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Class 9 NCERT Solutions 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.

Ex. 7.1

Ex. 7.2

Ex. 7.3

Ex. 7.4

Ex. 7.5

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 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

In

ADC and

ADB
AD = AD (Common)

ADC =

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

Solution 3

In

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

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

ACB +

ACB +

ACD +

ACD = 180^o

2(

ACB +

ACD) = 180^o

2(

BCD) = 180^o

BCD = 90^o

Solution 7

Given that

AB = AC

C =

B (angles opposite to equal sides are also equal)

In

ABC,

A +

B +

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

90^o +

B +

C = 180^o

90^o +

B +

B = 180^o

2

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 +

A = 180^o

3

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

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

(common)

B =

C (by using CPCT)

Triangles Exercise Ex. 7.4

Solution 1

Let us consider a right angled triangle ABC, right angle at B.
In

ABC

A +

B +

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

A + 90^o +

C = 180^o

A +

C = 90^o
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.

Solution 2

In the given figure,

ABC +

PBC = 180^p (linear pair)

ABC = 180^o -

PBC ... (1)
Also,

ACB +

QCB = 180^o

ACB = 180^o -

QCB ... (2)
As

PBC <

QCB

180 -

PBC > 180^o -

QCB.

ABC >

ACB [From equations (1) and (2)]

AC > AB (side opposite to larger angle is larger)

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

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

Solution 5

As PR > PQ

PS is the bisector of

QPR

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 = 90^o
Now,

P +

N +

M = 180^o (Angle sum property of a triangle)

P +

M = 90^o
Clearly,

M is an acute angle

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.

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.

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.

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.

Solution 4

We may observe that hexagonal shaped rangoly is having 6 equilateral triangles in it.

Area of

OAB =

(side)² =

(5)²

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