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. Here you can find the NCERT Solutions for Chapter 7 Triangles Ex 7.2.

• NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1
• NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2
• NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3
• NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4
• NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5

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

 

Ex 7.2 Class 9 Maths Question 2.
In ΔABC, 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 ΔADB and ΔADC,
AD = AD                       [Common]
BD = DC                       [Given]
and ∠ADB = ∠ADC     [Each= 90°]
ΔADB ≅ ΔADC            [By SAS congruence rule]
⇒ AB = AC                   [By CPCT]
So, ΔABC 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: ΔABC is an isosceles triangle in which AB = AC, BE ⊥ AC and CF ⊥ AB
To prove: BE = CF
Proof: In ΔABE and ΔACF,
AB = AC                [given]
∠AEB = ∠AFC      [each 90 °, BE ⊥ AC and CF ⊥ AB]
∠BAE = ∠CAF      [common angle]
∴ ΔABE ≅ ΔACF   [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 ≅ ΔACF
(ii) AB = AC, i.e., ΔABC is an isosceles triangle.

Solution:
Given: ΔABC in which BE ⊥ AC and CF⊥ AB, such that BE = CF.
To prove: (i) ΔABE ≅ ΔACF
(ii) AB = AC
Proof: (i) In ΔABC and ΔACF
∠AEB= ∠AFC                      [each 90 °]
∠BAE = ∠CAF                     [common angle]

and BE = CF                        [Given]
ΔABE ≅ ΔACF                     [by AAS congruence rule]

(ii) From part (i),  ΔABE ≅ ΔACF
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 ΔABC, 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 ΔABC, AB = AC                      (Given)
⇒ ∠ACB = ∠ABC          ...(i)      (∵ Angles opposite to  equal sides are equal)
Now, AB = AD                            (Given)
∴ AC = AD                                    (∵ AB = AC)
Now, in ΔADC, 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
⇒ 2∠BCD = 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, ΔABC 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)]
⇒ 2∠B = 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: ΔABC 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.

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