Class 9 NCERT Solutions Maths Chapter 7: Triangles

Ex. 7.1

Ex. 7.2

Ex. 7.3

Triangles Exercise Ex. 7.1

Solution 1

ABC and

ABD
AC = AD (given)

CAB =

DAB (given)
AB = AB (common)

So, BC and BD are of equal length.

Solution 2

ABD and

BAC
AD = BC (given)

DAB =

CBA (given)
AB = BA (common)

And

ABD =

BAC (by CPCT)

Solution 3

BOC and

AOD

BOC =

AOD (vertically opposite angles)

CBO =

DAO (each 90^o)
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 = 180^o (co-interior angles)

DBC + 90^o = 180^o

DBC + 90^o= 180⁰

DBC = 90^o

(iii) Now in

DBC and

ACB
DB = AC (Already proved)

DBC =

ACB (each 90^o )
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

ADC and

ADB
AD = AD (Common)

ADC =

ADB (each 90^o)
CD = BD (AD is the perpendicular bisector of BC)

Solution 3

AEB and

AFC

AEB =

AFC (each 90^o)

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

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 = 180^o (angle sum property of a triangle)

ACB +

ACD +

ACD = 180^o

ACB +

ACD) = 180^o

BCD) = 180^o

BCD = 90^o

Solution 7

Given that

AB = AC

C =

B (angles opposite to equal sides are also equal)

ABC,

A +

B +

C = 180^o (angle sum property of a triangle)

90^o +

B +

C = 180^o

90^o +

B +

B = 180^o

B = 90^o

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 = 180^o

A +

A = 180^o

A = 60^o

A =

B =

C = 60^o
Hence, in an equilateral triangle all interior angles are of 60^o.

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 = 180^o (linear pair angles)

BPD + BPD = 180^o

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

BPD = 90^o ...(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 90^o as AD is an altitude)
       AB = AC                                            (given)
       AD = AD                                                  (common)

(ii) Also by CPCT,

BAD =

CAD
Hence, AD bisects

(ii) Also by CPCT,

ÐBAD = ÐCAD

Hence, AD bisects ÐA.

Solution 3

(i) In

ABC, AM is median to BC

BM =

PQR, PN is median to QR

QN =

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

BEC and

CFB

BEC =

CFB                                              (each 90^o )
BC = CB                                                         (common)
BE = CF                                                         (given)

(Sides opposite to equal angles of a triangle are equal)

Hence,

ABC is isosceles.

Solution 5

APB and

APC

APB =

APC                (each 90^o)
AB =AC                            (given)
AP = AP

(common)

B =

C (by using CPCT)

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Class 9 NCERT Solutions Maths Chapter 7: Triangles

Triangles Exercise Ex. 7.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Triangles Exercise Ex. 7.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Triangles Exercise Ex. 7.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

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