Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3

Question 1.

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

Solution:

(i) Using the angle sum property of a ‘triangle’ we have

50° + 60° + x = 180°

or 110° + x = 180°

or x = 180° – 110° = 70°

Thus, the required value of x is 70°.

(ii) Using the ‘angle sum property of a triangle’,

we have

30° + 90° + x = 180°

[the ∆ is right angled at P.]

or 120° + x = 180°

or x = 180° – 120° = 60°

Thus, the required value of x is 60°.

(iii) Using the ‘angle sum property of a triangle’, we have

30° + 110° + x = 180°

or 140° + x = 180°

or x = 180° – 140° = 40°

Thus, the required value of x is 40°.

(iv) Using the ‘angle sum property of a triangle’,

we have

x + x + 50° = 180°

∴ 2x + 50° = 180°

or 2x = 180° – 50° = 130°

or \(\frac { 2x }{ 2 }\) = \(\frac { 130° }{ 2 }\)

or x = 65°

(v) Using the ‘angle sum property of a triangle’, we have

x + x + x = 180°

or 3x = 180°

or \(\frac { 3x }{ 3 }\) = \(\frac { 180° }{ 3 }\)

or x = 60°

(vi) Using the ‘angle sum property of a triangle’, we have

x + 2x + 90° =180°

or 3x + 90° = 180°

or 3x = 180° – 90° = 90°

or \(\frac { 3x }{ 3 }\) = \(\frac { 90° }{ 3 }\)

[Dividing both sides by 3]

or x = 30°

Question 2.

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

Solution:

(i) ∵ Angles y and 120° form a linear pair.

∴ y + 120° = 180°

or y = 180° – 120° = 60°

Now, using the angle sum property of a triangle, we have

x + y + 50° = 180°

or x + 60° + 50° = 180°

or x + 110° =180°

or x = 180° – 110° = 70°

Thus, \(\left.\begin{array}{l}

x=70^{\circ} \\

y=60^{\circ}

\end{array}\right\}\)

(ii) ∵ y and 80° angle are vertically opposite angles, then y = 80°

Now x + y + 50° = 180°

[Using angle sum property]

or x + 80° + 50° = 180°

or x + 130° = 180°

or x = 180° – 130° = 50°

Thus, \(\left.\begin{array}{l}

x=50^{\circ} \\

y=80^{\circ}

\end{array}\right\}\)

(iii) Using the angle sum property of triangle, we have

50° + 60° + y = 180°

or y + 110° = 180°

or y = 180° – 110° = 70°

Again, x and y form a linear pair.

∴ x + y = 180°

or x + 70° = 180°

or x = 180° – 70° = 110°

Thus, \(\left.\begin{array}{l}

x=110^{\circ} \\

y=70^{\circ}

\end{array}\right\}\)

(iv) ∵ x and 60° angle are vertically opposite angles

∴ x = 60°

Now, using the angle sum property of triangle, we have

x + y + 30° = 180°

or 60 + y + 30° = 180°

or y + 90° = 180°

or y = 180° – 90° =

Thus, \(\left.\begin{array}{l}

x=60^{\circ} \\

y=90^{\circ}

\end{array}\right\}\)

(v) ∵ y and 90° are vertically opposite angles, then y = 90°

Now, using the angle sum property of triangles, we have

x + x + y = 180°

2x + y = 180°

or 2x + 90° = 180°

or 2x = 180° – 90° = 90°

or \(\frac { 2x }{ 2 }\) = \(\frac { 90° }{ 2 }\) or x = 45°

Thus, \(\left.\begin{array}{l}

x=45^{\circ} \\

y=90^{\circ}

\end{array}\right\}\)

(vi) One angle of the triangle = y

Each of the other two angles is equal to their vertically opposite angle x.

∴ Using the angle sum property

x + x + y = 180°

or 2x+ y = 180°

or 2x + x = 180°

[x = y vertically opposite angles]

or 3x = 180°

or \(\frac { 3x }{ 3 }\) = \(\frac { 180° }{ 3 }\)

∴ x = 60°

But y = x

∴ y = 60°

Thus, \(\left.\begin{array}{l}

x=60^{\circ} \\

y=60^{\circ}

\end{array}\right\}\)