NCERT Solution for Class 9 Mathematics Chapter 7 - Triangles, RR ACADEMY MEERUT

NCERT Solution for Class 9 Mathematics Chapter 7 - Triangles Page/Excercise 7.1

Question 1

In quadrilateral ACBD, AC = AD and AB bisects A (See the given figure). Show that ABC  ABD. What can you say about BC and BD? 

Solution 1

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

Question 2

ABCD is a quadrilateral in which AD = BC and DAB = CBA (See the given figure). Prove that 
(i)    ABD  BAC
(ii)    BD = AC                
(iii)    ABD = BAC.   

Solution 2

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

Question 3

AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.  

Solution 3

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

Question 4

l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ABC  CDA. 

Solution 4

Question 5

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of A (see the given figure). Show that:
(i)    APB  AQB
(ii)    BP = BQ or B is equidistant from the arms of A. 

Solution 5

Question 6

In the given figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE. 

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) 

Question 7

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB (See the given figure). Show that 
(i) DAP  EBP
(ii) AD = BE 

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) 

Question 8

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure).  Show that:
    (i)    AMC  BMD
    (ii)    DBC is a right angle.
    (iii)    DBC  ACB
    (iv)    CM =  AB 

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  

NCERT Solution for Class 9 Mathematics Chapter 7 - Triangles Page/Excercise 7.2

Question 1

In an isosceles triangle ABC, with AB = AC, the bisectors of B and C intersect each other at O. Join A to O. Show that: 
(i) OB = OC        (ii) AO bisects A

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.)

Question 2

In ABC, AD is the perpendicular bisector of BC (see the given figure).  Show that ABC is an isosceles triangle in which AB = AC.

Solution 2

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

Question 3

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

Solution 3

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

Question 4

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure) show that 
    (i)  ABE  ACF 
    (ii) AB = AC, i.e., ABC is an isosceles triangle. 

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)

Question 5

ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ABD = ACD.

Solution 5

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

Question 6

ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure).  Show that BCD is a right angle.

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

Question 7

ABC is a right angled triangle in which A = 90o and AB = AC.  Find B and C.

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� 

Question 8

Show that the angles of an equilateral triangle are 60o each.

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.

NCERT Solution for Class 9 Mathematics Chapter 7 - Triangles Page/Excercise 7.3

Question 1

ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that 
(i)    ABD  ACD
(ii)    ABP  ACP                 
(iii)    AP bisects A as well as D.
(iv)    AP is the perpendicular bisector of BC.  

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.

      

Question 2

AD is an altitude of an isosceles triangles ABC in which AB = AC.  Show that  (i) AD bisects BC    (ii) AD bisects A.

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.

Question 3

Two side AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR (see the given figure). Show that: 
(i)    ABM  PQN
(ii)    ABC  PQR 

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)  

Question 4

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

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.

Question 5

ABC is an isosceles triangle with AB = AC. Drawn AP  BC to show that B = C.

Solution 5

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

NCERT Solution for Class 9 Mathematics Chapter 7 - Triangles Page/Excercise 7.4

Question 1

Show that in a right angled triangle, the hypotenuse is the longest side.

Solution 1

Let us consider a right angled triangle ABC, right angle at B. 
In ABC
A + B + C = 180o            (angle sum property of a triangle) 
A + 90o + C = 180o
A + C = 90o
Hence, the other two angles have to be acute (i.e. less than 90�).  [In any triangle, the side opposite to the larger (greater) angle is longer]
So, AC is the largest side in ABC.
But AC is the hypotenuse of ABC. Therefore, hypotenuse is the longest side in a right angled triangle.

Question 2

In the given figure sides AB and AC of ABC are extended to points P and Q respectively. Also, PBC  QCB. Show that AC > AB.

Solution 2

In the given figure,
ABC + PBC = 180p            (linear pair)
 ABC = 180o - PBC             ... (1)
Also,
ACB + QCB = 180o
ACB = 180o - QCB                    ... (2)
As PBC < QCB
  180� - PBC > 180o - QCB.
 ABC > ACB                [From equations (1) and (2)]
 AC > AB                           (side opposite to larger angle is larger)

Question 3

In the given figure, B < A and C < D. Show that AD < BC.

Solution 3

In AOB
B < A
 AO < BO     (side opposite to smaller angle is smaller)        ... (1)
Now in COD
C < D
 OD < OC     (side opposite to smaller angle is smaller)        ... (2)
On adding equations (1) and (2), we have
AO + OD < BO + OC
AD < BC

Question 4

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that A > C and B > D.

Solution 4

Let us join AC.
In ABC
AB < BC           (AB is smallest side of quadrilateral ABCD)  (1) In ADC
AD < CD          (CD is the largest side of quadrilateral ABCD)  (2) On adding equations (1) and (2), we have
2 + 4 < 1 + 3
 C < A
 A > C
Let us join BD.  In ABD
AB < AD            (AB is smallest side of quadrilateral ABCD)  (3) In BDC
 BC < CD         (CD is the largest side of quadrilateral ABCD)  On adding equations (3) and (4), we have
8 + 7 < 5 + 6
 D < B
 B > D

Question 5

In the given figure, PR > PQ and PS bisects QPR. Prove that PSR >PSQ.

Solution 5

  As PR > PQ    PS is the bisector of QPR       

Question 6

Show that of all line segments drawn form a given point not on it, the perpendicular line segment is the shortest.

Solution 6

  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 PNM
N = 90o
Now, P + N + M = 180o    (Angle sum property of a triangle)    
P + M = 90o             
Clearly, M is an acute angle 

NCERT Solution for Class 9 Mathematics Chapter 7 - Triangles Page/Excercise 7.5

Question 1

ABC is a triangle. Locate a point in the interior of ABC which is equidistant from all the vertices of ABC.

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.As here in ABC 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 ABC.

Question 2

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

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.Here in ABC 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 ABC.

Question 3

In a huge park people are concentrated at three points (see the given figure)      A:    where there are different slides and swings for children,
B:    near which a man-made lake is situated, C:    which is near to a large parking and exit.   Where should an icecream parlour be set up so that maximum number of persons can approach it?

 (Hint: The parlor should be equidistant from A, B and C)

Solution 3

Ice-cream parlour should be set up at the circumcentre O of ABC.    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.

Question 4

Complete the hexagonal and star shaped Rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?  

Solution 4

We may observe that hexagonal shaped rangoly is having 6 equilateral triangles in it.Area of OAB =  (side)2 =  (5)2