**NCERT Solutions for Class 7 Maths**

Chapter 6 The Triangle and its Properties

Exercise 6.3

Chapter 6 The Triangle and its Properties

Exercise 6.3

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

**Answer**

(i) We know that,

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

Then,

= ∠BAC + ∠ABC + ∠BCA = 180^{o
}= x + 50^{o} + 60^{o} = 180^{o
}= x + 110^{o} = 180^{o
}By transposing 110^{o} from LHS to RHS it becomes – 110^{o
}= x = 180^{o} – 110^{o
}= x = 70^{o}

(ii) We know that,

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

The given triangle is a right angled triangle. So the ∠QPR is 90^{o}.

Then,

= ∠QPR + ∠PQR + ∠PRQ = 180^{o
}= 90^{o} + 30^{o} + x = 180^{o
}= 120^{o} + x = 180^{o
}By transposing 110^{o} from LHS to RHS it becomes – 110^{o
}= x = 180^{o} – 120^{o
}= x = 60^{o}

(iii) We know that,

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

Then,

= ∠XYZ + ∠YXZ + ∠XZY = 180^{o
}= 110^{o} + 30^{o} + x = 180^{o
}= 140^{o} + x = 180^{o
}By transposing 140^{o} from LHS to RHS it becomes – 140^{o
}= x = 180^{o} – 140^{o
}= x = 40^{o}

(iv) We know that,

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

Then,

= 50^{o} + x + x = 180^{o
}= 50^{o} + 2x = 180^{o
}By transposing 50^{o} from LHS to RHS it becomes – 50^{o
}= 2x = 180^{o} – 50^{o
}= 2x = 130^{o
}= x = 130^{o}/2

= x = 65^{o}

(v) We know that,

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

Then,

= x + x + x = 180^{o
}= 3x = 180^{o
}= x = 180^{o}/3

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

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

**Answer**

(i) We Know that, an exterior angle of a triangle is equal to the sum of its interior opposite angles.

Then,

= 50^{o} + x = 120^{o
}By transposing 50^{o} from LHS to RHS it becomes – 50^{o
}= x = 120^{o} – 50^{o
}= x = 70^{o
}We also know that,

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

Then,

= 50^{o} + x + y = 180^{o
}= 50^{o} + 70^{o }+ y = 180^{o
}= 120^{o} + y = 180^{o
}By transposing 120^{o} from LHS to RHS it becomes – 120^{o
}= y = 180^{o }– 120^{o
}= y = 60^{o}

(ii) From the rule of vertically opposite angles,

= y = 80^{o
}Then, we know that,

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

Then,

= 50^{o} + 80^{o} + x = 180^{o
}= 130^{o }+ x = 180^{o
}By transposing 130^{o} from LHS to RHS it becomes – 130^{o
}= x = 180^{o }– 130^{o
}= x = 50^{o}

(iii) We know that, the sum of all the interior angles of a triangle is 180^{o}.

Then,

= 50^{o} + 60^{o} + y = 180^{o
}= 110^{o }+ y = 180^{o
}By transposing 110^{o} from LHS to RHS it becomes – 110^{o
}= y = 180^{o }– 110^{o
}= y = 70^{o
}Now, from the rule of linear pair,

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

= x = 110^{o}

(iv) From the rule of vertically opposite angles,

= x = 60^{o
}Then, we know that, the sum of all the interior angles of a triangle is 180^{o}.

Then,

= 30^{o} + x + y = 180^{o
}= 30^{o} + 60^{o} + y = 180^{o
}= 90^{o }+ y = 180^{o
}By transposing 90^{o} from LHS to RHS it becomes – 90^{o
}= y = 180^{o }– 90^{o
}= y = 90^{o}

(v) From the rule of vertically opposite angles,

= y = 90^{o
}Then, we know that, the sum of all the interior angles of a triangle is 180^{o}.

Then,

= x + x + y = 180^{o
}= 2x + 90^{o} = 180^{o
}By transposing 90^{o} from LHS to RHS it becomes – 90^{o
}= 2x = 180^{o }– 90^{o
}= 2x = 90^{o
}= x = 90^{o}/2

= x = 45^{o}

(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 180^{o}.

Then,

= x + x + x = 180^{o
}= 3x = 180^{o
}= x = 180^{o}/3

= x = 60^{o}