Students can Download Chapter 6 The Triangles and Its Properties Ex 6.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3

Question 1.

Find the value of the unknown x in the following diagrams :

Solution:

i) ∠A + ∠B + ∠C = 180°

x° + 50° + 60° = 180° (By Angle sum property of a triangle)

∠x + 110° = 180°

∠x = 180° – 110°

∠x = 70°

ii) ∠P + ∠Q + ∠R = 180°

90° + 30° + ∠x = 180° (By Angle sum property of a triangle)

120° + ∠x = 180°

∠x = 180° – 120° = 60°

iii) ∠X + ∠Y + ∠Z = 180°

Solution:

30° + 110° + ∠x = 180° (By Angle sum property of a triangle)

140° + ∠x = 180°

∴ ∠x = 180° – 140° = 40°

iv) 50° + x + x = 180°

50° + 2x = 180°

(By Angle sum property of a triangle)

2x = 180° – 50° = 130°

x = \(\frac{180^{\circ}}{2}\)

∴ ∠x = 60°

v) x + x + x = 180°

(By Angle sum property of a triangle)

3x = 180°

x = \(\frac{180^{\circ}}{2}\)

∴ x = 60°

vi) x + 2x + 90° = 180° (ByAngle sum property of a triangle)

3x = 180°

Question 2.

Find the values of the unknowns x and y in the following diagrams:

Solution:

i) x + 50° = 120° [By the exterior angle property of a triangle]

x= 120° – 50° = 7o°

By Angle sum property of a triangle

x + y + 50° = 180°

x + y = 180° – 50° = 130°

70° + y = 130°

y = 130° – 70° = 60°

∴ x = 70°, y = 60°

ii) 50° + y° + x° = 180°

∠y = 80° [vertically opposite angles]

50° + 80° + x° = 180° (Angle sum property of a triangle)

130° + x = 180°

x° = 180° – 130° = 50°

[∴ x = 50°, y = 80°]

iii) ∠x = 50° + 60° = 110° [By exterior – angle property of a triangle]

y + 50° + 60° = 180° ( By Angle sum property of a triangle)

y + 110° = 180°

∴ y = 180° – 110° = 70°

∴ y = 70°, x = 110°

iv) ∠x = 60° [vertically opposite angles]

30° + x° + y° = 180°

(By Angle sum property of a triangle)

30° + 60° + y° = 180°

90° + y° = 180°

∴ y°= 180° – 90° = 90°

∴ x = 60° y = 90°

v) ∠y = 90° [vertically opposite angles]

∠x + ∠x + ∠y = 180°

(Angle sum property of a triangle)

2x + y° = 180°

2x + 90° = 180°

2x = 180° – 90°= 90°

∴ ∠x = 45° & ∠y = 90°

vi) ∠x ∠y [vertically opposite angles]

∠x + ∠x + ∠y = 180°

(Angle sum property of a triangle)

2x + ∠y = 180°

2x + x° = 180° (∵ y = x)

3x = 180°

∴ x = 60° & y = 60°