Well-designed AP 7th Class Maths Guide Chapter 6 The Triangle and its Properties Exercise 6.3 offers step-by-step explanations to help students understand problem-solving strategies.
The Triangle and its Properties Class 7 Exercise 6.3 Solutions – 7th Class Maths 6.3 Exercise Solutions
Question 1.
Find the value of the unknown x in the following diagrams :
Solution:
The property used for all the questions is “The sum of the angles in a triangle is 180°.
i) x° + 50° + 60° = 180°
x° + 110° = 180°
x° = 180° – 110° ⇒ x° = 70°
ii) 90° + x° + 30° = 180°
x° + 120° = 180°
x° = 180° – 120° ⇒ x° = 60°
iii) 110° + 30° + x° = 180°
140° + x° = 180°
x° = 180° – 140° ⇒ x° = 40°
iv) 50° + x° + x° = 180°
2x° + 50° = 180°
2x° = 180° – 50°
2x° = 130° ⇒ x° = \(\frac{130^{\circ}}{2}\) ⇒ x° = 65°
v) x° + x° + x° = 180°
3x° = 180°
x° = \(\frac{180^{\circ}}{3}\) ⇒ x° = 60°
vi) x° + 2x° + 90° = 180°
3x° = 180° – 90°
3x° = 90° ⇒ x° = \(\frac{90^{\circ}}{3}\) ⇒ x° = 30°
Question 2.
Find the values of the unknowns x and y in the following diagrams :
Solution:
1) In the figure
120° = x° + 50° (Exterior angle property)
x° = 120° – 50°
x° = 70°
also, x° + 50° + y° = 180° (Angle sum property)
70° + 50° + y° = 180°
y° + 120° = 180°
y° = 180° – 120° ⇒ y° = 60°
ii) y° = 80° (Vertically opposite angles)
y° + x° + 50° = 180°
80° + x° + 50° = 180°
130° + x° = 180°
x° = 180° – 130° ⇒ x° = 50°
iii) x° = 50° + 60° (Exterior angle property)
x° = 110°
y + x = 180°
y° + 50° + 60° = 180° (Angle sum property)
y° + 110° = 180°
y° = 180° – 110° ⇒ y° = 70°
iv) x° = 60° (Vertically opposite angle)
30° + y° + x° = 180° (Angle sum property)
30° + y° + 60° = 180°
y° + 90° = 180°
y° = 180° – 90° ⇒ y° = 90°
v) y° = 90° (Vertically opposite angles)
x° + x° + 90° = 180° (Angle sum property)
2x° + 90° = 180°
2x° = 180° – 90°
2x° = 90° ⇒ x° = \(\frac{90^{\circ}}{2}\) ⇒ x° = 45°
vi) y° = x° (Vertically opposite angles) Let us draw the figure in the following way
x° = b° (Vertically opposite angles)
x° = a° (Vertically opposite angles) In the triangle
a + b + y = 180° (From above)
x° + x° + x° = 180°
3x° = 180° ⇒ x° = \(\frac{180^{\circ}}{3}\) ⇒ x° = 60°