC, DR = 2 RC
and, DS = 2 SA
In ADC, we have
DS 2SA
DR 2RC
2 and ,
2
SA SA
RC RC
DS DR
SA RC
S and R divide the sides DA and DC respectively in the same ratio.
..(i)
SR  AC
[By the converse of Thale’s Theorem]
In ABC, we have
BP 2PA
BQ 2QC
2 and
2
PA PA
QC QC
BP BQ
PA QC
P and Q divide the sides BA and BC respectively in the same ratio.
PQ  AC
..(ii) [By the converse of Thale’s Theorem]
From equations (i) and (ii), we have
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SR  AC and PQ  AC
SR  PQ
Similarly, by joining BD, we can prove that
QR  PS
Hence, PQRS is a parallelogram.
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