NCERT Solutions for Class 9 Maths Chapter 7 - Triangles

Do you struggle to prove that the two triangles given in a textbook question are congruent? Do you find it difficult to explain the concept of SSS congruence in a Maths problem? If yes, then our NCERT Solutions for CBSE Class 9 Mathematics Chapter 7 Triangles will be of immense help. These are model answers that Maths experts have developed to support students for conceptual clarity. 

TopperLearning’s NCERT textbook solutions are easily available online for students trying to practise and understand the properties of triangles. You can use these chapter solutions along with our practice tests, concept videos and CBSE Class 9 sample papers to effectively revise your Maths lessons.

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Chapter 7 - Triangles Exercise Ex. 7.1

Solution 1
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABD
AC = AD                                            (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCAB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAB                                 (given)
AB = AB                                             (common)
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
So, BC and BD are of equal length.

Solution 2
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABD and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC
    AD = BC                                         (given)
   Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCBA                              (given)
    AB = BA                                         (common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
And Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABD = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC                          (by CPCT)
Solution 3
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBOC and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAOD
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBOC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAOD                                 (vertically opposite angles)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCBO = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAO                                 (each 90o)
BC = AD                                             (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 4
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 5
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 6
Given that Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesEAC
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAD + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesEAC + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAC
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAE
Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAE
AB = AD                                             (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAE                                 (proved above)
AC = AE                                             (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 7
Given that Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesEPA = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDPB
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesEPA + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDPE = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDPB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDPE
 Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesNcert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDPA = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesEPB
     Now in  Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAP and Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles EBP
    Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDAP = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesEBP                               (given)
    AP = BP                                          (P is mid point of AB)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDPA = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesEPB                              (from above)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 8
 (i)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAMC and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBMD
AM = BM                                              (M is mid point of AB)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAMC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBMD                                  (vertically opposite angles)
CM = DM                                             (given)
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
(ii) We have Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACM = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBDM
But Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACM and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBDM are alternate interior angles
Since alternate angles are equal.  
Hence, we can say that DB || AC
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBC + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB = 180o                   (co-interior angles) Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBC + 90o = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBC + 90o = 1800 
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesNcert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBC          =  90o
 
(iii) Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBC and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB
DB = AC                                             (Already proved)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB                                 (each 90o )
BC = CB                                             (Common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
(iv) We have Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesDBC Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles

Chapter 7 - Triangles Exercise Ex. 7.2

Solution 1
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
(i)    It is given that in triangle ABC, AC = AB             
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles  Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC     (angles opposite to equal sides of a triangle are equal)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesNcert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesOBC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesOBC
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesOB = OC                  (sides opposite to equal angles of a triangle are also equal)
 
(ii) Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesOAB and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesOAC
    AO =AO                               (common)
   AB = AC                                (given)
   OB = OC                               (proved above)    
   So, Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesOAB Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesOAC         (by SSS congruence rule)
   Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAO = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCAO             (C.P.C.T.)
Solution 2
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADC and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADB
    AD = AD                                         (Common)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADC =Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADB                              (each 90o)
    CD = BD                                        (AD is the perpendicular bisector of BC)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 3
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAEB and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAFC
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAEB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAFC                                             (each 90o)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA                                                     (common angle)
AB = AC                                                        (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 4
(i) In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAEB and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAFC
    Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAEB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAFC                          (each 90)
    Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA                                      (common angle)
     BE = CF                                        (given)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
(ii) We have already proved
    Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAEB Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAFC
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles AB = AC                                      (by CPCT)

Solution 5
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
Let us join AD
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABD and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD
AB = AC                                    (Given)
BD = CD                                    (Given)
AD = AD                                    (Common side)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 6
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC
AB = AC                                                (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC                               (angles opposite to equal sides of a triangle are also equal)
Now In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD
AC = AD
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD                              (angles opposite to equal sides of a triangle are also equal)
Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBCD
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBCD + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADC = 180o          (angle sum property of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesNcert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB +Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles 2(Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD) = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles 2(Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBCD) = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBCD = 90o
Solution 7
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Given that
AB = AC
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB                      (angles opposite to equal sides are also equal)
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC,
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o     (angle sum property of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles 90o + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles 90o + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles 2 Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 90o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = 45
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 8
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
Let us consider that ABC is an equilateral triangle.
So, AB = BC = AC
Now, AB = AC
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB         (angles opposite to equal sides of a triangle are equal)
    
We also have
AC = BC    
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA             (angles opposite to equal sides of a triangle are equal)
    
So, we have
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC
    Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles 3Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = 60o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 60o
Hence, in an equilateral triangle all interior angles are of 60o.

Chapter 7 - Triangles Exercise Ex. 7.3

Solution 1
(i)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABD and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACD
      AB = AC                                             (given)
      BD = CD                                            (given)
      AD = AD                                            (common)
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
(ii)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABP and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACP
       AB = AC                                            (given).
      Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAP = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCAP                                  [from equation (1)]
       AP = AP                                             (common)
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
(iii)   From equation (1)
        Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAP = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCAP            
        Hence, AP bisect Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA
        Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBDP and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCDP
        BD = CD                                            (given)
        DP = DP                                            (common)
        BP = CP                                            [from equation (2)]
        Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
(iv)   We have Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBDP Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCDP
       Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
       

         Now, Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBPD + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCPD = 180o             (linear pair angles)

         Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBPD + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBPD = 180o 

         2Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBPD = 180o                                    [from equation (4)]

        Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBPD = 90o                                                                    ...(5)

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

      

Solution 2
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
(i)   In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAD and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCAD
        Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADC                                      (each 90o as AD is an altitude)
        AB = AC                                                  (given)
        AD = AD                                                  (common)
  Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
(ii)     Also by CPCT,
         Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBAD = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCAD
          Hence, AD bisects Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA.

(ii)              Also by CPCT,

          ÐBAD = ÐCAD

         Hence, AD bisects ÐA.

 

Solution 3
(i)  In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC, AM is median to BC
     Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBM = Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles BC
     In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPQR, PN is median to QR
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles QN = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesQR
     But BC = QR
 
      Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles BN = QN                                                     ...(i)
      Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABM and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPQN
     AB = PQ                                                       (given)
     BM = QN                                                       [from equation (1)]
     AM = PN                                                        (given)
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesNcert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
 (ii)  Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPQR

      AB = PQ                                                        (given)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPQR                                             [from equation (2)]
     BC = QR                                                        (given)
     Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPQR                                  (by SAS congruence rule)  

Solution 4
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesIn Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBEC and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCFB
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBEC = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCFB                                              (each 90o )
BC = CB                                                         (common)
BE = CF                                                         (given)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles                                                 (Sides opposite to equal angles of a triangle are equal)
 
Hence, Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC is isosceles.
Solution 5
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAPB and Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAPC
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAPB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAPC                                             (each 90o)
AB =AC                                                          (given)
AP = AP                                                   
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
     (common)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB = Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC                                                (by using CPCT)

Chapter 7 - Triangles Exercise Ex. 7.4

Solution 1
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Let us consider a right angled triangle ABC, right angle at B.
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o            (angle sum property of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + 90o + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC = 90o
Hence, the other two angles have to be acute (i.e. less than 90).
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
[In any triangle, the side opposite to the larger (greater) angle is longer]
So, AC is the largest side in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC.
But AC is the hypotenuse of Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC. Therefore, hypotenuse is the longest side in a right angled triangle.

Solution 2
In the given figure,
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPBC = 180p            (linear pair)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC = 180o - Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPBC             ... (1)
Also,
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesQCB = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB = 180o - Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesQCB                    ... (2)
As Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPBC < Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesQCB
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles 180 - Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPBC > 180o - Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesQCB.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC > Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesACB                [From equations (1) and (2)]
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles AC > AB                           (side opposite to larger angle is larger)
Solution 3
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesAOB
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB < Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles AO < BO     (side opposite to smaller angle is smaller)        ... (1)
Now in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesCOD
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC < Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesD
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles OD < OC     (side opposite to smaller angle is smaller)        ... (2)
On adding equations (1) and (2), we have
AO + OD < BO + OC
AD < BC

Solution 4
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Let us join AC.
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC
AB < BC           (AB is smallest side of quadrilateral ABCD)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
(1)
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesADC
AD < CD          (CD is the largest side of quadrilateral ABCD)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
(2)
On adding equations (1) and (2), we have
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles2 + Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles4 < Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles1 + Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles3
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC < Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesA > Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesC
Let us join BD.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABD
AB < AD            (AB is smallest side of quadrilateral ABCD)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
(3)
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesBDC
 BC < CD         (CD is the largest side of quadrilateral ABCD)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
On adding equations (3) and (4), we have
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles8 + Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles7 < Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles5 + Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles6
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesD < Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesB > Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesD

Solution 5
  As PR > PQ
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
  PS is the bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesQPR
  Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
   Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Solution 6
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
Let us take a line l and from point P (i.e. not on line l) we have drawn two line segments PN and PM. Let PN be perpendicular to line l and PM is drawn at some other angle.
In Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesPNM
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesN = 90o
Now, Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesP + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesN + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesM = 180o    (Angle sum property of a triangle)    
Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesP + Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesM = 90o            
Clearly, Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesM is an acute angle
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles

Chapter 7 - Triangles Exercise Ex. 7.5

Solution 1
Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors, of all the sides of triangles meet together.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
As here in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. So O is point which is equidistant from all the vertices of Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC.
Solution 2
The point which is equidistant from all the sides of a triangle is incenter of triangle. Incentre of triangle is the intersection point of angle bisectors of interior angles of that triangle.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Here in Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC we can find the incentre of this triangle by drawing the angle bisectors of interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. So, I is the point, equidistant from all the sides of Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC.
Solution 3
Ice-cream parlour should be set up at the circumcentre O of Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesABC.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
In this situation maximum number of persons can approach to it. We can find circumcentre O of this triangle by drawing perpendicular bisectors of the sides of this triangle.

Solution 4
We may observe that hexagonal shaped rangoly is having 6 equilateral triangles in it.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - TrianglesOAB = Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles (side)2 = Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles (5)2
                          Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
    Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles

 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Triangles