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Home » NCERT Solutions » Class 7 » Maths » NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.3

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

Last Updated on February 16, 2023 By Mrs Shilpi Nagpal

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

  • Class 7 Maths Triangle and its Properties Exercise 6.1
  • Class 7 Maths Triangle and its Properties Exercise 6.2
  • Class 7 Maths Triangle and its Properties Exercise 6.3

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

Class 7 Maths Chapter 6 Triangle and its Properties Exercise 6.3

Answer

(i) We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= ∠BAC + ∠ABC + ∠BCA = 180o
= x + 50o + 60o = 180o
= x + 110o = 180o
By transposing 110o from LHS to RHS it becomes – 110o
= x = 180o – 110o
= x = 70o

(ii) We know that,
The sum of all the interior angles of a triangle is 180o.
The given triangle is a right angled triangle. So the ∠QPR is 90o.
Then,
= ∠QPR + ∠PQR + ∠PRQ = 180o
= 90o + 30o + x = 180o
= 120o + x = 180o
By transposing 110o from LHS to RHS it becomes – 110o
= x = 180o – 120o
= x = 60o

(iii) We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= ∠XYZ + ∠YXZ + ∠XZY = 180o
= 110o + 30o + x = 180o
= 140o + x = 180o
By transposing 140o from LHS to RHS it becomes – 140o
= x = 180o – 140o
= x = 40o

(iv) We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= 50o + x + x = 180o
= 50o + 2x = 180o
By transposing 50o from LHS to RHS it becomes – 50o
= 2x = 180o – 50o
= 2x = 130o
= x = 130o/2

= x = 65o

(v) We know that,
The sum of all the interior angles of a triangle is 180o.
Then,
= x + x + x = 180o
= 3x = 180o
= x = 180o/3

= x = 60o
∴ The given triangle is an equiangular triangle.

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

Class 7 Maths Chapter 6 Triangle and its Properties Exercise 6.3 -2

Class 7 Maths Chapter 6 Triangle and its Properties Exercise 6.3 -3

Answer

(i) We Know that, an exterior angle of a triangle is equal to the sum of its interior opposite angles.
Then,
= 50o + x = 120o
By transposing 50o from LHS to RHS it becomes – 50o
= x = 120o – 50o
= x = 70o
We also know that,

The sum of all the interior angles of a triangle is 180o.
Then,
= 50o + x + y = 180o
= 50o + 70o + y = 180o
= 120o + y = 180o
By transposing 120o from LHS to RHS it becomes – 120o
= y = 180o – 120o
= y = 60o

(ii) From the rule of vertically opposite angles,
= y = 80o
Then, we know that,

The sum of all the interior angles of a triangle is 180o.
Then,
= 50o + 80o + x = 180o
= 130o + x = 180o
By transposing 130o from LHS to RHS it becomes – 130o
= x = 180o – 130o
= x = 50o

(iii) We know that, the sum of all the interior angles of a triangle is 180o.
Then,
= 50o + 60o + y = 180o
= 110o + y = 180o
By transposing 110o from LHS to RHS it becomes – 110o
= y = 180o – 110o
= y = 70o
Now, from the rule of linear pair,

= x + y = 180o
= x + 70o = 180o
By transposing 70o from LHS to RHS it becomes – 70o
= x = 180o – 70

= x = 110o

(iv) From the rule of vertically opposite angles,
= x = 60o
Then, we know that, the sum of all the interior angles of a triangle is 180o.

Then,
= 30o + x + y = 180o
= 30o + 60o + y = 180o
= 90o + y = 180o
By transposing 90o from LHS to RHS it becomes – 90o
= y = 180o – 90o
= y = 90o

(v) From the rule of vertically opposite angles,
= y = 90o
Then, we know that, the sum of all the interior angles of a triangle is 180o.

Then,
= x + x + y = 180o
= 2x + 90o = 180o
By transposing 90o from LHS to RHS it becomes – 90o
= 2x = 180o – 90o
= 2x = 90o
= x = 90o/2

= x = 45o

(vi)
From the rule of vertically opposite angles,
= x = y
Then, we know that, the sum of all the interior angles of a triangle is 180o.
Then,
= x + x + x = 180o
= 3x = 180o
= x = 180o/3

= x = 60o

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