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  1. Asked: In:

    AM is a median of a triangle ABC. Is AB + BC + CA > 2AM? (Consider the sides of triangles ∆ABM and ∆AMC.)

    Aagar

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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  2. Asked: In:

    Take any point O in the interior of a triangle PQR. Is: (i) OP + OQ > PQ ? (ii) OQ + OR > QR ? (iii) OR + OP > RP ?

    Aagar

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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  3. Asked: In:

    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

    Aagar

    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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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  4. Asked: In:

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

    Aagar

    (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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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  6. Asked: In:

    Find the value of unknown x in the following diagrams:

    Aagar

    (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:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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  7. Asked: In:

    Find the value of the unknown interior angle x in the following figures:

    Aagar

    Since, Exterior angle = Sum of interior opposite angles, therefore (i) x + 50° = 115° ⇒ x = 115° - 50° = 65° (ii) 70° + x = 100° ⇒ x = 100° - 70° = 30° (iii) x + 90° = 125° ⇒ x = 120° - 90° = 35° (iv) 60° + x = 120° ⇒ x = 120° - 60° = 60° (v) 30° + x = 80° ⇒ x = 80° - 30° = 50° (vi) x + 35° = 75° ⇒Read more

    Since, Exterior angle = Sum of interior opposite angles, therefore
    (i) x + 50° = 115° ⇒ x = 115° – 50° = 65°
    (ii) 70° + x = 100° ⇒ x = 100° – 70° = 30°
    (iii) x + 90° = 125° ⇒ x = 120° – 90° = 35°
    (iv) 60° + x = 120° ⇒ x = 120° – 60° = 60°
    (v) 30° + x = 80° ⇒ x = 80° – 30° = 50°
    (vi) x + 35° = 75° ⇒ x = 75° – 35° = 40°

    Class 7 Maths Chapter 6 Exercise 6.2

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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    • 7
  8. Asked: In:

    Find the value of the unknown exterior angle x in the following diagrams:

    Aagar

    Since, Exterior angle = Sum of interior opposite angles, therefore (i) x = 50° + 70° = 120° (ii) x = 65° + 45° = 110° (iii) x = 30° + 40° = 70° (iv) x = 60° + 60° = 120° (v) x = 50° + 50° =100° (vi) x = 60° + 30° = 90° Class 7 Maths Chapter 6 Exercise 6.2 for more answers vist to: https://www.tiwariRead more

    Since, Exterior angle = Sum of interior opposite angles, therefore
    (i) x = 50° + 70° = 120°
    (ii) x = 65° + 45° = 110°
    (iii) x = 30° + 40° = 70°
    (iv) x = 60° + 60° = 120°
    (v) x = 50° + 50° =100°
    (vi) x = 60° + 30° = 90°

    Class 7 Maths Chapter 6 Exercise 6.2

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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  9. Asked: In:

    Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same.

    Aagar

    Isosceles triangle means any two sides are same. Take ∆ABC and draw the median when AB = AC. AL is the median and altitude of the given triangle. Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

    Isosceles triangle means any two sides are same.
    Take ∆ABC and draw the median when AB = AC.
    AL is the median and altitude of the given triangle.

    Class 7 Maths Chapter 6 Exercise 6.1

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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  11. Asked: In:

    Draw rough sketches for the following: (a) In ∆ABC, BE is a median. (b) In ∆PQR, PQ and PR are altitudes of the triangle. (c) In ∆XYZ, YL is an altitude in the exterior of the triangle.

    Aagar

    follow the link for more answers and video: Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

    follow the link for more answers and video:

    Class 7 Maths Chapter 6 Exercise 6.1

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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  12. Asked: In:

    In ∆ PQR, D is the mid-point of QR. PM is _______________. PD is ________________ Is QM = MR?

    Aagar

    Given: QD = DR ∴ PM is altitude. PD is median. No, QM ≠ MR as D is the mid-point of QR. Class 7 Maths Chapter 6 Exercise 6.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

    Given: QD = DR
    ∴ PM is altitude.
    PD is median.
    No, QM ≠ MR as D is the mid-point of QR.

    Class 7 Maths Chapter 6 Exercise 6.1

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/

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