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# 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 180o.

(i)

Solution:

∠BAC + ∠ABC + ∠BCA = 180o

x + 50o + 60o = 180o

x + 110o = 180o

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

x = 180o – 110o

x = 70o

(ii)

Solution:

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

Now,

∠QPR + ∠PQR + ∠PRQ = 180o

90o + 30o + x = 180o

120o + x = 180o

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

x = 180o – 120o

x = 60o

(ii)

Solution:

∠XYZ + ∠YXZ + ∠XZY = 180o

110o + 30o + x = 180o

140o + x = 180o

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

x = 180o – 140o

x = 40o

Solution:

50o + x + x = 180o

50o + 2x = 180o

Transposing 50o from LHS to RHS it becomes – 50o

2x = 180o – 50o

2x = 130o

x = 130o/2

x = 65o

(v)

Solution:

x + x + x = 180o

3x = 180o

x = 180o/3

x = 60o

Hence, the given triangle is an equilateral triangle.

(vi)

Solution:

90o + 2x + x = 180o

90o + 3x = 180o

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:

Solution:

(i)

Solution:

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

Now,

50o + x = 120o

By transposing 50o from LHS to RHS it becomes – 50o

x = 120o – 50o

x = 70o

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

Now,

50o + x + y = 180o

50o + 70o + y = 180o

120o + y = 180o

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

y = 180o – 120o

y = 60o

(ii)

Solution:

According to the rule of vertically opposite angles,

y = 80o

Now,

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

50o + 80o + x = 180o

130o + x = 180o

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

x = 180o – 130o

x = 50o

(iii)

Solution:

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

50o + 60o + y = 180o

110o + y = 180o

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

y = 180o – 110o

y = 70o

Now,

According to the rule of linear pair,

x + y = 180o

x + 70o = 180o

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

x = 180o – 70o

x = 110o

(iv)

Solution:

According the rule of vertically opposite angles,

x = 60o

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

30o + x + y = 180o

30o + 60o + y = 180o

90o + y = 180o

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

y = 180o – 90o

y = 90o

(v)

Solution:

According to the rule of vertically opposite angles,

y = 90o

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

Now,

x + x + y = 180o

2x + 90o = 180o

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

2x = 180o – 90o

2x = 90o

x = 90o/2

x = 45o

(vi)

Solution:

According to the rule of vertically opposite angles,

x = y

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

Now,

x + x + x = 180o

3x = 180o

x = 180o/3

x = 60o

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