Chapter 6 The Triangles and its Properties Ex 6.2 Solutions
Question - 1 : - Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
Answer - 1 : -
Draw a Line segment PS ⊥ BC. It is an altitude for this triangle. Here we observe that length of QS and SR is also same. So PS is also a median of this triangle.
Question - 2 : - Find the value of the unknown exterior angle x in thefollowing diagrams:
Answer - 2 : -
(i) ∠x = 50° + 70° = 120° (Exterior angle is equal to sum of its interior opposite angles)
(ii) ∠x = 65°+ 45° = 110° (Exterior angle is equal to sum of its interior opposite angles)
(iii) ∠x = 30° + 40° = 70° (Exterior angle is equal to sum of its interior opposite angles)
(iv) ∠x = 60° + 60° = 120° (Exterior angle is equal to sum of its interior opposite angles)
(v) ∠x = 50° + 50° =100° (Exterior angle is equal to sum of its interior opposite angles)
(vi) ∠x = 30° + 60° = 90° (Exterior angle is equal to sum of its interior opposite angle)
Question - 3 : - Find the value of the unknown interior angle x in the following figures:
Answer - 3 : -
(i) ∠x + 50° = 115° (Exterior angle of a triangle)
∴ ∠x = 115°- 50° = 65°
(ii) ∠x + 70° = 110° (Exterior angle of a triangle)
∴ ∠x = 110° – 70° = 40°
(iii) ∠ x + 90° = 125° (Exterior angle of a right triangle)
∴ ∠x = 125° – 90° = 35°
(iv) ∠x + 60° = 120° (Exterior angle of a triangle)
∴ ∠x = 120° – 60° = 60°
(v)∠ X + 30° = 80° (Exterior angle of a triangle)
∴ ∠x = 80° – 30° = 50°
(vi) ∠ x + 35° = 75° (Exterior angle of a triangle)
∴ ∠ x = 75° – 35° = 40°