In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.6 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 11.6 pdf, free KSEEB SSLC Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.6 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 11.6**

Question 1.

Suppose ABCD is a rectangle. Using RHS theorem prove that the triangles ABC and ADC are congruent.

Answer:

In ∆ ABC and ∆ ADC

∠ABC = ∠ ADC = 90°

[Angles of a rectangle]

AC = AC [Common side]

AB = DC [Opposite side rectangle]

∴∆ ABC = ∆ ADC [RHS theorem]

Question 2.

Suppose ABC is a triangle and D is the midpoint of BC. Assume that the perpendiculars from D to AB and AC are of equal length. Prove that ABC is isosceles.

Answer:

DF ⊥ AB and DE ⊥ AC

In ∆BFD and ∆CED

∠BFD = ∠ CEP [= 90° by data]

BD = DC [D is the midpoint of BC]

DF = DE[data]

∴ ∆ BFD ≅ ∆CED [RHS theorem]

∴∠B =∠C [Corresponding angles]

∴AC = AB [Theorem 2]

Question 3.

Suppose ABC is a triangle in which BE and CF are respectively the perpendiculars to the sides AC and AB.

If BE = CF. Prove that triangle ABC is isosceles.

Answer:

In ∆ BFC and ∆ BEC

∠BFC = ∠BEC [= 90° data]

BC = BC [Common side]

CF = BE [data]

∴ ∆BFC ≅ ∆BEC [RHStheorem]

∠EBC = ∠FCB [Corresponding angle]

i. e., AB = AC

∴ ∆ ABC is isosceles

**All Chapter KSEEB Solutions For Class 8 maths**

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