NCERT Solutions For Class 9 Math Chapter – 7 Exercise – 7.2

NCERT Solutions For Class 9 Math Chapter – 7 Exercise – 7.2

 

Q1. 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:

• OB = OC      (ii) AO bisects ∟A.

Solution:-

 

 

 

 

 

 

 

In ∆ABC , AB = AC

∟ABC = ∟ACB  [ Angles opposite to equal sides are equal].

= 1/2 ∟ABC = 1/2∟ACB

∟OBC = ∟OCB [ OB and OC are bisectors of ∟ABC and ∟ACB respectively.]

OC = OB [Sides opposite to equal angles of a triangle are equal.]

• Consider triangles AOB and AOC,

OC = OB [From (I)]

AO = OA [Common]

∆AOB  ∆AOC [SSS rule]

∟OAB = ∟OAC [CPCT]

= AO bisects ∟BAC.

 

Q2. In ∆ABC , AD is the perpendicular bisector of BC Show that ∆ABC is an isosceles triangle in which AB = AC.

Solution:-

In triangles ADB and ADC,

BD = BC

∟ADB = ∟ADC

AD = AD

∆ADB  ∆ADC [SAS Rule]

AB = AC.

Therefore ∆ABC is an isosceles triangle with AB = AC. [CPCT]

 

Q3. ABC is an isosceles  triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively Show that these altitudes are equal.

Solution:-

In ∆BFC and BCE.

∟FBC = ∟ECB.

BC = CB

∟BFC = ∟CEB

∆BCF  ∆CBE.

CF = BE.

 

Q4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal Show that:

• ∆ABE ∆ACF
• AB = AC ABC is an isosceles triangle.

Solution:-

 

 

 

 

 

 

• In triangle ABE and ACF,

BE = CF

∟A = ∟A

∟AEB = ∟AFC

∆ABE  ∆ACF [AAS Rule]

 

• Since ∆ABE ∆ACF

Hence AB = AC

∆ABC is a isosceles.

 

Q5. ABC and DBC are two isosceles triangles on same base BC Show that ∟ABD = ∟ACD

Solution:-

In triangle ABD and ACD,

AB = AC

BD = CD

AD = AD

∆ABD  ∆ACD

∟ABD = ∟ACD

 

Q6. ∆ABC is an isosceles triangle in which AB = AC Side BA is produced to D Such that AD = AB Show that ∟BCD is a right angle.

Solution:-

AB = AC  [Given]

∟1 = ∟2  (I)

[Angles opposite to equal sides of a triangle are equal.]

AB = AD (Given)

And

AC = AD

∟3 = ∟4

[Angles opposite to equal sides of a triangle are equal.]

In ∆ DBC,

∟DBC + ∟BCD + ∟CDB = 180⁰

[ Sum of angles of a triangle is 180⁰]

= ∟1 + ∟2 + ∟3 + ∟4 = 180⁰

= ∟2 + ∟2 + ∟3 + ∟3 = 180⁰

2(∟2 + ∟3) = 180⁰

∟2 + ∟3 = 90⁰

 

Q7. ABC is a right angled triangle in which ∟A = 90⁰ and AB = AC Find ∟B and ∟C.

Solution:-

 

 

 

 

 

As AB = AC

∟B = ∟C     [ Angels opposite to equal sides of a triangle are equal].

In ∆ABC ,

∟A + ∟B + ∟C = 180⁰

2∟B = 90⁰

∟B = 45⁰

∟B = ∟C = 45⁰

 

Q8. Show that the angels of an equilateral triangle are 60⁰ each.

Solution:-

In ∆ABC is equilateral

AB = BC = CA

AB = BC

∟A = ∟C

[ Angles opposite to equal sides of a triangle are equal].

BC = AC = ∟A = ∟B

∟A =  ∟B = ∟C

In ∆ABC  ,

∟A + ∟B + ∟C = 180⁰

[ Sum of angels of a triangle is 180⁰.]

= ∟A + ∟A + ∟A = 180⁰

3∟A = 180⁰

∟A = 60⁰

∟A = ∟B = ∟C = 60⁰