SChand CLASS 9 Chapter 8 Triangles Exercise 8(B)

 Exercise 8(B)

Question 1

The vertical angle of an isosceles triangle is 90°. Find each of its base angles.

Sol :

Sum of angles of a triangle=180°

Vertical Angles of an isosceles triangle=90°

So sum of its base angles =180°-90°=90°

If Base angle are equal to each other

So Each angle will be $=\frac{90^{\circ}}{2}=45^{\circ}$

Hence each base angle =45°

Question 2

Prove that each angle of an equilateral triangle is 60°. Hence show that every equilateral triangle is equiangular. What is the measure of each of the exterior angle of an equilateral triangle ?

Sol :

Given : An equilateral ΔABC

To prove : 

(i) Each angle is of 60°

(ii) What is each exterior angle

Proof : ΔABC

If AB=AC

So , ∠C=∠B

Similarly, AC=CB

So, ∠B=∠A

From (i) and (ii)

∠A=∠B=∠C

But ∠A+∠B+∠C=180°

So, ∠A=∠B=∠C$=\frac{180^{\circ}}{3}=60^{\circ}$

Hence proved

If each equilateral triangle has 60° each

So , every equilateral triangle is equiangular 

If ∠ABC=60°

and ∠ABC+∠CBZ=180°

60°+∠CBZ=180°

∠CBZ=180°-60°=120°

So, Each exterior angle of an equilateral triangle measures 120°

Question 3

Prove that if the base of an isosceles triangle is produced at both ends, the exterior angles so formed are equal to each other?

Sol :

ABC is an isosceles triangle in which AB=AC. BC is produced to both sides to D and E forming exterior angles ACD and ABE respectively.

To prove : ∠ACD=∠ABE

Proof : In ΔABC 

If AB=AC

So ∠C=∠B

or ∠ACB=∠ABC

But ∠ABE+∠ABC=180° [Linear pair....(i)]

Similarly ∠ACD+∠ACB=180°...(ii)

From (i) and (ii)

∠ABE+∠ABC=∠ACD+∠ACB

But ∠ABC=∠ACB (proved)

∠ABE=∠ACD or ∠ACD=∠ABE

Hence proved

Question 4

In a ∆ABC, AD bisects ∠BAC and AD = DC. If ∠BDA = 70°, calculate ∠ACD and ∠ABD.

Sol :

ΔABC , AD is the bisector of ∠BAC which meets BC at AD AD=DC and ∠BDA=70°

If AD is the bisector of ∠BAC

So, ∠1=∠2

In ΔADC

If AD=DC

So, ∠2=∠C

But exterior ∠ADB=∠2+∠C

70°=∠2+∠C=∠C+∠C=2∠C

So ∠C$=\frac{70^{\circ}}{2}$=35°

In ΔABD

∠1+∠B+∠ADB=180°

∠2+∠B+70°=180°

∠C+∠B+70°=180°

35°+∠B+70°=180°

B=180°-105°=75°

Hence ∠ACD=35°

and ∠ABD=75°

Question 5

What are the measures of the angles of an isosceles triangle in which each of the base angle is:

(i) Double the vertical angle ?

(ii) Thrice the vertical angle ?

Sol :

(i) Vertical angle of an isosceles triangle=x

Then each equal angle=2x

So, x+2x+2x=180°

5x=180°

$x=\frac{180^{\circ}}{5}=36^{\circ}$

So, vertical angle=36°

and each of base angles=36°×2°=72°

Hence angles are 36°, 72°, 72°

(ii) Vertical angle of an isosceles triangle=x

Then each base angle =3x

x+3x+3x=180°

7x=180°

$x=\frac{180^{\circ}}{7}$

So vertical angle$=\frac{180^{\circ}}{7}=25 \frac{5^{\circ}}{7}$

and each base angle$=\frac{180^{\circ}}{7} \times 3=\frac{540^{\circ}}{7}$

$=77 \frac{1^{\circ}}{7}$

So angles are $25 \frac{5^{\circ}}{7}, 77 \frac{1}{7}$ and $77 \frac{1^{\circ}}{7}$

Question 6

Find the lettered angles in each of the following figures :

Sol :

(i) 

If AB||CD BC is its transversal 

So, ∠y=∠1

In ΔECD

Ext. ∠x=∠C+∠D=∠1+40°

In ΔACE

AE=AC

So ∠ACE=∠AEC=x

If AB||CD

So, ∠BAC+∠ACB=180°

84°+x+∠1=180°

x+∠1=180°-84°=96°

∠1+40°+∠1=96°

2∠1=96°-40°=56°

So, ∠1$=\frac{56^{\circ}}{2}=28^{\circ}$

∠y=28°

But x=∠1+40°=28°+40°=68°

Hence x=63°, y=23°

(ii) 

In the figure in ΔABC

AB=AC

So, ∠B=∠ACB=2x

But ∠BAC+∠ABC+∠ACB=180°

x+2x+2x=180°

5x=180°

$x=\frac{180^{\circ}}{5}=36^{\circ}$

If AB||CD

So ∠ACD=∠BAC

=x=36°

Now in ΔACD

∠CAD+∠ACD+∠ADC=180°

y+x+88°=180°

y+36°+88°=180°

y+124°=180°

y=180°-124°=56°

Hence ,x=36°,  y=56°

(iii) 

In the figure

AB=BD=DC

∠BDC=180°

In ΔBCD

If BD=DC

So, ∠DCB=∠DBC=x

But ∠DCB+∠DBC+∠BDC=180°

x+x+108°=180°

2x=180°-108°=72°

$x=\frac{72^{\circ}}{2}=36^{\circ}$

In ΔABD

BA=BD

But ∠BDA=180°-108°=72°

So, ∠BAD=∠BDA=72°

and ext.∠BDC=∠BAD+∠ABD

108°=72°+y²

Hence x=36° and y=36°

(iv) 

In ΔABC, AB=AC

and BC is produced to D such that AC=CD

AD is joined 

In ΔACD

If AC=CD

So ∠CAD=∠CDA=2x

and ext.∠ACB=∠CAD+∠CDA=2x+2x=4x

If In ΔABC, AB=AC

So ∠ABC=∠ACB=4x

So but ∠BAC+∠ABC+∠ACB=180

4x+4x+x=180°

9x=180°

$x=\frac{180^{\circ}}{9}=20^{\circ}$

Hence x=20°

Question 7

In figure, LM = LN, ∠FLN = 110°.

Calculate:

(i) ∠LMN

(ii) ∠MLN.

Sol :
LM=LN

∠FN=110° , ∠F=∠Q=90°

In quad. FLNQ
∠F+∠Q+∠QNL+∠FLN=360°

90°+90°+∠QN+110°=360°

290°+∠QNL=360°
∠QNL=360°-290°=70°

In ΔLMN

If LM=LN

(i) So ∠LMN=∠LNM or ∠LNQ=70°

(ii) ∠LMN=∠MLN=180°-∠LMN+∠LNM

=180°-70°-70°=180°-140°=40°

Hence ∠QNL=70° and ∠MLN=40°

Question 8

In figure, O is the centre of the circular arc ABC. Find the angles of AABC.

Sol :

OABC is a sector

∠AOB=40° and ∠BOC=30°

AB.BC and CA are joined

If OA=OB=OC

So, In ΔAOB

∠OAB=∠OBA and ∠OBC=∠OCB

But ∠AOB=40° and ∠BOC=30°

So, ∠OAB=∠OBA$=\frac{180^{\circ}-40^{\circ}}{2}=\frac{140^{\circ}}{2}=70^{\circ}$

and ∠OBC=∠OCB$=\frac{180^{\circ}-30^{\circ}}{2}=\frac{150^{\circ}}{2}=75^{\circ}$

So, 

∠ABC=70°+75°=145°

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle

$\angle ACB=\frac{1}{2} \quad \angle A O B=\frac{1}{2} \times 40^{\circ}=20^{\circ}$

Similarly $\angle C A B=\frac{1}{2} \angle B O C=\frac{1}{2} \times 30^{\circ}=15^{\circ}$

Hence angles of ΔABC are $15^{\circ} 120^{\circ}$ and $1+5^{\circ}$

Question 9

In figure, ABC is an isosceles triangle with AB equal to AC. AC is produced to point D and CE is drawn parallel to BA. If ∠CBA = 52°, find ∠DCE.

Sol :

In the figure ΔABC in which AB=AC, CF||BA and AC is produced to D ∠ABC=52°

Now we have to find ∠DCE In ΔABC , AB=AC

∠ACB=∠ABC=52°

If AB||EC

If ∠A=∠ACE

But ∠A=180°-(∠ABC+∠ACB)

=180°-(52°+52°)=180°-104°=76°

So, ∠ACE=∠A=76°

But ∠ACE+∠ECD=180°

=180°-(52°+52°)=180°-104°=76°

So, ∠ACE=∠A=76°

But ∠ACE+∠ECD=180°

76°+∠ECD=180°

∠ECD=180°-76°=104°

or ∠DCE=104°

Question 10

In figure, it is given that AB = AC and DA is parallel to BC.

∠DAB = 70°. Find ∠BAC.

Sol :

In the figure ΔABC, AB=AC and AD||BC is drawn ∠DAB=70°

We have to find ∠BAC

If DA||BC

So, ∠ABC=∠DAB

But in ΔABC, AB=AC

So, ∠ABC=∠ACB

So, ∠ACB=∠ABC=70°

Now ∠BAC+∠ABC+∠ACB=180°

∠BAC+70°+70°=180°

∠BAC=140°=180°

∠BAC=180°-140°=140°

Question 11

In figure, angles ACB is a right angle, AC = CD, CDEF is a rectangle, and ∠BAC = 50°.

Sol :

∠ACB is a right angle 

AC=CD, CDEF is a rectangle and ∠BAC=50°

Join DF

We have to find 

(i) ∠BDE

(ii) The angle between the diagonals CE and DF or rectangle CDEF

In ΔADC, AC=CD

So ∠CDA=∠DAC or ∠BAC

∠CDA=50°

(i) Now ∠CDA+∠CDE+∠BDE=180°

50°+90°+∠BDE=180°

140°+∠BDE=180°

∠BDE=180°-140°=40°

(ii) So DE=DB

In ΔDAC

∠ACD=180°-(∠A+∠ADC)

=180°-(50°+50°)

=180°-100°=80°

So ∠DCO=∠ACB-∠ACD=90°-80°=10°

In ΔOCD

If OC=OD

So, ∠ODC=∠OCD

So, ∠COD=180°-(10°+10°)

=180°-20°=160°

Hence ∠BDE=40° and ∠COD=160°

Question 12

ABC is a triangle. The bisector of the angle BCA meets AB in X. A point Y lies on CX such that AX = AY. Prove that ∠CAY = ∠ABC.

Sol :

In ΔABC, bisector of ∠BCA meets AB in X. A point Y is on CX. Such that AX=AY

To prove: ∠CAY=∠ABC

Proof : In ΔABC

If CX is the bisector of ∠C

So, ∠1=∠2

In  ΔAXY

If AX=AY

So, ∠3=∠4

In ΔBCX

Ext.∠4=∠2+∠5

=∠1+∠5....(i)

Similarly in ΔCAY

Ext.∠3=∠1+∠6....(ii)

From (i) and (ii)

If ∠3=∠4

So, ∠1+∠5=∠1+∠6

∠5=∠6

∠CAY=∠ABC

Hence proved

Question 13

In figure, PS = PR, ∠TPS = ∠QPR. Prove that PT = PQ.

Sol :

In the figure,

PS=PR , ∠TPS=∠QPR

To prove : PT=PQ

Proof : In ΔPRS

If PS=PR

So, ∠PSR=∠PRS

But ∠PSR+∠PST=180°

Similarly ∠PRS+∠PRQ=180°

So, ∠PRS+∠PST=∠PSR+∠PRT

But ∠PSR=∠PRS

Now in ΔPST and ΔPRQ

PS=PR

∠PST=∠PRQ

∠TPS=∠QPR

So, ΔPST≅ΔPRQ

So PT=PQ 

Hence proved

Question 14

ABC is an isosceles triangle with AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE.

Sol :

In an isosceles ΔABC, AB=AC, BD and CF are the medians of sides AC and AB respectively.

To prove : BD=CE

Proof : In ΔBDC and ΔBEC

BC=BC

∠C=∠B

So ΔBDC≅ΔBEC

So BD=CE

Hence proved

Question 15

Prove that the triangle formed by joining the midpoints of an isosceles triangles is also an isosceles triangle.

Sol :

In an isosceles ΔABC in which AB=AC, D, E and F are the mid-point of the sides BC, CA and AB respectively. DE, EF and FD are joined

To prove : ΔDEF is an isosceles triangle

Proof : If D and F are the mid-points of BC and BA respectively

So, DF||AC and $\frac{1}{2}$AC

Similarly D and E are the mid-point of BC and CA respectively.

So, DE||AB and $\frac{1}{2}$AB

But AB=AC

So DE=DF

So ΔDEF is an isosceles triangle. 

Hence proved

Question 16

In the figure, AD = BC, AC = BD. Prove that PAB is an isosceles triangle.

Sol :

In the figure AD=BC, AC=BD prove that PAB is an isosceles triangle.

Given : In the figure AD=BC, AC=BD

To prove : ΔPAB is an isosceles triangle.

Proof : In ΔADB and ΔACB

AB=AB

BD=AC

AD=BC

So, ΔADB≅ΔACB

So, ∠D=∠C

Now in ΔAPD and ΔBPC

AD=BC

∠APD=∠BPC

∠D=∠C

So, ΔAPD≅ΔBPC

So AP=BP 

So ΔPAB is an isosceles triangle 

Hence proved

Question 17

In the figure, (i) and (ii), AB = AC and BD = DC. Prove that ∠ABD = ∠ACD.

Sol :

In the figure (i) and (ii) AB=AC=DC

To prove : ∠ABD=∠ACD

Proof : ΔABD and ΔACD

AB=AC

DB=DC

AD=AD

So ΔABD≅ΔACD

So ∠ABD=∠ACD

Hence proved

Question 18

In figure, ABC is an isosceles triangle in which AB = AC. The side BA is produced to D such that AB = AD. Prove that ∠BCD is a right angle.

Sol :

In ΔABC , AB=AC

BA is produced to D such that BA=AD. DC is joined 

To prove : ∠BCD=⏊ right angle

Proof : In ΔABC

If AB=AC

So, ∠ABC=∠ACB

Similarly in ΔACD

AB=AD

So, ∠1=∠2

Now adding 

∠ABC+∠1=∠ACB+∠2

∠ABC+∠1=∠BCD

 

But ∠ABC+∠1+∠BCD=180°

So, ∠BCD+∠BCD=180°

2∠BCD=180°

So ∠BCD=90° which means ∠BCD is right angle

Hence proved

Question 19

In figure, ABC is an isosceles triangle in which AB = AC. The side BA is produced to D such that AB = AD. Prove that ∠BCD is a right angle.

Sol :

In ΔABC, AB=AC Bisector of ∠B and ∠C meets AC and AB is D and E respectively.

To prove : BD=CE

Proof : In ΔBDC≅ΔBEC

BC=BC

∠C=∠B

∠DBC=∠ECB

So, ΔBDC≅ΔBEC

So, BD=CE

Question 20

If the bisector of the vertical angle of a triangle bisects the base also, the triangle is isosceles.

Sol :

In ΔABC the bisector of vertical angle A bisects the side BC at D

So, x₁=x₂ and BC=DC

To prove : ΔABC is an isosceles 

i.e. AB=AC

Construction : Produce AD to E such that AD=DE

Join EC

Proof : In ΔABD and ΔECD

AD=DE

BD=DC

∠ADB=∠CDE

So ΔABD≅ΔECD

So AB=CE

and  ∠x₁=∠x₃

But ∠x₁=∠x₂

So ∠x₂=∠x₃

So AC=CE

But AB=CE

So AB=AC

Hence ΔABC is an isosceles triangle