1. Let each adjacent angle be x. ∴ x+x=180° ⇒ 2x=180° ⇒ x=180°/2 =90° Hence, each adjacent angle is 90° ∴ x+x+x=180° ⇒ 3x=180° ⇒ x=60° Class 8 Maths Chapter 3 Exercise 3.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/

    Let each adjacent angle be x.
    ∴ x+x=180° ⇒ 2x=180° ⇒ x=180°/2 =90°
    Hence, each adjacent angle is 90°
    ∴ x+x+x=180° ⇒ 3x=180° ⇒ x=60°

    Class 8 Maths Chapter 3 Exercise 3.3 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/

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  2. Let two adjacent angles be 3x and 2 . x Since the adjacent angles in a parallelogram are supplementary. ∴ 3x+2x=180° ⇒ 5x=180°/5=36° ∴ One angle = 3x=3x36°=108° and another angle = 2x= 2x36° =72° Class 8 Maths Chapter 3 Exercise 3.3 Solution in Video for more answers vist to: https://www.tiwariacadeRead more

    Let two adjacent angles be 3x and 2 . x
    Since the adjacent angles in a parallelogram are supplementary.
    ∴ 3x+2x=180°
    ⇒ 5x=180°/5=36°
    ∴ One angle = 3x=3×36°=108°
    and another angle = 2x= 2×36° =72°

    Class 8 Maths Chapter 3 Exercise 3.3 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/

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  3. ABCD is a quadrilateral in which angles ∠A = ∠C = 110°.Therefore, it could be a kite. Class 8 Maths Chapter 3 Exercise 3.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/

    ABCD is a quadrilateral in which angles ∠A = ∠C = 110°.Therefore, it could be a kite.

    Class 8 Maths Chapter 3 Exercise 3.3 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/

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  4. (i) ∠ D + ∠ B = 180° It can be, but here, it needs not to be. (ii) No, in this case because one pair of opposite sides are equal and another pair of opposite sides are unequal. So, it is not a parallelogram. (iii) No. ∠A ≠ ∠C. Since opposite angles are equal in parallelogram and here opposite anglesRead more

    (i) ∠ D + ∠ B = 180° It can be, but here, it needs not to be.

    (ii) No, in this case because one pair of opposite sides are equal and another pair of opposite sides are unequal. So, it is not a parallelogram.

    (iii) No. ∠A ≠ ∠C. Since opposite angles are equal in parallelogram and here opposite angles are not equal in quadrilateral ABCD. Therefore it is not aparallelogram.

    Class 8 Maths Chapter 3 Exercise 3.3 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/

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  5. (i) AD = BC [Since opposite sides of a parallelogram are equal] (ii) ∠DCB = ∠DAB [Since opposite angles of a parallelogram are equal] (iii) OC = OA [Since diagonals of a parallelogram bisect each other] (iv) m ∠DAB+m∠CDA = 180° [Adjacent angles in a parallelogram are supplementary] Class 8 Maths ChaRead more

    (i) AD = BC [Since opposite sides of a parallelogram are equal]
    (ii) ∠DCB = ∠DAB [Since opposite angles of a parallelogram are equal]
    (iii) OC = OA [Since diagonals of a parallelogram bisect each other]
    (iv) m ∠DAB+m∠CDA = 180°
    [Adjacent angles in a parallelogram are supplementary]

    Class 8 Maths Chapter 3 Exercise 3.3 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/

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