Mathematics – Class 7 – Chapter 6 – The Triangle And Its Properties – Exercise 6.3 – NCERT Exercise Solution

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

NOTE: We know that the sum of all the interior angles of a triangle is 180^o.

(i)

Solution:

∠BAC + ∠ABC + ∠BCA = 180^o

x + 50^o + 60^o = 180^o

x + 110^o = 180^o

Transposing 110^o from LHS to RHS it becomes – 110^o

x = 180^o – 110^o

x = 70^o

(ii)

Solution:

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

Now,

∠QPR + ∠PQR + ∠PRQ = 180^o

90^o + 30^o + x = 180^o

120^o + x = 180^o

Transposing 110^o from LHS to RHS it becomes – 110^o

x = 180^o – 120^o

x = 60^o

(ii)

Solution:

∠XYZ + ∠YXZ + ∠XZY = 180^o

110^o + 30^o + x = 180^o

140^o + x = 180^o

Transposing 140^o from LHS to RHS it becomes – 140^o

x = 180^o – 140^o

x = 40^o

Solution:

50^o + x + x = 180^o

50^o + 2x = 180^o

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)

Solution:

x + x + x = 180^o

3x = 180^o

x = 180^o/3

x = 60^o

Hence, the given triangle is an equilateral triangle.

(vi)

Solution:

90^o + 2x + x = 180^o

90^o + 3x = 180^o

Transposing 90^o from LHS to RHS it becomes – 90^o

3x = 180^o – 90^o

3x = 90^o

x = 90^o/3

x = 30^o

Then,

2x = 2 × 30^o = 60^o

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

Solution:

(i)

Solution:

An exterior angle of a triangle is equal to the sum of its interior opposite angles.

Now,

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

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

Now,

50^o + x + y = 180^o

50^o + 70^o + y = 180^o

120^o + y = 180^o

Transposing 120^o from LHS to RHS it becomes – 120^o

y = 180^o – 120^o

y = 60^o

(ii)

Solution:

According to the rule of vertically opposite angles,

y = 80^o

Now,

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

50^o + 80^o + x = 180^o

130^o + x = 180^o

Transposing 130^o from LHS to RHS it becomes – 130^o

x = 180^o – 130^o

x = 50^o

(iii)

Solution:

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

50^o + 60^o + y = 180^o

110^o + y = 180^o

Transposing 110^o from LHS to RHS it becomes – 110^o

y = 180^o – 110^o

y = 70^o

Now,

According to the rule of linear pair,

x + y = 180^o

x + 70^o = 180^o

Transposing 70^o from LHS to RHS it becomes – 70^o

x = 180^o – 70^o

x = 110^o

(iv)

Solution:

According the rule of vertically opposite angles,

x = 60^o

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

30^o + x + y = 180^o

30^o + 60^o + y = 180^o

90^o + y = 180^o

Transposing 90^o from LHS to RHS it becomes – 90^o

y = 180^o – 90^o

y = 90^o

(v)

Solution:

According to the rule of vertically opposite angles,

y = 90^o

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

Now,

x + x + y = 180^o

2x + 90^o = 180^o

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)

Solution:

According to the rule of vertically opposite angles,

x = y

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

Now,

x + x + x = 180^o

3x = 180^o

x = 180^o/3

x = 60^o