Students can Download Maths Chapter 11 Congruency of Triangles Ex 11.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.3

Question 1.

In a Δ ABC, AB = AC andlA.= 50° find ∠B and ∠C.

Answer:

∠A+ ∠B + ∠C = 180°

(Sum of the angles of a triangle is 180° )

50 +∠[B + ∠B = 180°

∠B = ∠C Base angles of an isosceles triangle

50 + 2∠B= 180°

2∠B = 180 – 50

2∠B = 130°

∠B = \(\frac{130^{\circ}}{2}\)

∠B = 65°

∠B = ∠C = 65°

Question 2.

In AABC,AB = BCand|B = 64°find|£,

Answer:

AB = BC [data]

∴ ∠C = ∠A [Theorem 1]

∠A + ∠B + ∠C = 180°

(Sum of the angles of a triangle is 180°)

∠C + 64 + ∠C = 180° [∠A = ∠C]

64 + 2∠C = 180°

2∠C = 180 – 64

2∠C = 116

∠C = \(\frac { 116 }{ 2 }\) = 58°

Question 3.

In each of the following figure find the value of x :

i.

Answer:

In ∆ABC, AB = AC

∴ ∠ABC = ∠ACB

∴ ∠BAC + ∠ABC + ∠ACB = 180°

(Sum of the angles of a triangle is 180°)

40 + ∠ABC + ∠ABC = 180°

(∠ABC =∠ACB)

40 + 2∠ABC = 180°

2∠ABC = 180 – 40

2∠ABC = 140°

∠ABC = 70°

∠ACB = ∠ ABC = 70°

∠ACB + ∠ACD = 180°

70 + x = 180°

x = 180 – 70

x = 110°

ii

AC = CD

∠CAD = ∠CDA

∠CAD = ∠CAD = 30°

∠ACD + ∠CAD + ∠CDA = 180°

(Sum of the angles of a triangle is 180°)

∠ACD + 30 + 30=180°

∠ACD + 60=180°

∠ACD = 180-60

∠ACD = 120°

∠ACD = ∠BAC + ∠ABC

120 = 65° + x

120 – 65 = x

55 = x

x=55°

iiii

Answer:

AB = AC

∠ABC = ∠ACB = 55° [Theorem l]

Exterior ∠APB = ∠DAC +∠ACD

75 = x + 55

75 – 55 = x

20 = x

x = 20°

iv.

BD = DC = Ad

BD = DC = AD & ∠ABD = 50 °

∠ABD = ∠BAD(Th. l)

∠BAD = 50°

∠ABD + ∠BAD + ∠ADB = 180°

50 + 50 +∠ADB = 180°

∠ADB = 180 – 100 = 80°

∠APB = 80°

∴ ∠APB +∠ADC = 180°

80 +∠ADC = 180°

∠ADC = 100°

Now AD = DC

∴∠DAC = ∠DCA = x°

∴ x + x +∠ADC = 180°

2x + 100 = 180°

2x = 180-100=80°

x=40°

Question 4.

Suppose ABC is an equilateral triangle. Its base BC is produced to D such that BC = CD.

Calculate: l.∠ACD

2.∠ADC

Answer:

∠ABC = ∠ACB = ∠BAC = 60°

(ABC is an equilateral triangle)

∠ACB +∠ACD = 180° (Linearpoint)

60 + ∠ACD = 180°

∠ACD = 180 – 60

∠ACD = 120°

In ∆ACD,AC=CD

∠ CAD = ∠CPA (Theorem l)

∠ACB +∠ACD = 180° [linear pair]

60° +∠ACD = 180°

∠ACD = 180° – 60° = 120°

∠ACA +∠CAD + ∠CDA = 180°

2∠CDA = 180 – 120°

2∠CDA = 60°

∠CDA = \(\frac { 60 }{ 2 }\) = 30°

∠CDA = 30

Question 5.

Show that the perpendicular drawn from the vertices of the base of an isosceles triangle to the opposite sides are equal. ,

Answer:

Data : In ∆ABC, AB = AC,

BD ⊥AC & CE⊥ AB

To prove : BD = CE

Proof:

In ∆ABC, AB = AC [data]

∠ABC = ∠ACB [Theorem l]

In ∆EBC and ∆DCB

∠EBC =∠DCB( Base angles )

∠BEC = ∠CDB [= 90° ]

BC = BC (Common side)

∆EBC = ∆DCB [ASA postulate]

BD = CE [Corresponding sides]

Question 6.

Prove that an ∆ABC is an isosceles triangle if the altitude AD from A on BC bisects BC.

Answer:

In ∆ADB and ∆ADC

AD = AD [Common side]

∠APB =∠ ADC [90° ]

BD = DC [AD bisects BC]

∴∆ADB ≅ ∆ADC [SAS postulate]

∴AB = AC [Correspondingsides]

∴∆ ABC is an isosceles triangle

Question 7.

Suppose a triangle is equilateral, prove that it is equiangular.

Answer:

To prove: ∠A = ∠B = ∠C

Proof: In ∆ABC, AB = BC

∠C = ∠B [Theorem l]….(i)

BC = AC

∠A =∠B [Theorem l]…(ii)

From (i) and (ii)

∠A =∠B = ∠C

∆ABC is equiangular