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Class 9 NCERT Solutions Maths Chapter 7 - Triangles

Triangles Exercise Ex. 7.1

Solution 1

In ABC and ABD
AC = AD                                            (given)
CAB = DAB                                 (given)
AB = AB                                             (common)
 
 
So, BC and BD are of equal length.

Solution 2

In ABD and BAC
    AD = BC                                         (given)
   DAB = CBA                              (given)
    AB = BA                                         (common)

And ABD = BAC                          (by CPCT)

Solution 3

In BOC and AOD
BOC = AOD                                 (vertically opposite angles)
CBO = DAO                                 (each 90o)
BC = AD                                             (given)

Solution 4

Solution 5

Solution 6

Given that BAD = EAC
BAD + DAC = EAC + DAC
BAC = DAE
Now in BAC and DAE
AB = AD                                             (given)
BAC = DAE                                 (proved above)
AC = AE                                             (given)

Solution 7

Given that EPA = DPB
EPA + DPE = DPB + DPE
 DPA = EPB
     Now in  DAP and  EBP
    DAP = EBP                               (given)
    AP = BP                                          (P is mid point of AB)
    DPA = EPB                              (from above)

Solution 8

 (i)  In AMC and BMD
AM = BM                                              (M is mid point of AB)
AMC = BMD                                  (vertically opposite angles)
CM = DM                                             (given)
 
 
(ii) We have ACM = BDM
But ACM and BDM are alternate interior angles
Since alternate angles are equal.  
Hence, we can say that DB || AC
DBC + ACB = 180o                   (co-interior angles) DBC + 90o = 180o
DBC + 90o = 1800 
DBC          =  90o
 
(iii) Now in DBC and ACB
DB = AC                                             (Already proved)
DBC = ACB                                 (each 90o )
BC = CB                                             (Common)
(iv) We have DBC  ACB
 

Triangles Exercise Ex. 7.2

Solution 1

(i)    It is given that in triangle ABC, AC = AB             
  ACB = ABC     (angles opposite to equal sides of a triangle are equal)
OBC = OBC
OB = OC                  (sides opposite to equal angles of a triangle are also equal)
 
(ii) Now in OAB and OAC
    AO =AO                               (common)
   AB = AC                                (given)
   OB = OC                               (proved above)    
   So, OAB  OAC         (by SSS congruence rule)
    BAO = CAO             (C.P.C.T.)

Solution 2

In ADC and ADB
    AD = AD                                         (Common)
    ADC =ADB                              (each 90o)
    CD = BD                                        (AD is the perpendicular bisector of BC)

Solution 3

In AEB and AFC
AEB = AFC                                             (each 90o)
A = A                                                     (common angle)
AB = AC                                                        (given)

Solution 4

(i) In AEB and AFC
    AEB = AFC                          (each 90)
    A = A                                      (common angle)
     BE = CF                                        (given)
    
 
(ii) We have already proved
    AEB  AFC
     AB = AC                                      (by CPCT)

Solution 5

 
Let us join AD
In ABD and ACD
AB = AC                                    (Given)
BD = CD                                    (Given)
AD = AD                                    (Common side)

Solution 6

In ABC
AB = AC                                                (given)
ACB = ABC                               (angles opposite to equal sides of a triangle are also equal)
Now In ACD
AC = AD
ADC = ACD                              (angles opposite to equal sides of a triangle are also equal)
Now, in BCD
ABC + BCD + ADC = 180o          (angle sum property of a triangle)
ACB + ACB +ACD + ACD = 180o
 2(ACB + ACD) = 180o
 2(BCD) = 180o
BCD = 90o

Solution 7

Given that
AB = AC
C = B                      (angles opposite to equal sides are also equal)
In ABC,
A + B + C = 180o     (angle sum property of a triangle)
90o + B + C = 180o
90o + B + B = 180o
 2 B = 90o
B = 45

Solution 8

 
Let us consider that ABC is an equilateral triangle.
So, AB = BC = AC
Now, AB = AC
C = B         (angles opposite to equal sides of a triangle are equal)
    
We also have
AC = BC    
B = A             (angles opposite to equal sides of a triangle are equal)
    
So, we have
A = B = C
    Now, in ABC
A + B + C = 180o
A + A + A = 180o
 3A = 180o
A = 60o
A = B = C = 60o
Hence, in an equilateral triangle all interior angles are of 60o.

Triangles Exercise Ex. 7.3

Solution 1

(i)  In ABD and ACD
      AB = AC                                             (given)
      BD = CD                                            (given)
      AD = AD                                            (common)
     
 
(ii)  In ABP and ACP
       AB = AC                                            (given).
      BAP = CAP                                  [from equation (1)]
       AP = AP                                             (common)
      
 
(iii)   From equation (1)
        BAP = CAP            
        Hence, AP bisect A
        Now in BDP and CDP
        BD = CD                                            (given)
        DP = DP                                            (common)
        BP = CP                                            [from equation (2)]
       
(iv)   We have BDP  CDP
       
       

         Now, BPD + CPD = 180o             (linear pair angles)

         BPD + BPD = 180o 

         2BPD = 180o                                    [from equation (4)]

        BPD = 90o                                                                    ...(5)

        From equations (2) and (5), we can say that AP is perpendicular  bisector of BC.

      

Solution 2

(i)   In BAD and CAD
        ADB = ADC                                      (each 90o as AD is an altitude)
        AB = AC                                                  (given)
        AD = AD                                                  (common)
 
(ii)     Also by CPCT,
         BAD = CAD
          Hence, AD bisects A.

(ii)              Also by CPCT,

          ÐBAD = ÐCAD

         Hence, AD bisects ÐA.

 

Solution 3

(i)  In ABC, AM is median to BC
     BM = BC
     In PQR, PN is median to QR
     QN = QR
     But BC = QR
 
     
     BN = QN                                                     ...(i)
      Now, in ABM and PQN
     AB = PQ                                                       (given)
     BM = QN                                                       [from equation (1)]
     AM = PN                                                        (given)
 
 
 (ii)  Now in ABC and PQR

      AB = PQ                                                        (given)
     ABC = PQR                                             [from equation (2)]
     BC = QR                                                        (given)
      ABC  PQR                                  (by SAS congruence rule)  

Solution 4

In BEC and CFB
BEC = CFB                                              (each 90o )
BC = CB                                                         (common)
BE = CF                                                         (given)
                                                 (Sides opposite to equal angles of a triangle are equal)
 
Hence, ABC is isosceles.

Solution 5

In APB and APC
APB = APC                                             (each 90o)
AB =AC                                                          (given)
AP = AP                                                   
     (common)
B = C                                                (by using CPCT)
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