NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2 are the part of NCERT Solutions for Class 9 Maths. In this post, you will find the NCERT Solutions for Chapter 7 Triangles Ex 7.2.



NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2


Ex 7.2 Class 9 Maths 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:


(i) In Î”ABC, AB = AC            (Given)
B = C                           (Angles opposite to equal sides are equal)
½
B = ½ C
OBC = OCB
and 
OBA = OCA       …(i)  ( OB and OC are bisectors of B and C, respectively)
 OBC = OCB
OB = OC                     …(ii)  ( Sides opposite to equal angles are equal)

(ii) In Î”ABO and Î”ACO,
AB = AC                                 (Given)
OBA = OCA                     [From eq. (i)]
OB = OC                           [From eq. (ii)]
ΔABO
 Î”ACO                     (By SAS congruence rule)
⇒ ∠BAO = CAO                (By CPCT)
AO is the bisector of BAC.

 

Ex 7.2 Class 9 Maths Question 2.
In Î”ABC, AD is the perpendicular bisector of BC (see figure). Show that ΔABC is an isosceles triangle in which AB = AC.

Solution:
Given:
AD is the perpendicular bisector of BC.
To prove: ΔABC is an isosceles triangle, i.e.,
AB = AC
Proof: In Î”ADB and Î”ADC,
AD = AD                       [Common]
BD = DC                       [Given]
and 
ADB = ADC     [Each= 90°]
ΔADB 
 Î”ADC            [By SAS congruence rule]
AB = AC                   [By CPCT]
So, Î”ABC is an isosceles triangle.

 

Ex 7.2 Class 9 Maths Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB, respectively (see figure). Show that these altitudes are equal.


Solution:

Given: Î”ABC is an isosceles triangle in which AB = AC, BE  AC and CF  AB
To prove: BE = CF
Proof: In Î”ABE and Î”ACF,
AB = AC                [given]
AEB = AFC      [each 90 °, BE AC and CF AB]
BAE = CAF      [common angle]
ΔABE  Î”ACF   [by AAS congruence rule]

Then BE = CF       [by CPCT]                Hence proved.

 

Ex 7.2 Class 9 Maths Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that


(i) ΔABE  Î”ACF
(ii) AB = AC, i.e., Î”ABC is an isosceles triangle.

Solution:
Given:
 Î”ABC in which BE AC and CF AB, such that BE = CF.
To prove: (i) Î”ABE 
 Î”ACF
(ii) AB = AC
Proof: (i) In Î”ABC and Î”ACF
AEB= AFC                      [each 90 °]
BAE = CAF                     [common angle]

and BE = CF                        [Given]
ΔABE 
 Î”ACF                     [by AAS congruence rule]

(ii) From part (i),  ΔABE 
 Î”ACF
AB = AC                              [by CPCT]

Therefore, ΔABC is an isosceles triangle.

 

Ex 7.2 Class 9 Maths Question 5.
ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that ABD = ACD.


Solution:
In Î”ABC, we have
AB = AC
ABC = ACB  …(i)       [ angles opposite to equal sides are equal]
In ΔDBC, we have
BD = CD
DBC = DCB  …(ii)      [ angles opposite to equal sides are equal]
Adding equations. (i) and (ii), we get
ABC + DBC = ACB + DCB
ABD = ACD               Hence, proved.

 

Ex 7.2 Class 9 Maths Question 6.
ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that BCD is a right angle.

Solution:


In Î”ABC, AB = AC                      (Given)
ACB = ABC          ...(i)      (Angles opposite to  equal sides are equal)
Now, AB = AD                            (Given)
AC = AD                                    ( AB = AC)
Now, in Î”ADC, we have
AD = AC                                        (from above)
ACD = ADC             ...(ii)    ( Angles opposite to equal sides are equal)
On adding equations. (i) and (ii), we have

ACB + ACD = ABC + ADC
BCD = ABC + BDC            ( ADC = BDC)
Adding 
BCD on both sides, we have
BCD + BCD = ABC + BDC + BCD
2BCD = 180°                          (By angle sum property)
BCD = 90°

 

Ex 7.2 Class 9 Maths Question 7.
ABC is a right-angled triangle in which A = 90° and AB = AC. Find B and C.

Solution:
Given, Î”ABC is a right-angled triangle in which A = 90° and AB = AC.
Then, 
C = B  …(i)     [since, angles opposite to equal sides of a triangle are equal]
Now, 
A + B + C = 180°[since, sum of three angles of a triangle is 180°]
90 °+ B + B = 180°     [from eq. (i)]
2B = 90°                        [ B = 45°]
Hence, 
B = 45° and C = 45°

 

Ex 7.2 Class 9 Maths Question 8.
Show that the angles of an equilateral triangle are 60° each.

Solution:

Given: Î”ABC is an equilateral triangle.


To prove:  A = B = C = 60°
Proof: Since, all the three angles of an equilateral triangle are equal,

i.e., A = B = C    …(i)
 A + B + C = 180°                 [since, sum of three angles of a triangle is 180°]
A + A + A = 180°                [from eq. (i)]
3
A = 180°
A = 180/3
A = 60°
Hence, 
A = B = C = 60°
Hence proved.


Related Links:

NCERT Solutions for Maths Class 10

NCERT Solutions for Maths Class 11

NCERT Solutions for Maths Class 12

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