# 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.

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## Chapter 7 - Triangles Exercise Ex. 7.1

Solution 1
In ABC and ABD
CAB = DAB                                 (given)
AB = AB                                             (common)

So, BC and BD are of equal length.

Solution 2
In ABD and BAC
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)
Solution 4
Solution 5
Solution 6
BAD + DAC = EAC + DAC
BAC = DAE
Now in BAC and DAE
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
DBC = ACB                                 (each 90o )
BC = CB                                             (Common)
(iv) We have DBC  ACB

## Chapter 7 - 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
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)

AEB  AFC
AB = AC                                      (by CPCT)

Solution 5

In ABD and ACD
AB = AC                                    (Given)
BD = CD                                    (Given)
Solution 6
In ABC
AB = AC                                                (given)
ACB = ABC                               (angles opposite to equal sides of a triangle are also equal)
Now In ACD
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.

## Chapter 7 - Triangles Exercise Ex. 7.3

Solution 1
(i)  In ABD and ACD
AB = AC                                             (given)
BD = CD                                            (given)

(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
AB = AC                                                  (given)

(ii)     Also by CPCT,

(ii)              Also by CPCT,

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)

## Chapter 7 - Triangles Exercise Ex. 7.4

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.

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)
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

Solution 4
Let us join AC.
In ABC
AB < BC           (AB is smallest side of quadrilateral ABCD)
(1)
(2)
On adding equations (1) and (2), we have
2 + 4 < 1 + 3
C < A
A > C
Let us join BD.

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

## 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.
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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