GSEB Solutions Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.3

Gujarat BoardĀ 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 ātriangleā we have
50Ā° + 60Ā° + x = 180Ā°
or 110Ā° + x = 180Ā°
or x = 180Ā° – 110Ā° = 70Ā°
Thus, the required value of x is 70Ā°.

(ii) Using the āangle sum property of a triangleā,
we have
30Ā° + 90Ā° + x = 180Ā°
[the ā is right angled at P.]
or 120Ā° + x = 180Ā°
or x = 180Ā° – 120Ā° = 60Ā°
Thus, the required value of x is 60Ā°.

(iii) Using the āangle sum property of a triangleā, we have
30Ā° + 110Ā° + x = 180Ā°
or 140Ā° + x = 180Ā°
or x = 180Ā° – 140Ā° = 40Ā°
Thus, the required value of x is 40Ā°.

(iv) Using the āangle sum property of a triangleā,
we have
x + x + 50Ā° = 180Ā°
ā“ 2x + 50Ā° = 180Ā°
or 2x = 180Ā° – 50Ā° = 130Ā°
or $$\frac { 2x }{ 2 }$$ = $$\frac { 130Ā° }{ 2 }$$
or x = 65Ā°

(v) Using the āangle sum property of a triangleā, we have
x + x + x = 180Ā°
or 3x = 180Ā°
or $$\frac { 3x }{ 3 }$$ = $$\frac { 180Ā° }{ 3 }$$
or x = 60Ā°

(vi) Using the āangle sum property of a triangleā, we have
x + 2x + 90Ā° =180Ā°
or 3x + 90Ā° = 180Ā°
or 3x = 180Ā° – 90Ā° = 90Ā°
or $$\frac { 3x }{ 3 }$$ = $$\frac { 90Ā° }{ 3 }$$
[Dividing both sides by 3]
or x = 30Ā°

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

Solution:
(i) āµ Angles y and 120Ā° form a linear pair.
ā“ y + 120Ā° = 180Ā°
or y = 180Ā° – 120Ā° = 60Ā°
Now, using the angle sum property of a triangle, we have
x + y + 50Ā° = 180Ā°
or x + 60Ā° + 50Ā° = 180Ā°
or x + 110Ā° =180Ā°
or x = 180Ā° – 110Ā° = 70Ā°
Thus, $$\left.\begin{array}{l} x=70^{\circ} \\ y=60^{\circ} \end{array}\right\}$$

(ii) āµ y and 80Ā° angle are vertically opposite angles, then y = 80Ā°
Now x + y + 50Ā° = 180Ā°
[Using angle sum property]
or x + 80Ā° + 50Ā° = 180Ā°
or x + 130Ā° = 180Ā°
or x = 180Ā° – 130Ā° = 50Ā°
Thus, $$\left.\begin{array}{l} x=50^{\circ} \\ y=80^{\circ} \end{array}\right\}$$

(iii) Using the angle sum property of triangle, we have
50Ā° + 60Ā° + y = 180Ā°
or y + 110Ā° = 180Ā°
or y = 180Ā° – 110Ā° = 70Ā°
Again, x and y form a linear pair.
ā“ x + y = 180Ā°
or x + 70Ā° = 180Ā°
or x = 180Ā° – 70Ā° = 110Ā°
Thus, $$\left.\begin{array}{l} x=110^{\circ} \\ y=70^{\circ} \end{array}\right\}$$

(iv) āµ x and 60Ā° angle are vertically opposite angles
ā“ x = 60Ā°
Now, using the angle sum property of triangle, we have
x + y + 30Ā° = 180Ā°
or 60 + y + 30Ā° = 180Ā°
or y + 90Ā° = 180Ā°
or y = 180Ā° – 90Ā° =
Thus, $$\left.\begin{array}{l} x=60^{\circ} \\ y=90^{\circ} \end{array}\right\}$$

(v) āµ y and 90Ā° are vertically opposite angles, then y = 90Ā°
Now, using the angle sum property of triangles, we have
x + x + y = 180Ā°
2x + y = 180Ā°
or 2x + 90Ā° = 180Ā°
or 2x = 180Ā° – 90Ā° = 90Ā°
or $$\frac { 2x }{ 2 }$$ = $$\frac { 90Ā° }{ 2 }$$ or x = 45Ā°
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.
ā“ Using the angle sum property
x + x + y = 180Ā°
or 2x+ y = 180Ā°
or 2x + x = 180Ā°
[x = y vertically opposite angles]
or 3x = 180Ā°
or $$\frac { 3x }{ 3 }$$ = $$\frac { 180Ā° }{ 3 }$$
ā“ x = 60Ā°
But y = x
ā“ y = 60Ā°
Thus, $$\left.\begin{array}{l} x=60^{\circ} \\ y=60^{\circ} \end{array}\right\}$$