Download 9th CBSE {SA - 1} Revision Pack Booklet-6

Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Revision Question Bank
Triangles
1.
2.
In the given figure, find the values of x and y.
Solution.
Since, AB = AC
C =B
[angles opposite to the equal sides are equal]
x = 50°
Also, the sum of all angles of a triangle is 180°.
50° +x + y=180°
[put x = 50°]
 50° + 50° + y = 180°
 y= 180° – 100° = 80°
Hence, x = 50° and y = 80°
In the given figure, PQ > PR, QS and RS are the bisectors of  Q and  R, respectively.
Then, prove that SQ > SR
Solution.
Given, PQ > PR
  PRQ >  PQR
[angle opposite to the longer angle is longer]
1
1
 PRO >  PQR

2
2
[dividing both sides by 2]
  SRQ >  SQR
 SQ > SR
[side opposite to the longer angle is longer]
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 3
Pioneer Education {The Best Way To Success}
3.
4.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
In the given figure, BA  AC,DE  DF such that BA=DE and BF = EC. Show that
ABC  DEF
Solution.
In  ABC and  DEF,
BA =DE
[given]
BF = EC
[given]
 BF + FC = EC + CF
[adding both sides by FC]
 BC = EF
Also,  CAB =  FDF [each 90°]
[by RHS congruence rule]
  ABC   DEF
In figure, AB = DE, BC=EF and median AP = median DQ. Prove that  B =  E.
Solution.
Given  ABC and  DEF in which AB = DE, BC = EF and median AP = median DQ.
To prove  B =  E
Proof BC = EF
[given]
1
1
BC  EF
[dividing both sides by 2]
2
2
 BP = EQ
[ AP and DQ are medians]
In  ABP and  DEQ,
AB = DE
[given]
AP=DQ
[given]
and
BP = EQ
[proved above]
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 4
Pioneer Education {The Best Way To Success}
 ABP  DEQ
5.
6.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
[by SSS congruence rule]
[by CPCT]
 B =E
Hence proved.
Show that a median of a triangle divides it into two triangles of equal areas.
Solution.
Given  ABC in which AD is the median i.e., BD = CD
To prove ar(  ABD) = ar(  ACD)
Construction Draw AE  BC
1
Proof ar(  ABD) × BD × AE
...(i)
2
1
and ar (  ACD) = CD × AE
2
1
= BD × AE
...(ii)
2
[ CD = BD, given]
From Eqs. (i) and (ii), we get
ar(  ABD) = ar(  ACD)
Hence proved.
In the given figure,  XYZ and  PYZ are two isosceles triangles on the base YZ with
XY = XZ and PY= PZ. If  P=120° and  XYP=40°, then find  YXZ.
Solution.
Given, in  PXZ, PY = PZ
 PZY =  PYZ
[opposite angles of equal sides are equal]
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 5
Pioneer Education {The Best Way To Success}
7.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Let  PYZ =  PZY =x
 P +  PYZ +  PZY = 180°
[ the sum of all angles of a triangle is 180°]
 120°+x + x = 180°
 2x = 180° – 120°
 2x = 60°
 x = 30°
  PYZ =  PZY = 300
Now,  XYZ =  XYP +  PYZ
= 40° + 30° = 70°
In  XYZ,  YXZ +  XYZ +  XZY = 180°
[ sum of all angles of a triangle is 180°]
  YXZ + 70° + 70° = 180°
  YXZ = 180° – 140° = 40°
In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is
double the smallest side.
Solution.
We have, AABC in which  B is right angle and  C is double  A.
Produce CB upto D such that BC = BD and join AD. In  ABD and  ABC,
BD = BC
[by construction]
[each 90°]
 ABD =  ABC
and AB = AB
[common side]
 ABD  ABC
[ by SAS Congruence rule]
 AD = AC
and  DAB =  CAB = x [CPCT]
[by CPCT]
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 6
Pioneer Education {The Best Way To Success}
8.
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Now,  DAC =  DAB +  CAB
  DAC = x + x = 2x
  DAC =  ACB [  ACB = 2x]
 DC = AD
[ sides opposite to equal angles are equal]
[ BC = DB]
 2BC = AD = AC
Hence proved.
In the given figure, RS = QT and QS=RT. Prove that PQ =PR.
Solution.
Given In  PQR, RS = QT and QS = RT
To prove PQ = PR
Proof In  RESQ and  QTR,
RS = QT
.. (i) [given]
SQ = RT
..(ii) [given]
and
RQ = RQ
...(iii) [common side]
 RSQ  QTR
[by SSS congruence rule]
...(i)
  QRS =  RQT
[by CPCT]
and
[by CPCT] ...(ii)
 QRT =  RQS
On subtracting Eq. (ii) from Eq. (i), we get
 QRS –  QRT =  RQT –  RQS
  TRS =  SQT
Also,  RSQ =  QTR
On multiplying both sides by -1 and then adding 180°, we get
180° –  RSQ =180°–  QTR
  QSP =  RTP
 RPT  QPS
[by ASA congruence rule]
9.
PQ = PR
Hence proved.
ABC is a triangle in which  B = 2  C. D is a point on side BC such that AD bisects  BAC
and AB = CD. Prove that  BAC=72°.
Solution.
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 7
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Given  B = 2  C
AD is the bisector of  A and AB = DC
To prove  BAC =72°
Construction Draw BE as angle bisector of  B.
Join DE. If
 C = x, then
 ABE =  EBC = x
Proof Since, AD is the bisector of  A.
[let]
  BAD =  CAD = y
Now,  EBC is. an isosceles triangle.
So,  EBC =  ECB= x°  EB = EC
Consider  ABE   DCE
[by SAS congruence rule]
[by CPCT]
...(i)
 EDC = 2y
Consider isosceles
 AED as EA = ED
EA = ED
[by CPCT]
... (ii)
 LDA =  EAD = y
From Eqs. (i) and (ii), we get
... (iii)
 ADC = y + 2y =3y
or
... (iv)
 ADC = 2x + y
[exterior angles of AABD ]
From Eqs. (iii) and (iv), we get
3y = 2x + y
...(v)
x = y
Now, in  ABC,
 A +  B +  C=180°
[ sum of all angles of a triangle is 180°)
 2y + 2x + x =180°
[from Eq. (v)]
 2y + 2y + y = 180°
 5y =180°
1800
 360
 y
5
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 8
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
  A = 2y = 2 × 36° = 72°
Hence,  BAC = 72°
Hence proved.
10. In the given figure, if  E >  A,  C >  D, then AD > EC. Is it true?
Solution.
True, In  AEB,  E >  A
...(i)
 AB > BE
In  BDC,  C >  D
...(ii)
 BD > BC
On adding Eqs. (i) and (ii), we get
AB + BD > BE + BC  AD > EC
Chapter Test {Triangles}
M: Marks: 40
1.
M: Time: 40
BE and CF are two equal altitudes of a ABC . Using RHS congruence rule, prove that
[4]
ABC is an isosceles triangle.
Solution.
Given In a  ABC, BE and CF are two equal altitudes of a  ABC.
To prove  ABC is an isosceles triangle.
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 9
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Proof In  ABC and  CFB,
BE=CF
 BEC =  CFB
and BC = BC
 BEC  CFB
[given, equal altitudes]
[each 90°]
[common side]
[by RHS congruence rule]
[by CPCT]
  ECB =  FBC
or
 ACB =  ABC Hence proved.
2.
In figure, S is any point on the side QR of  PQR. Prove that PQ + QR+ RP>2PS.
[4]
Solution.
We know that, sum of the two sides of a triangle is greater than the third side.
In  PQS,
PQ + QS>PS
...(i)
and in  PRS,
PR + SR>PS
...(ii)
On adding Eqs. (i) and (ii), we get
(PQ + QS) + (PR + SR)> 2PS
 PQ + (QS + SR) +PR > 2PS
 PQ + QR + RP>2PS
Hence proved.
3.
In figure, AB = AC, D is a point in the interior of  ABC such that  DBC =  DCB.
Prove that AD bisects  BAC of  ABC.
[4]
Solution.
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 10
Pioneer Education {The Best Way To Success}
In  BDC, we have
 DBC =  DCB
 DC = DB
..(i)
IIT – JEE /AIPMT/NTSE/Olympiads Classes
[Given]
 Sides opposite to equal 
angles of DBC are equal 


Now, in  ABD and  ACD, we have
AB = AC
[Given]
BD = CD
[From (i)]
and, AD = AD
[Common side]
So, by SSS congruency criterion, we have
ABD  ACD
  BAD =  CAD
Hence, AD is the bisector of  BAC.
4.
[c.p.c.t.]
In a  ABC, AB = AC and AD is bisector of  BAC.
Anju has to prove that:  ABD   ACD, which as follows:
Here, AB = AC
[given]
[given]
 BAD =  CAD
and  B =  C [corresponding angles of a equal side are equal)
  ABD   ACD
[by ASA congruence rule]
Further Anju shows this proof to his classmate Amita and she find that there is some
errors in the proof.
(i) Write the correct proof.
(ii) What is the mistake in Anju's proof?
(iii) Which value is depicted from this action?
[4]
Solution.
(i) Given In  ABC, AB = AC and AD is bisector of  BAC.
To prove  ABD   ACD
Proof In  ABD and  ACD
AB = AC
[given]
 BAD =  CAD
[given]
and AD = AD
[common side]
  ABD   ACD
[by SAS congruence rule]
(ii) Anju has used the result  B =  C for proving  ABD =  ACD, which is
wrong.
Because for proving  B =  C, firstly we prove that  ABD   ACD.
(iii) The value depicted from this action, it is cooperative and learning among
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 11
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
students without any religion bias.
5.
In the adjoining figure, ABC is a triangle and D is any point in its interior. Show that
BD + DC < AB + AC.
[4]
Produced BD to meet AC in E.
In  ABE, AB + AE>BE
[ sum of two sides is greater than the third side]
...(i)
 AB+AE > BD + DE
[ BE =BD + DE]
In  CDE, DE + EC > DC
...(ii)
[ sum of two sides is greater than the third side]
On adding Eqs. (i) and (ii), we get
AB + AE +DE +EC > BD + DE + DC
 AB + (AE +EC)>BD + DC
 AB + AC>BD + DC
Hence,
BD + DC < AB + AC
6.
In  ABC, if AD is the bisector of  A, then show that AB > BD and AC > DC.
[4]
Solution :
In  ABC, AD is the bisector of  A
..(i)
 1 = 2
Since exterior angle of a triangle is greater than each of interior opposite angle.
Therefore, in  ADC, we have
Ext.  ADC >  2
 3 > 2
 3 > 1
[Using (i)]
Thus, in  ABD, we have
3 > 1
[Side opp, to greater angle is larger]
 AB >BD
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 12
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Hence, AB > BD.
Similarly, we can prove that AC < CD.
7.
In the given figure, PQR is an equilateral triangle and QRST is a square. Prove that
(i) PT = PS
(ii)  PSR = 15°
Solution
Given : QRST is a square and PQR is an equilateral triangle.
To prove : (i) PS = PS
(ii)  PSR = 150
Proof : Since QRST is a square and  PQR is an equilateral triangle.
 TQR = 900 and  PQR = 600
  TQR +  PQR = 900 + 600
  TQP = 1500
Thus, in  TQP and  SRP, we have
TQ = RS
 TQP =  SRP = 1500
and QP = RP
(by SAS congruence criteria)
 TQP  SRP
(by cpct)
 PT = PS
(ii) In  PSR
SR = RP
( QR = SR)
  RPS =  PSR = x (say)
Now,  RPS +  PSR +  SRP = 1800
 x + x + 1500 = 1800
 2x + 1500 = 1800
 2x = 1800 – 1500
[4]
(by angle sum property)
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 13
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
 2x =
 x = 150
  PSR = 150
300
8.
A plot is in the form of  AEF. Owner of this plot wants to build old age home,
health centre and dispensary for elderly people as shown in figure. In which ABCD is a
parallelogram and AB = BE and AD = DF.
(i) Prove that BEC  DCF .
(ii) What values are depicted here?
[4]
Solution :
Given : ABCD is a parallelogram
AB = BE and AD = DF
To prove : – BEC  DCF
Proof :
Since ABCD is a parallelogram
 AB = DC
But AB = BE
(given)
 DC = BE
Also, AD = BC
but AD = DF (given)
 BC = DF
As ABCD is a parallelogram ,
..(i)
..(ii)
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 14
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
 AD BC and AE as a transversal,
..(iii)
 1 =  2 (corresponding angles)
Also, AB DC and AF as a transversal,
..(iv)
 1 =  3 (corresponding angles)
from (iii) and (iv), we get
..(v)
2 = 3
Therefore, in  BEC and  DCF, we have
BE = DC
(from (i)
BC = DF
(from (ii)
(from (v)
2 = 3
(by SAS congruence rule)
 BEC  DCF
(ii) Kindness and respectful for elder people.
9.
In the figure, ABCD is a quadrilateral in which AD = BC and  DAB =  CBA.
Prove that
(i)  ABD   BAC
(ii) BD = AC
[4]
Solution :
Given : AD = BC and  DAB =  CBA
To prove : (i) ABD  ABC
(ii)BD = AC
Proof : In  ABD and  BAC, we have :
AD = BC
(Given)
 DAB =  CBA (Given)
AB = AB
(common)
Therefore, ABD  BAC (by SAS congruence rule)
(ii) BD = AC
(by cpct)
10. ABCD is a square. P and Q are points on DC and BC respectively, such that
AP = DQ, prove that
(i)  ADP ≅  DCQ
(ii)  DMP = 90°
[4]
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 15
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
Solution :
Given : ABCD is a square, AP = AQ
To prove : (i) ADP  DCQ
(ii)  DMP = 900
Proof :
(i) In  ADP and  DCQ
 ADP =  DCQ = 900
AP = DQ
(Given)
AD = CD
( ABCD is a square)
 ADP  DCQ A (by RHS congruence rule)
(ii) Since ADP  DCQ (by (i))
 DAP +  DAP +  DPA = 1800
(by angle sum property)
 900 +  DAP +  DPA = 1800
  DAP +  DPA = 900
  QDC +  DPA = 900
  MDP +  DPM = 900
 In  MDP,
 MDP +  DPM +  PMD = 180
 900 +  PMD = 1800
  PMD = 1800 – 900
  PMD = 900
(by (i))
.. (ii)
(by angle sum property by ii)
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 16
Pioneer Education {The Best Way To Success}
IIT – JEE /AIPMT/NTSE/Olympiads Classes
www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721 17