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Properties of Triangles – Maharashtra Board Class 6 Solutions for Mathematics

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Properties of Triangles – Maharashtra Board Class 6 Solutions for Mathematics (English Medium)

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise-76

Solution 1:
By the property of exterior angle of a triangle,
m∠XZP = m∠XYZ + m∠YXZ
∴ 90° = m∠XYZ + 55°
∴ m∠XYZ = 90° – 55°
∴ m∠XYZ = 35°

Solution 2:
By the property of exterior angle of a triangle,
m∠BCD = m∠BAC + m∠ABC
∴ 130° = m∠BAC + 60°
∴ m∠BAC = 130° – 60°
∴ m∠BAC = 70°

Solution 3:
The sum of the measures of the three angles of a triangle is 180°.
Thus, in ∆PST,
m∠P + m∠S + m∠PTS = 180°
∴ 60° + 80° + m∠PTS = 180°
∴ 140° + m∠PTS = 180°
∴ m∠PTS = 180° – 140°
∴ m∠PTS = 40°
Now, By the property of exterior angle of a triangle,
m∠PTN = m∠P + m∠S
∴ m∠PTN = 60° + 80°
∴ m∠PTN = 140°

Solution 4:
The sum of the measures of the three angles of a triangle is 180°.
Thus, in ∆RST,
m∠R + m∠S + m∠T = 180°
∴ 70° + 30° + m∠T = 180°
∴ 100° + m∠T = 180°
∴ m∠T = 180° – 100°
∴ m∠T = 80°

Solution 5:
By the property of exterior angle of a triangle,
m∠TRM = m∠N + m∠T
∴ m∠TRM = 30° + 80°
∴ m∠TRM = 110°

Solution 6:
By the property of exterior angle of a triangle,
m∠ACD = m∠A + m∠B
∴ m∠ACD = m∠A + m∠A (∵ m∠A = m∠B)
∴ m∠ACD = 2 × m∠A
∴ 140° = 2 × m∠A
∴ m∠A = 70° (dividing both sides by 2)
∴ m∠A = m∠B = 70°

Exercise-77

Solution 1:
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

  1. l(AP) + l(BP) > l(AB)
  2. l(AP) + l(CP) >l(AC)
  3. l(BP)+ l(CP) > l(BC)
  4. l(AB) + l(AC)> l(BC)