(i) 126° + 44° = 170° l ||m because sum of interior opposite angles should be 180. (ii) 75° + 75° = 150° l ||m because sum of angles does not obey the property of parallel lines. (iii) 57° + 123° = 180° l ||m due to supplementary angles property of parallel lines. (iv) 98° + 72° + 170° l is not parRead more
(i) 126° + 44° = 170°
l ||m because sum of interior opposite angles should be 180.
(ii) 75° + 75° = 150°
l ||m because sum of angles does not obey the property of parallel lines.
(iii) 57° + 123° = 180°
l ||m due to supplementary angles property of parallel lines.
(iv) 98° + 72° + 170°
l is not parallel to m because sum of angles does not obey the property of
parallel lines.
(i) Given, AB || DE and BC is a transversal line and ∠ABC = 70° ∵ ∠ABC = ∠DGC [Corresponding angles] ∴ ∠DGC = 70° ……….(i) (ii) Given, BC || EF and DE is a transversal line and ∠DGC = 70° ∵ ∠DGC = ∠DEF [Corresponding angles] ∴ ∠DEF = 70° [From equation (i)] Class 7 Maths Chapter 5 Exercise 5.2 for moRead more
(i) Given, AB || DE and BC is a transversal line and ∠ABC = 70°
∵ ∠ABC = ∠DGC [Corresponding angles]
∴ ∠DGC = 70° ……….(i)
(ii) Given, BC || EF and DE is a transversal line and ∠DGC = 70°
∵ ∠DGC = ∠DEF [Corresponding angles]
∴ ∠DEF = 70° [From equation (i)]
(i) Given, l||m and t is transversal line. ∴ Interior vertically opposite angle between lines l and t = 110° ∴ 110° + x = 180° [Supplementary angles] ⇒ x = 180° - 110° = 70° (ii) Given, l||m and t is transversal line. x + 2x = [Interior opposite angles] ⇒ 3x = 180° ⇒ x= 180°/3 = 60° (iii) Given, l||Read more
(i) Given, l||m and t is transversal line.
∴ Interior vertically opposite angle between lines l and t = 110°
∴ 110° + x = 180° [Supplementary angles]
⇒ x = 180° – 110° = 70°
(ii) Given, l||m and t is transversal line.
x + 2x = [Interior opposite angles]
⇒ 3x = 180°
⇒ x= 180°/3 = 60°
(iii) Given, l||m and a||b.
x=100° [Corresponding angles]
Given, p||q and cut by a transversal line. ∵ 125° + e = 180° [Linear pair] ∴ e = 180° - 125° = 55° ……….(i) Now e = f = 55° [Vertically opposite angles] Also a = f = 55° [Alternate interior angles] a = b + 180° [Linear pair] ⇒ 55° + b = 180° [From equation (i)] ⇒ b = 180° - 55° = 125° Now a = c = 55°Read more
Given, p||q and cut by a transversal line.
∵ 125° + e = 180° [Linear pair]
∴ e = 180° – 125° = 55° ……….(i)
Now e = f = 55° [Vertically opposite angles]
Also a = f = 55° [Alternate interior angles]
a = b + 180° [Linear pair]
⇒ 55° + b = 180° [From equation (i)]
⇒ b = 180° – 55° = 125°
Now a = c = 55° and b = d = 125° [Vertically opposite angles]
Thus, a = 55°,b = 125°, c = 55°, d = 125°, e = 55° and f = 55°.
(i) The pairs of corresponding angles: ∠1, ∠5; ∠2, ∠6; ∠4, ∠8 and ∠3, ∠7 (ii) The pairs of alternate interior angles are: ∠3, ∠5 and ∠2, ∠8 (iii) The pair of interior angles on the same side of the transversal: ∠3, ∠8 and ∠2, ∠5 (iv) The vertically opposite angles are: ∠1, ∠3; ∠2, ∠4; ∠6, ∠8 and ∠5,Read more
(i) The pairs of corresponding angles:
∠1, ∠5; ∠2, ∠6; ∠4, ∠8 and ∠3, ∠7
(ii) The pairs of alternate interior angles are:
∠3, ∠5 and ∠2, ∠8
(iii) The pair of interior angles on the same side of the transversal:
∠3, ∠8 and ∠2, ∠5
(iv) The vertically opposite angles are:
∠1, ∠3; ∠2, ∠4; ∠6, ∠8 and ∠5, ∠7
In the given figures below, decide whether l is parallel to m.
(i) 126° + 44° = 170° l ||m because sum of interior opposite angles should be 180. (ii) 75° + 75° = 150° l ||m because sum of angles does not obey the property of parallel lines. (iii) 57° + 123° = 180° l ||m due to supplementary angles property of parallel lines. (iv) 98° + 72° + 170° l is not parRead more
(i) 126° + 44° = 170°
l ||m because sum of interior opposite angles should be 180.
(ii) 75° + 75° = 150°
l ||m because sum of angles does not obey the property of parallel lines.
(iii) 57° + 123° = 180°
l ||m due to supplementary angles property of parallel lines.
(iv) 98° + 72° + 170°
l is not parallel to m because sum of angles does not obey the property of
parallel lines.
Class 7 Maths Chapter 5 Exercise 5.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
In the given figure, the arms of two angles are parallel. If ∆ABC = 70°, then find:, (i) ∠DGC (ii) ∠ DEF
(i) Given, AB || DE and BC is a transversal line and ∠ABC = 70° ∵ ∠ABC = ∠DGC [Corresponding angles] ∴ ∠DGC = 70° ……….(i) (ii) Given, BC || EF and DE is a transversal line and ∠DGC = 70° ∵ ∠DGC = ∠DEF [Corresponding angles] ∴ ∠DEF = 70° [From equation (i)] Class 7 Maths Chapter 5 Exercise 5.2 for moRead more
(i) Given, AB || DE and BC is a transversal line and ∠ABC = 70°
∵ ∠ABC = ∠DGC [Corresponding angles]
∴ ∠DGC = 70° ……….(i)
(ii) Given, BC || EF and DE is a transversal line and ∠DGC = 70°
∵ ∠DGC = ∠DEF [Corresponding angles]
∴ ∠DEF = 70° [From equation (i)]
Class 7 Maths Chapter 5 Exercise 5.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
Find the values of x in each of the following figures if l||m
(i) Given, l||m and t is transversal line. ∴ Interior vertically opposite angle between lines l and t = 110° ∴ 110° + x = 180° [Supplementary angles] ⇒ x = 180° - 110° = 70° (ii) Given, l||m and t is transversal line. x + 2x = [Interior opposite angles] ⇒ 3x = 180° ⇒ x= 180°/3 = 60° (iii) Given, l||Read more
(i) Given, l||m and t is transversal line.
∴ Interior vertically opposite angle between lines l and t = 110°
∴ 110° + x = 180° [Supplementary angles]
⇒ x = 180° – 110° = 70°
(ii) Given, l||m and t is transversal line.
x + 2x = [Interior opposite angles]
⇒ 3x = 180°
⇒ x= 180°/3 = 60°
(iii) Given, l||m and a||b.
x=100° [Corresponding angles]
Class 7 Maths Chapter 5 Exercise 5.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
In the adjoining figure, p||q. Find the unknown angles.
Given, p||q and cut by a transversal line. ∵ 125° + e = 180° [Linear pair] ∴ e = 180° - 125° = 55° ……….(i) Now e = f = 55° [Vertically opposite angles] Also a = f = 55° [Alternate interior angles] a = b + 180° [Linear pair] ⇒ 55° + b = 180° [From equation (i)] ⇒ b = 180° - 55° = 125° Now a = c = 55°Read more
Given, p||q and cut by a transversal line.
∵ 125° + e = 180° [Linear pair]
∴ e = 180° – 125° = 55° ……….(i)
Now e = f = 55° [Vertically opposite angles]
Also a = f = 55° [Alternate interior angles]
a = b + 180° [Linear pair]
⇒ 55° + b = 180° [From equation (i)]
⇒ b = 180° – 55° = 125°
Now a = c = 55° and b = d = 125° [Vertically opposite angles]
Thus, a = 55°,b = 125°, c = 55°, d = 125°, e = 55° and f = 55°.
Class 7 Maths Chapter 5 Exercise 5.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/
In the adjoining figure, identify:(i) the pairs of corresponding angles. (ii) the pairs of alternate interior angles. (iii) the pairs of interior angles on the same side of the transversal. (iv) the vertically opposite angles.
(i) The pairs of corresponding angles: ∠1, ∠5; ∠2, ∠6; ∠4, ∠8 and ∠3, ∠7 (ii) The pairs of alternate interior angles are: ∠3, ∠5 and ∠2, ∠8 (iii) The pair of interior angles on the same side of the transversal: ∠3, ∠8 and ∠2, ∠5 (iv) The vertically opposite angles are: ∠1, ∠3; ∠2, ∠4; ∠6, ∠8 and ∠5,Read more
(i) The pairs of corresponding angles:
∠1, ∠5; ∠2, ∠6; ∠4, ∠8 and ∠3, ∠7
(ii) The pairs of alternate interior angles are:
∠3, ∠5 and ∠2, ∠8
(iii) The pair of interior angles on the same side of the transversal:
∠3, ∠8 and ∠2, ∠5
(iv) The vertically opposite angles are:
∠1, ∠3; ∠2, ∠4; ∠6, ∠8 and ∠5, ∠7
Class 7 Maths Chapter 5 Exercise 5.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-5/