NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3

# NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 are the part of NCERT Solutions for Class 7 Maths. Here you can find the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3.

### Ex 6.3 Class 7 Maths Question 1.

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

Solution:
(i) Using angle sum property of a triangle, we get
x + 50° + 60° = 180°
x + 110° = 180°
x = 180° – 110° = 70°

(ii) Using angle sum property of a triangle, we get
x + 90° + 30 = 180°            [The triangle is a right-angled triangle]
x + 120° = 180°
x = 180° – 120° = 60°

(iii) Using angle sum property of a triangle, we get
x + 30° + 110° = 180°
x + 140° = 180°
x = 180° – 140° = 40°

(iv) Using angle sum property of a triangle, we get
x + x + 50° = 180°
2x + 50° = 180°
2x = 180° – 50°
2x = 130°
x = 130°/2 = 65°

(v) Using angle sum property of a triangle, we get
x + x + x = 180°
3x = 180°
x = 180°/3 = 60°

(vi) Using angle sum property of a triangle, we get
x + 2x + 90° = 180°         (Triangle is a right-angled triangle)
3x + 90° = 180°
3x = 180° – 90°
3x = 90°
x = 90°/3 = 30°

### Ex 6.3 Class 7 Maths Question 2.

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

Solution:
(i)
x + 50° = 120°      (Using exterior angle property of a triangle)
x = 120° 50° = 70°
x + y + 50° = 180°   (Using angle sum property of a triangle)
70° +
y + 50° = 180°
y + 120° = 180°
y = 180° – 120°
y = 60°
Thus,
x = 70 and y = 60°

(ii) y = 80°                  (Vertically opposite angles are same)
x + y + 50° = 180°   (Using angle sum property of a triangle)
x + 80° + 50° = 180°
x + 130° = 180°
x = 180° – 130° = 50°
Thus,
x = 50° and y = 80°

(iii) y + 50° + 60° = 180°      (Using angle sum property of a triangle)
y + 110° = 180°
y = 180° – 110° = 70°
x + y = 180°             (Linear pairs)
x + 70° = 180°
x = 180° – 70° = 110°
Thus,
x = 110° and y = 70°

(iv) x = 60°               (Vertically opposite angles)
x + y + 30° = 180° (Using angle sum property of a triangle)
60° + y + 30° = 180°
y + 90° = 180°
y = 180° – 90° = 90°
Thus,
x = 60° and y = 90°

(v) y = 90°                (Vertically opposite angles)
x + x + y = 180°  (Using angle sum property of a triangle)
2x + 90° = 180°
2x = 180° – 90°
2x = 90°
x = 90°/2 = 45°
Thus,
x = 45° and y = 90°

(vi) From the given figure, we have

y + 1 + 2 = 3x
180° = 3x             (Using angle sum property of a triangle)
x = 180°/3 = 60°
Thus,
x = 60°, y = 60°

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