Students can Download Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.7

Question 1.

In a triangle ABC, ∠B = 28° and ∠C= 56°. Find the largest and smallest side.

Answer:

∠A+∠B + ∠C = 180°

[Sum of the angles of a triangle]

∠A + 28 + 56 = 180°

∠A + 84° = 180°

∠A = 180 – 84

∠A = 96°

BC is the largest side

[Opposite to largest angle ∠A]

AC is the smallest side

[Opposite to smalles angle∠B]

Question 2.

In a triangle ABC, we have AB = 4cm, BC = 5.6 cm and CA = 7.6 cm. Write the angles of the triangle in ascending order of measures.

Answer:

∠C < ∠A < ∠B

Question 3.

Let Δ ABC be a triangle such that ∠B = 70° and ∠C = 40°. suppose D is a point on BC such that AB = AD. Prove that AB > CD.

Answer:

In Δ ABD, AB = AD

∴∠ABD = ∠ APB = 70° [Theorem l]

∠APB +∠ ADC = 180° [Linear pair]

70 +∠ADC = 180°

∠ADC = 180 – 70°

∠ADC = 110°

In Δ ADC,

∠DAC = 180 – ( 110 + 40)

= 180-150

∠DAC = 30°

In Δ ADC,

∠ACD > ∠DAC

AD > CD

but AB = AD

∴ AB > CD

Question 4.

Let ABCD be a quadrilateral in which AD is the largest side and BC is the smallest side. Prove that ∠A < ∠C. [Hint: Join AC]

Answer:

Join AC

In Δ ABC AB > BC [BC is the smallest side]

∴ ∠ACB > ∠BAC …(i)

In Δ ADC, AD > DC [AD is the largest side]

∠ACD > ∠PAC …(ii)

By adding (i) and (ii)

∠ACB + ∠ACD >∠BAC + ∠DAC

∠BCD > ∠BAD or

∠BAD < ∠BCD

∴∠A < ∠C

Question 5.

Let ABC be a triangle and P be an interior point, prove that AB + BC + CA < 2(PA + PB + PC).

Answer:

In ΔPBA, AB < PA + PB ….(i)

In ΔPBC, BC < PB + PC ….(ii)

In ΔPCA, AC < PC + PA ….(iii)

By adding (i), (ii), and (iii)

AB + BC + AC < PA + PB + PB + PC + PC + PA > 2PA + 2PB + 2PC

AB + BC + AC <2 (PA+ PB + PC).