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Given: Quadrilateral ABCD has vertices A(-5,6),B(6,6),C(8,-3), and D(-3,-3). Prove: Quadrilateral ABCD is a parallelogram but is neither a rhombus nor a rectangle.

User Mastiff
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Final answer:

Using the distance formula to calculate the lengths of the sides of quadrilateral ABCD, we can confirm that it is a parallelogram because opposite sides are equal in length. It is not a rhombus as not all sides are equal, and it is not a rectangle since there is no evidence of all angles being right angles.

Step-by-step explanation:

To prove that quadrilateral ABCD with vertices A(-5,6), B(6,6), C(8,-3), and D(-3,-3) is a parallelogram but neither a rhombus nor a rectangle, we need to check the properties that define these shapes.

The definition of a parallelogram is a quadrilateral with both pairs of opposite sides parallel. To establish this, we can demonstrate that both pairs of opposite sides are equal in length using the distance formula: distance = √((x2-x1)² + (y2-y1)²). Calculating the distances of AB, BC, CD, and DA:

  • AB = √((6 - (-5))² + (6 - 6)²) = √(11² + 0²) = 11
  • BC = √((8 - 6)² + (-3 - 6)²) = √(2² + (-9)²) = √85
  • CD = √((-3 - 8)² + (-3 - (-3))²) = √(-11² + 0²) = 11
  • DA = √((-5 - (-3))² + (6 - (-3))²) = √(-2² + 9²) = √85

Since AB = CD and BC = DA, ABCD is a parallelogram. A rhombus has all sides equal in length; as AB and BC are not equal, ABCD is not a rhombus. A rectangle has all angles at 90 degrees; since we do not have enough information to conclude the angles are all right angles, we cannot prove it's a rectangle.

User Jinyoung Kim
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