7th mathematics NCERT exercise questions with solutions, Mathematics

The Triangle and its Properties Exercise 6.3 – Class 7

The Triangle and its Properties Exercise 6.3

Questions:

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

(i)

The Triangle and its Properties Exercise 6.3

(ii)

The Triangle and its Properties Exercise 6.3

(iii)

The Triangle and its Properties Exercise 6.3

(iv)

The Triangle and its Properties Exercise 6.3

(v)

The Triangle and its Properties Exercise 6.3

(vi)

The Triangle and its Properties Exercise 6.3

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

(i)

The Triangle and its Properties Exercise 6.3

(ii)

The Triangle and its Properties Exercise 6.3

(iii)

The Triangle and its Properties Exercise 6.3

(iv)

The Triangle and its Properties Exercise 6.3

(v)

The Triangle and its Properties Exercise 6.3

(vi)

The Triangle and its Properties Exercise 6.3


The Triangle and its Properties Exercise 6.3

Solutions:

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

(i)

The Triangle and its Properties Exercise 6.3

Solution:

We know, sum of all the interior angles is equal to 180º.

x + 50º + 60º = 180º

x + 110º = 180º

x = 180º110º

x = 70º

 

(ii)

The Triangle and its Properties Exercise 6.3

Solution:

We know, sum of all the interior angles is equal to 180º.

x + 90º + 30º = 180º

x + 120º = 180º

x = 180º120º

x = 60º

 

(iii)

The Triangle and its Properties Exercise 6.3

Solution:

We know, sum of all the interior angles is equal to 180º.

x + 30º + 110º = 180º

x + 140º = 180º

x = 180º140º

x = 40º

 

(iv)

The Triangle and its Properties Exercise 6.3

Solution:

We know, sum of all the interior angles is equal to 180º.

x + 50º + xº = 180º

2x + 50º = 180º

2x = 180º50º

2x = 130º

x = 75º

 

(v)

The Triangle and its Properties Exercise 6.3

We know, sum of all the interior angles is equal to 180º.

x + x + x = 180º

3x = 180º

3x = 180º

x = 60º

 

(vi)

The Triangle and its Properties Exercise 6.3

We know, sum of all the interior angles is equal to 180º.

x + 90º + 2x = 180º

3x + 90º = 180º

3x = 180º – 90º

2x = 90º

x = 45º


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

(i)

The Triangle and its Properties Exercise 6.3

Solution:

We know, y + 120º = 180º (sum of adjacent angles is equal to 180º)

Therefore, y = 180º – 120º

y = 60º.

Also, sum of all the interior angles is equal to 180º.

x + 50º + y = 180º

x + 50º + 60º = 180º (Since y = 60º)

x = 180º – 110º

x = 70º

 

(ii)

The Triangle and its Properties Exercise 6.3

Solution:

We know, y = 80º (Vertically opposite angles are equal)

Therefore, y = 80º.

Also, sum of all the interior angles is equal to 180º.

x + 50º + y = 180º

x + 50º + 80º = 180º (Since y = 60º)

x = 180º – 130º

x = 50º

 

(iii)

The Triangle and its Properties Exercise 6.3

Solution:

We know, sum of all the interior angles is equal to 180º.

60º + 50º + y = 180º

y + 50º + 60º = 180º

y = 180º – 110º

y = 70º

We know, x + y = 180º (sum of adjacent angles is equal to 180º)

x + 70º = 180º

Therefore, x = 180º – 70º

x = 110º.

 

(iv)

The Triangle and its Properties Exercise 6.3

Solution:

We know, x = 60º (Vertically opposite angles are equal)

Therefore, y = 60º.

Also, sum of all the interior angles is equal to 180º.

x + 30º + y = 180º

x + 30º + 60º = 180º (Since y = 60º)

x = 180º – 90º

x = 90º

 

(v)

The Triangle and its Properties Exercise 6.3

Solution:

We know, y = 90º (Vertically opposite angles are equal)

Therefore, y = 90º.

Also, sum of all the interior angles is equal to 180º.

x + 90º + x = 180º

2x + 90º = 180º

2x = 180º – 90º

2x = 90º

x = 45º

 

(vi)

The Triangle and its Properties Exercise 6.3

Solution:

We know, vertically opposite angles are equal.

Therefore, y = x.

Thus, all the interior angles are x, x and x.

Also, sum of all the interior angles is equal to 180º.

x + x + x = 180º

3x = 180º

x = 60º

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