Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. Therefore, In ∆ABM, AB + BM > AM ... (i) In ∆AMC, AC + MC > AM ... (ii) Adding eq. (i) and (ii), AB + BM + AC + MC > AM + AM ⇒ AB + AC + (BM + MC) > 2AM ⇒ AB + AC + BC > 2AM HenceRead more
Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
Therefore, In ∆ABM, AB + BM > AM … (i)
In ∆AMC, AC + MC > AM … (ii)
Adding eq. (i) and (ii),
AB + BM + AC + MC > AM + AM
⇒ AB + AC + (BM + MC) > 2AM
⇒ AB + AC + BC > 2AM
Hence, it is true.
Join OR, OQ and OP. (i) Is OP + OQ > PQ ? Yes, POQ form a triangle. (ii) Is OQ + OR > QR ? Yes, RQO form a triangle. (iii) Is OR + OP > RP ? Yes, ROP form a triangle. Class 7 Maths Chapter 6 Exercise 6.4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/cRead more
Join OR, OQ and OP.
(i) Is OP + OQ > PQ ?
Yes, POQ form a triangle.
(ii) Is OQ + OR > QR ?
Yes, RQO form a triangle.
(iii) Is OR + OP > RP ?
Yes, ROP form a triangle.
Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side. (i) 2cm, 3cm, 5cm 2 + 3 > 5 No 2 + 5 > 3 Yes 3 + 5 > 2 Yes This triangle is not possible. (ii) 3 cm, 6 cm, 7 cm 3 + 6 > 7 Yes 6 + 7 > 3 Yes 3 + 7 > 6 Yes This triRead more
Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side.
(i) 2cm, 3cm, 5cm
2 + 3 > 5 No
2 + 5 > 3 Yes
3 + 5 > 2 Yes
This triangle is not possible.
(ii) 3 cm, 6 cm, 7 cm
3 + 6 > 7 Yes
6 + 7 > 3 Yes
3 + 7 > 6 Yes
This triangle is possible.
(iii) 6 cm, 3 cm, 2 cm
6 + 3 > 2 Yes
6 + 2 > 3 Yes
2 + 3 > 6 No
This triangle is not possible.
(i) 50°+ x =120° [Exterior angle property of a ∆ ] ⇒ x =120°- 50° = 70° Now, 50° + x + y =180° [Angle sum property of a ∆ ] ⇒ 50° + 70°+ y = 180° ⇒ 120° + y = 180° ⇒ y =180° - 120° = 60° (ii) y = 80° ……….(i) [Vertically opposite angle] Now, 50° + x + y =180° [Angle sum property of a ∆ ] ⇒ 50° + 80°Read more
(i) 50°+ x =120° [Exterior angle property of a ∆ ]
⇒ x =120°- 50° = 70°
Now, 50° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 70°+ y = 180°
⇒ 120° + y = 180°
⇒ y =180° – 120° = 60°
(ii) y = 80° ……….(i) [Vertically opposite angle]
Now, 50° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 80° + y = 180° [From equation (i)]
⇒ 130° + y =180°
⇒ y =180° – 130° = 50°
(iii) 50°+ 60°= x [Exterior angle property of a ∆ ]
⇒ x =110°
Now 50° + 60°+ y =180° [Angle sum property of a ∆ ]
⇒ 110° + y =180°
⇒ y = 180° – 110°
⇒ y = 70°
(iv) x = 60° ……….(i) [Vertically opposite angle]
Now, 30° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 60° + y =180° [From equation (i)]
⇒ 90° + y =180°
⇒ y =180° – 90° = 90°
(v) y = 90° ……….(i) [Vertically opposite angle]
Now, y + x + x =180° [Angle sum property of a ∆ ]
⇒ 90°+ 2x =180° [From equation (i)]
⇒ 2x = 180°- 90°
⇒ 2x = 90°
⇒ x = 90°/2 = 45°
(vi) x = y ……….(i) [Vertically opposite angle]
Now, x + x + y =180° [Angle sum property of a ]
⇒ 2x + x =180° [From equation (i)]
⇒ 3x = 180°
⇒ 180° / 3 = 60°
(i) In ∆ABC, ∠BAC + ∠ACB + ∠ABC = 180° [By angle sum property of a triangle] ⇒ x +50° + 60° =180° ⇒ x +110°+180° ⇒ x =180°-110° = 70° (ii) In ∆ PQR, ∠RPQ + ∠PQR + ∠RPQ = 180° [By angle sum property of a triangle] ⇒ 90° + 30° + x =180° ⇒ x = 120°= 180° ⇒ x =180°- 120° = 60° (iii) In ∆ XYZ, ∠ZXY + ∠XYRead more
(i) In ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180° [By angle sum property of a triangle]
⇒ x +50° + 60° =180°
⇒ x +110°+180°
⇒ x =180°-110° = 70°
(ii) In ∆ PQR,
∠RPQ + ∠PQR + ∠RPQ = 180° [By angle sum property of a triangle]
⇒ 90° + 30° + x =180°
⇒ x = 120°= 180°
⇒ x =180°- 120° = 60°
(iii) In ∆ XYZ,
∠ZXY + ∠XYZ + ∠YZX = 180° [By angle sum property of a triangle]
⇒ 30° +110° + x = 180°
⇒ x + 140°= 180°
⇒ x =180°- 140°= 40°
(iv) In the given isosceles triangle,
x + x + 50° = 180° [By angle sum property of a triangle]
⇒ 2x + 50° = 180°
⇒ 2x =180° – 50°
⇒ 2x =130°
⇒x=130°/2 = 65°
(v) In the given equilateral triangle,
x + x + x = 180° [By angle sum property of a triangle]
⇒ 3x = 180°
⇒ x = 180° / 3 = 60°
(vi) In the given right angled triangle,
X + 2x + 90° = 180° [By angle sum property of a triangle]
⇒ 3x + 90°=180°
⇒ 3x =180°-90°
⇒3x=90°
⇒ x = 90°/3 = 30°
AM is a median of a triangle ABC. Is AB + BC + CA > 2AM? (Consider the sides of triangles ∆ABM and ∆AMC.)
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. Therefore, In ∆ABM, AB + BM > AM ... (i) In ∆AMC, AC + MC > AM ... (ii) Adding eq. (i) and (ii), AB + BM + AC + MC > AM + AM ⇒ AB + AC + (BM + MC) > 2AM ⇒ AB + AC + BC > 2AM HenceRead more
Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
Therefore, In ∆ABM, AB + BM > AM … (i)
In ∆AMC, AC + MC > AM … (ii)
Adding eq. (i) and (ii),
AB + BM + AC + MC > AM + AM
⇒ AB + AC + (BM + MC) > 2AM
⇒ AB + AC + BC > 2AM
Hence, it is true.
Class 7 Maths Chapter 6 Exercise 6.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Take any point O in the interior of a triangle PQR. Is: (i) OP + OQ > PQ ? (ii) OQ + OR > QR ? (iii) OR + OP > RP ?
Join OR, OQ and OP. (i) Is OP + OQ > PQ ? Yes, POQ form a triangle. (ii) Is OQ + OR > QR ? Yes, RQO form a triangle. (iii) Is OR + OP > RP ? Yes, ROP form a triangle. Class 7 Maths Chapter 6 Exercise 6.4 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/cRead more
Join OR, OQ and OP.
(i) Is OP + OQ > PQ ?
Yes, POQ form a triangle.
(ii) Is OQ + OR > QR ?
Yes, RQO form a triangle.
(iii) Is OR + OP > RP ?
Yes, ROP form a triangle.
Class 7 Maths Chapter 6 Exercise 6.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Is it possible to have a triangle with the following sides? (i) 2 cm, 3 cm, 5 cm (ii) 3 cm, 6 cm, 7 cm (iii) 6 cm, 3 cm, 2 cm
Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side. (i) 2cm, 3cm, 5cm 2 + 3 > 5 No 2 + 5 > 3 Yes 3 + 5 > 2 Yes This triangle is not possible. (ii) 3 cm, 6 cm, 7 cm 3 + 6 > 7 Yes 6 + 7 > 3 Yes 3 + 7 > 6 Yes This triRead more
Since, a triangle is possible whose sum of the lengths of any two sides would be greater than the length of third side.
(i) 2cm, 3cm, 5cm
2 + 3 > 5 No
2 + 5 > 3 Yes
3 + 5 > 2 Yes
This triangle is not possible.
(ii) 3 cm, 6 cm, 7 cm
3 + 6 > 7 Yes
6 + 7 > 3 Yes
3 + 7 > 6 Yes
This triangle is possible.
(iii) 6 cm, 3 cm, 2 cm
6 + 3 > 2 Yes
6 + 2 > 3 Yes
2 + 3 > 6 No
This triangle is not possible.
Class 7 Maths Chapter 6 Exercise 6.4
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Find the values of the unknowns x and y in the following diagrams:
(i) 50°+ x =120° [Exterior angle property of a ∆ ] ⇒ x =120°- 50° = 70° Now, 50° + x + y =180° [Angle sum property of a ∆ ] ⇒ 50° + 70°+ y = 180° ⇒ 120° + y = 180° ⇒ y =180° - 120° = 60° (ii) y = 80° ……….(i) [Vertically opposite angle] Now, 50° + x + y =180° [Angle sum property of a ∆ ] ⇒ 50° + 80°Read more
(i) 50°+ x =120° [Exterior angle property of a ∆ ]
⇒ x =120°- 50° = 70°
Now, 50° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 70°+ y = 180°
⇒ 120° + y = 180°
⇒ y =180° – 120° = 60°
(ii) y = 80° ……….(i) [Vertically opposite angle]
Now, 50° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 80° + y = 180° [From equation (i)]
⇒ 130° + y =180°
⇒ y =180° – 130° = 50°
(iii) 50°+ 60°= x [Exterior angle property of a ∆ ]
⇒ x =110°
Now 50° + 60°+ y =180° [Angle sum property of a ∆ ]
⇒ 110° + y =180°
⇒ y = 180° – 110°
⇒ y = 70°
(iv) x = 60° ……….(i) [Vertically opposite angle]
Now, 30° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 60° + y =180° [From equation (i)]
⇒ 90° + y =180°
⇒ y =180° – 90° = 90°
(v) y = 90° ……….(i) [Vertically opposite angle]
Now, y + x + x =180° [Angle sum property of a ∆ ]
⇒ 90°+ 2x =180° [From equation (i)]
⇒ 2x = 180°- 90°
⇒ 2x = 90°
⇒ x = 90°/2 = 45°
(vi) x = y ……….(i) [Vertically opposite angle]
Now, x + x + y =180° [Angle sum property of a ]
⇒ 2x + x =180° [From equation (i)]
⇒ 3x = 180°
⇒ 180° / 3 = 60°
Class 7 Maths Chapter 6 Exercise 6.3
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/
Find the value of unknown x in the following diagrams:
(i) In ∆ABC, ∠BAC + ∠ACB + ∠ABC = 180° [By angle sum property of a triangle] ⇒ x +50° + 60° =180° ⇒ x +110°+180° ⇒ x =180°-110° = 70° (ii) In ∆ PQR, ∠RPQ + ∠PQR + ∠RPQ = 180° [By angle sum property of a triangle] ⇒ 90° + 30° + x =180° ⇒ x = 120°= 180° ⇒ x =180°- 120° = 60° (iii) In ∆ XYZ, ∠ZXY + ∠XYRead more
(i) In ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180° [By angle sum property of a triangle]
⇒ x +50° + 60° =180°
⇒ x +110°+180°
⇒ x =180°-110° = 70°
(ii) In ∆ PQR,
∠RPQ + ∠PQR + ∠RPQ = 180° [By angle sum property of a triangle]
⇒ 90° + 30° + x =180°
⇒ x = 120°= 180°
⇒ x =180°- 120° = 60°
(iii) In ∆ XYZ,
∠ZXY + ∠XYZ + ∠YZX = 180° [By angle sum property of a triangle]
⇒ 30° +110° + x = 180°
⇒ x + 140°= 180°
⇒ x =180°- 140°= 40°
(iv) In the given isosceles triangle,
x + x + 50° = 180° [By angle sum property of a triangle]
⇒ 2x + 50° = 180°
⇒ 2x =180° – 50°
⇒ 2x =130°
⇒x=130°/2 = 65°
(v) In the given equilateral triangle,
x + x + x = 180° [By angle sum property of a triangle]
⇒ 3x = 180°
⇒ x = 180° / 3 = 60°
(vi) In the given right angled triangle,
X + 2x + 90° = 180° [By angle sum property of a triangle]
⇒ 3x + 90°=180°
⇒ 3x =180°-90°
⇒3x=90°
⇒ x = 90°/3 = 30°
Class 7 Maths Chapter 6 Exercise 6.2
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/