**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.

**NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2****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 6.4****NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5**

**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°

⇒ 3∠x = 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)

⇒ 2∠x + 90° = 180°

⇒ 2∠x = 180° – 90°

⇒ 2∠x = 90°

∴ ∠x = 90°/2 = 45°

Thus, ∠x = 45° and ∠y = 90°

(vi)
From the given figure, we have

∠y + ∠1 + ∠2 = 3∠x

⇒ 180° = 3∠x (Using angle sum property of a triangle)

∴ ∠x = 180°/3 = 60°

Thus, ∠x = 60°, ∠y = 60°

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