KSEEB Solutions for Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.7

In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.7 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.7 pdf, free KSEEB SSLC Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.7 pdf download. Now you will get step by step solution to each question.

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:
KSEEB Solutions for Class 8 Maths Chapter 11 Congruency of Triangles Ex. 11.7 1
∠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:
KSEEB Solutions for Class 8 Maths Chapter 11 Congruency of Triangles Ex. 11.7 2
∠C <∠A <∠B

Question 3.
Let ΔABC 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.
KSEEB Solutions for Class 8 Maths Chapter 11 Congruency of Triangles Ex. 11.7 3
Answer:
In Δ ABD, AB = AD
∴ ∠ABD = ∠ APB = 70° [Theorem]
∠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]
KSEEB Solutions for Class 8 Maths Chapter 11 Congruency of Triangles Ex. 11.7 4
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).
KSEEB Solutions for Class 8 Maths Chapter 11 Congruency of Triangles Ex. 11.7 5
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).

All Chapter KSEEB Solutions For Class 8 maths

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All Subject KSEEB Solutions For Class 8

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