Mathematics – Class 7 – Chapter 6 – The Triangle And Its Properties – Exercise 6.2 – NCERT Exercise Solution

1. Find the value of the unknown exterior angle x in the following diagram:

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

Solution:

(i)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x = 50o + 70o

x = 120o

(ii)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x = 65o + 45o

x = 110o

(iii)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x = 30o + 40o

x = 70o

(iv)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x = 60o + 60o

x = 120o

(v)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x = 50o + 50o

x = 100o

(vi)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x = 30o + 60o

x = 90o

2. Find the value of the unknown interior angle x in the following figures:

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

Solution:

(i)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x + 50o = 115o

Transposing (50o) from LHS to RHS then it becomes (– 50o)

x = 115o – 50o

x = 65o

(ii)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

70o + x = 100o

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

x = 100o – 70o

x = 30o

(iii)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

Here the given triangle is a right-angled triangle. So the angle opposite to the x is 90o.

x + 90o = 125o

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

x = 125o – 90o

x = 35o

(iv)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

x + 60o = 120o

By transposing (60o) from LHS to RHS it becomes (– 60o)

x = 120o – 60o

x = 60o

(v)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

Here the given triangle is a right-angled triangle. So, the angle opposite to the x is 90o.

x + 30o = 80o

Transposing (30o) from LHS to RHS it becomes (– 30o)

x = 80o – 30o

x = 50o

(vi)

Mathematics - Class 7 - Chapter 6 - The Triangle And Its Properties - Exercise 6.2 - NCERT Exercise Solution

We know that an exterior angle of a triangle is equal to the sum of its interior opposite angles.

Here the given triangle is a right-angled triangle. So the angle opposite to the x is 90o.

x + 35o = 75o

By transposing (35o)from LHS to RHS it becomes (– 35o)

x = 75o – 35o

x = 40o

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