# RBSE Solutions For Class 7 Chapter 6 – The Triangle and its Properties

Exercise 6.1 Solutions

Exercise 6.2 Solutions

Exercise 6.4 Solutions

Exercise 6.5 Solutions

### Access answers to Maths NCERT Solutions for Class 7 Chapter 6 – The Triangle and its Properties Exercise 6.3

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

(i)

Solution:-

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)

Solution:-

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)

Solution:-

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)

Solution:-

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)

Solution:-

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.

(vi)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 90o + 2x + x = 180o

= 90o + 3x = 180o

By transposing 90o from LHS to RHS it becomes – 90o

= 3x = 180o – 90o

= 3x = 90o

= x = 90o/3

= x = 30o

Then,

= 2x = 2 × 30o = 60o

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

(i)

Solution:-

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 = 70

We also know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + x + y = 180o

= 50o + 70+ y = 180o

= 120o + y = 180o

By transposing 120o from LHS to RHS it becomes – 120o

= y = 180– 120

= y = 60o

(ii)

Solution:-

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

= 130+ x = 180o

By transposing 130o from LHS to RHS it becomes – 130o

= x = 180– 130

= x = 50o

(iii)

Solution:-

We know that,

The sum of all the interior angles of a triangle is 180o.

Then,

= 50o + 60o + y = 180o

= 110+ y = 180o

By transposing 110o from LHS to RHS it becomes – 110o

= y = 180– 110

= 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)

Solution:-

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

= 90+ y = 180o

By transposing 90o from LHS to RHS it becomes – 90o

= y = 180– 90

= y = 90o

(v)

Solution:-

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 = 180– 90

= 2x = 90o

= x = 90o/2

= x = 45o

(vi)

Solution:-

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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