NCERT solution class 7 chapter 6 The Triangle and its Properties exercise 6.3 mathematics

Exercise 6.3

QUESTION 1 

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

The sum of all interior angles of a triangle is 180°. By using this property, these problems can be
solved as follows.

(i) x + 50° + 60° = 180°

x + 110° = 180°

x = 180° -110°

x = 70°

 

(ii)  x + 90° + 30° = 180°

x + 120° = 180°

x = 180° – 120°

x = 60°

 

(iii) x + 30° + 110° = 180°

x + 140° = 180°

x = 180° – 140°

x = 40°

 

(iv) 50° + x + x = 180°

50° + 2x = 180°

2x = 180° – 50°

x=\dfrac{130°}{2}=65°

 

(v)  x + x + x = 180°

3x = 180° – 90°

x=\dfrac{90}{3}=30°

 

QUESTION 2 

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

Sol :

(i) y+ 120° = 180° (Linear pair)

y = 180° – 120°

y = 60°

x + y + 50° = 180° (Angle sum property)

x + 60° + 50° = 180°

x + 110° = 180°

x = 180° – 10°

x = 70°

 

(ii) y = 80° (Vertically opposite angles)

y + x + 50° = 180° (Angle sum property)

80° + x + 50° =180°

x + 130° = 180°

x = 180° – 130°

x = 50°

 

(iii) y + 50° +  60° = 180° (Angle sum property)

y = 180° – 60° – 50°

y = 70°

x +  y = 180° (Linear pair)

x = 180° – y

x = 180°-70°

x = 110°

 

(iv) x = 60° (Vertically opposite angles)

30° + x + y = 180°

30° + 60° + y = 180°

y = 180° – 30° – 60°

y = 90°

 

(v) y = 90° (Vertically opposite angles)

x + x + y = 180° (Angle sum property)

2x + y = 180°

2x + 90° = 180°

2x = 180° – 90°

x=dfrac{90°}{2}=45°

 

(vi)

y = x ( Vertically opposite angles )

a = x ( Vertically opposite angles )

b = x ( Vertically opposite angles )

a + b+ y=180° (Angle sum property)

x + x + x = 180°

3x = 180°

x=\dfrac{180°}{3}=60°

y = x = 60°

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