Practicing OP Malhotra Class 9 Maths Solutions Chapter 8 Triangles Ex 8(B) is the ultimate need for students who intend to score good marks in examinations.

## S Chand Class 9 ICSE Maths Solutions Chapter 8 Triangles Ex 8(B)

Question 1.

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

Solution:

Sum of angles of a triangle = 180°

Vertical angle of an isosceles triangle = 90°

∴ Sum of its base angles = 180° – 90° = 90°

∵ Base angles are equal to each other

∴ Each angle will be = \(\frac { 90° }{ 2 }\) = 45°

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 ?

Solution:

Given : An equilateral ∆ABC

To Prove:

(i) Each angle is of 60°

(ii) What is its each exterior angle.

Proof: In ∆ABC,

∵ AB = AC (sides of equilateral triangle)

∴ ∠C = ∠B (angles opposite to equal sides) … (i)

Similarly, AC = CB

∴ ∠B = ∠A … (ii)

From (i) and (ii)

∠A = ∠B = ∠C

But ∠A + ∠B + ∠C = 180°

(sum of angles of a triangle)

∴ ∠A = ∠B = ∠C = \(\frac { 180° }{ 3 }\) = 60°

Hence proved.

∵ Each equilateral triangle has 60° each

∴ Every equilateral triangle is equiangular

∵ ∠ABC = 60°

and ∠ABC + ∠CBZ = 180° (linear pair)

⇒ 60° + ∠CBZ = 180° ⇒ ∠CBZ =180°- 60° = 120°

∴ 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?

Solution:

Given : 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 ∆ABC,

∵ AB = AC (given)

∴ ∠C = ∠B (angles opposite to equal sides)

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.

Solution:

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

∵ AD is the bisector of ∠BAC

∴ ∠1 = ∠2

In ∆ADC,

∵ AD = DC (given)

∴ ∠2 = ∠C

But Ext. ∠ADB = ∠2 + ∠C

⇒ 70° = ∠2 + ∠C = ∠C + ∠C = 2 ∠C

∴ ∠C = \(\frac { 70° }{ 2 }\) = 35°

In ∆ABD,

∠1 + ∠B + ∠ADB = 180°

(sum of angles of a triangle)

⇒ ∠2 + ∠B + 70° = 180° (∵ ∠1 = ∠2 = ∠C)

⇒ ∠C + ∠B + 70° = 180°

⇒ 35° + ∠B + 70° = 180°

⇒ ∠B + 105° = 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 ?

Solution:

(i) Let the vertical angle of an isosceles triangle = x

Then each equal angle = 2x

∴ x + 2x + 2x = 180°

(sum of angles of a triangle)

⇒ 5x = 180°

⇒ x = \(\frac { 180° }{ 5 }\) = 36°

∴ Vertical angle = 36°

and each of base anlges = 36° x² = 12°

Hence angles are 36°, 12°, 12°

(ii) Let vertical angle of an isosceles triangle = x

Then each base angle = 3x

∴ x + 3x + 3x = 180°

(sum of angles of a triangle)

⇒ 7x = 180°

⇒ x = \(\frac { 180° }{ 7 }\)

∴ Vertical angle = \(\frac { 180° }{ 7 }\) = 25 \(\frac { 5° }{ 7 }\)

and each base angle = \(\frac { 180° }{ 7 }\) x 3 = \(\frac { 540° }{ 7 }\)

= 77\(\frac { 1° }{ 7 }\)

∴ Angle are 25\(\frac { 5° }{ 7 }\) , 77 \(\frac { 1° }{ 7 }\) and 77 \(\frac { 1° }{ 7 }\)

Question 6.

Find the lettered angles in each of the following figures :

Solution:

(i) ∵ AB || CD and BC is its transversal

∴ ∠y = ∠1 (alternate angle)

In ∆ECD,

Ext. ∠x = ∠C + ∠D = ∠1 + 40°

In ∆ACE,

AE = AC (given)

∴ ∠ACE = ∠AEC = x

∵ AB || CD

∴ ∠BAC + ∠ACB = 180° (colinear angles)

84° + x + ∠1 = 180°

⇒ x + ∠L = 180° – 84° = 96°

⇒ ∠1 + 40° + ∠L = 96° (∵ x = ∠1 + 40°)

⇒ 2 ∠1 = 96 – 40 = 56

⇒ ∠1 = \(\frac { 56° }{ 2 }\) = 28°

⇒ ∠y = 28° (∵ ∠1 = ∠y)

But x = ∠1 + 40° = 28° + 40° = 68°

Hence x = 68°, y = 28°

(ii) In the figure in ∆ABC,

AB = AC

∴ ∠B = ∠ACB = 2x

But ∠BAC + ∠ABC + ∠ACB = 180° (angles of a triangle)

⇒ x + 2x + 2x = 180°

⇒ 5x = 180°

x = \(\frac { 180° }{ 5 }\) = 36°

∵ AB || CD

∴ ∠ACD = ∠BAC (alternate angles)

= x = 36°

Now in ∆ACD,

∠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 = 108°

In ∆BCD,

∵ BD = DC (given)

∴ ∠DCB = ∠DBC = x

But ∠DCB + ∠DBC + ∠BDC = 180°

⇒ x + x + 108° = 180°

⇒ 2x = 180°- 108° = 72°

∴ x = \(\frac { 72° }{ 2 }\) = 36°

In ∆ABD,

BA = BD (given)

∴ ∠BAD = ∠BDA

(angles opposite to equal sides)

But ∠BDA = 180° – 108° = 72°

(∵ ∠BDC + BDA = 180°)

(linear pair)

∴ ∠BAD = ∠BDA = 72°

and Ext. ∠BDC = ∠BAD + ∠ABD

⇒ 108° = 72° + y²

⇒ y = 108°- 72° = 36°

Hence x = 36° and y = 36°

(iv) In ∆ABC, AB = AC

and BC is produced to D

Such that AC = CD

AD is joined

In ∆ACD,

∵ AC = CD (given)

∴ ∠CAD = ∠CDA = 2x

and Ext. ∠ACB = ∠CAD + ∠CDA = 2x + 2x = 4x

∵ In AABC, AB = AC

∴ ∠ABC = ∠ACB = 4x

∴ But ∠BAC +∠ABC + ∠ACB = 180° (angles of a triangle)

∴ 4x + 4x + x = 180° ⇒ 9x = 180°

⇒ x = \(\frac { 180° }{ 2 }\) = 20°

Hence x = 20°

Question 7.

In figure, LM = LN, ∠FLN = 110°.

Calculate:

(i) ∠LMN

(ii) ∠MLN.

Solution:

In the figure, LM = LN

∠FLN = 110°, ∠F = ∠Q = 90°

In quad. FLNQ,

∠F + ∠Q + ∠QNL + ∠FLN = 360°

⇒ 90° + 90° + ∠QNL + 110° = 360°

⇒ 290° + ∠QNL = 360°

⇒ ∠QNL = 360° – 290° = 70°

In ∆LMN

∵ LM = LN

(i) ∴ ∠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.

Solution:

OABC is a sector

∠AOB = 40° and ∠BOC = 30°

AB, BC and CA are joined

∵ OA = OB = OC (radii of the arc ABC)

∴ In ∆AOB,

∠OAB = ∠OBA and ∠OBC = ∠OCB

(angles opposite to equal sides)

But ∠AOB = 40° and ∠BOC = 30°

∴ ∠OAB = ∠OBA = \(\frac{180^{\circ}-40^{\circ}}{2}=\frac{140^{\circ}}{2}\) = 70°

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

∴ ∠ABC = 70° + 75° = 145°

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

∴ ∠ACB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) x 40° = 20°

Similarly ∠CAB = \(\frac { 1 }{ 2 }\) ∠BOC = \(\frac { 1 }{ 2 }\) x 30° = 15°

Hence angles of ∆ABC are 15°, 20° and 145°

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.

Solution:

In the figure, ∆ABC in which AB = AC, CE || M

and AC is produced to D

∠ABC = 52°

Now we have to find ∠DCE

In AABC,

AB =AC (given)

∠ACB = ∠ABC = 52°

(angles opposite to equal sides) AB || EC (given)

∠A = ∠ACE (alternate angles)

But ∠A = 180°- (∠ABC + ∠ACB)

= 180° – (52° + 52°) = 180° – 104° = 76° ∠ACE = ∠A = 76°

But ∠ACE + ∠ECD = 180° (linear pair) ⇒ 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.

Solution:

In the figure, in ∆ABC,

AB = AC and AD || BC is drawn

∠DAB = 70°,

We have to find ∠BAC

∵ DA || BC

∴ ∠ABC = ∠DAB (alternate angles)

But in ∆ABC, AB = AC

∴ ∠ABC = ∠ACB

(angles opposite to equal sides)

∴ ∠ACB = ∠ABC = 70°

Now ∠BAC + ∠ABC + ∠ACB = 180° (angles of a triangle)

⇒ ∠BAC + 70° + 70° = 180°

⇒ ∠BAC + 140° = 180°

⇒ ∠BAC = 180° – 140° = 40°

Hence ∠BAC = 40°

Question 11.

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

(i angle BDE;

(ii) the angle between the diagonals CE, DF of the rectangle.

Solution:

∠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 of rectangle CDEF

In AADC, AC = CD (given)

∠CDA = ∠DAC or ∠BAC

(angles opposite to equal sides)

∠CDA = 50°

(i) Now ∠CDA + ∠CDE + ∠BDE = 180°

50°+ 90° +∠BDE = 180°

⇒ 140° +∠BDE = 180°

⇒ ∠BDE = 180° – 140°

⇒ ∠BDE = 40°

(ii) ∴ DE = EB

In ∆DAC,

∠ACD = 180° – (∠A + ∠ADC)

= 180° – (50° + 50°) = 180° – 100° = 80°

∴ ∠DCO = ∠ACB – ∠ACD = 90° – 80° = 10°

In ∆OCD,

∵ OC = OD

∴ ∠ODC = ∠OCD

∴ ∠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. Solution:

Given : In ∆ABC, bisector of ∠BCA meets AB in X. A point Y lies on CX such that AX = AY

To prove : ∠CAY = ∠ABC

Proof: In ∆ABC,

∵ CX is the bisector of ∠C

∴ ∠1 = ∠2

In ∆AXY,

∵ AX = AY

∴ ∠3 = ∠4

In ∆BCX

Ext. ∠4 = ∠2 + ∠5

= ∠1 + ∠5 … (i) (∵ ∠1 = ∠2)

Similarly in ∆CAY

Ext. ∠3 = ∠1 + ∠6 … (ii)

From (i) and (ii)

∠3 = ∠4

∴ ∠1 + ∠5 = ∠1 + ∠6

⇒ ∠5 = ∠6

⇒ ∠CAY = ∠ABC

Hence proved.

Question 13.

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

Solution:

Given : In the figure,

PS = PR, ∠TPS = ∠QPR

To prove : PT = PQ

Proof: In ∆PRS,

∵ PS = PR (given)

∠PSR = ∠PRS

(angles opposite to equal sides)

But ∠PSR + ∠PST = 180° (linear pair)

Similarly ∠PRS + ∠PRQ = 180°

∠PSR + ∠PST = ∠PRS + ∠PRT

But ∠PSR = ∠PRS (proved)

∴ ∠PST = ∠PRQ

Now in APST and ∆PRQ,

PS = PR (given)

∠PST = ∠PRQ (proved)

∠TPS = ∠QPR (given)

∴ ∆PST ≅ ∆PRQ (ASA axiom)

∴ PT = PQ (c.p.c.t.)

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.

Solution:

Given : In an isosceles ∆ABC, AB = AC

BD and CF are the medians of sides AC and AB respectively

To prove : BD = CE

Proof: In ∆BDC and ∆BEC

BC = BC (common)

∠C = ∠B

(base angles of an isosceles triangle)

CD = BE (half of equal sides)

∴ ∆BDC ≅ ∆BEC (SAS axiom)

∴ BD = CE (c.p.c.t.)

Hence proved.

Question 15.

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

Solution:

Given : In an isosceles ∆ABC in which AB = AC.

D, E and F are the mid-points of the sides BC, CA and AB respectively

DE, EF and FD are joined

To prove : ∆DEF is an isosceles triangle

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

∴ DF || AC and \(\frac { 1 }{ 2 }\) AC

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

∴ DE || AB and \(\frac { 1 }{ 2 }\) AB

But AB = AC (given)

∴ DE = DF

∴ ∆DEF is an isosceles triangle

Hence proved.

Question 16.

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

Solution:

Given : In the figure,

AD = BC, AC = BD

To prove : ∆PAB is an isosceles triangle

Proof: In ∆ADB and ∆ACB

AB = AB (common)

BD = AC (given)

AD = BC (given)

∴ ∆ADB ≅ ∆ACB (SSS axiom)

∴ ∠D = ∠C (c.p.c.t.)

Now in ∆APD and ∆BPC AD = BC (given)

∠APD = ∠BPC (vertically opposite angle) ∠D = ∠C (proved)

∴ ∆APD ≅ ∆BPC (AAS axiom)

∴ AP = BP (c.p.c.t.)

∴ ∆PAB is an isosceles triangle

Hence proved.

Question 17.

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

Solution:

Given : In the figure (i) and (ii)

AB = AC, BD = DC

To prove : ∠ABD = ∠ACD

Proof:

∆ABD and ∆ACD

AB = AC (given)

DB = DC (given)

AD = AD (common)

∴ ∆ABD ≅ ∆ACD (SSS axiom)

∴ ∠ABD = ∠ACD (c.p.c.t)

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.

Solution:

Given : In ∆ABC, AB = AC

BA is produced to D

Such that BA = AD

DC is joined

To prove : ∠BCD = 1 right angle

Proof: In ∆ABC,

∵ AB = AC (given)

∴ ∠ABC = ∠ACB

(angles opposite to equal sides)

Similarly in ∆ACD

AB = AD (given)

⇒ AC = AD (∵ AB = AC)

∴ ∠1 = ∠2

Now adding

∠ABC + ∠1 = ∠ACB + ∠2

⇒ ∠ABC + ∠1 = ∠BCD

But ∠ABC + ∠1 + ∠BCD = 180°

(angles of a triangle)

∴ ∠BCD + ∠BCD = 180°

⇒ 2 ∠BCD = 180° ⇒ ∠BCD = \(\frac { 180° }{ 2 }\) = 90°

∴ ∠BCD is a right angle

Hence proved.

Question 19.

In an isosceles triangle, prove that the bisectors of the base angles meeting the opposite sides are equal.

Solution:

Given : In ∆ABC, AB = AC

Bisectors of ∠B and ∠C meets AC and AB in D and E respectively

To prove : BD = CE

Proof: In ∆BDC and ∆BEC

BC = BC (common)

∠C = ∠B

(base angles of an isosceles triangle)

∠DBC = ∠ECB (half of equal angles)

∴ ∆BDC ≅ ∆BEC (ASA axiom)

∴ BD = CE (c.p.c.t.)

Question 20.

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

Solution:

Given : In ∆ABC, the bisector of vertical angle A bisects the sides BC at D

∴ x_{1} = x_{2} and

BD = DC

To prove : AABC is an isosceles

i.e., AB = AC

Construction : Produce AD to E such that

AD = DE

Join EC

Proof: In ∆ABD and ∆ECD,

AD = DE (construction)

BD = DC (D is mid point of BC)

∠ADB = ∠CDE(vertically opposite angles)

∴ ∆ABD ≅ ∆ECD (SAS axiom)

∴ AB = CE (c.p.c.t.)

and ∠x_{1} = ∠x_{2}

But x_{1} = x_{2}

∴ ∠x_{2} = ∠x_{3}

∴ AC = CE (sides opposite to equal angles)

But AB = CE (proved)

∴ AB = AC

Hence ∆ABC is an isosceles triangle

Hence proved.