A quadrilateral with vertices K(0,4), A(3,6), T(6,2) and E(0,-2) is a ________.
Question
A quadrilateral with vertices K(0,4), A(3,6), T(6,2) and E(0,-2) is a ________.
Solution 1
To determine the type of the quadrilateral KATE with vertices K(0,4), A(3,6), T(6,2) and E(0,-2), we need to calculate the lengths of the sides and diagonals.
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Calculate the lengths of the sides:
- KA = sqrt[(3-0)² + (6-4)²] = sqrt[9 + 4] = sqrt[13]
- AT = sqrt[(6-3)² + (2-6)²] = sqrt[9 + 16] = sqrt[25] = 5
- TE = sqrt[(0-6)² + (-2-2)²] = sqrt[36 + 16] = sqrt[52]
- EK = sqrt[(0-0)² + (-2-4)²] = sqrt[0 + 36] = sqrt[36] = 6
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Calculate the lengths of the diagonals:
- KT = sqrt[(6-0)² + (2-4)²] = sqrt[36 + 4] = sqrt[40]
- AE = sqrt[(0-3)² + (-2-6)²] = sqrt[9 + 64] = sqrt[73]
From the above calculations, we can see that the lengths of the opposite sides are not equal (KA ≠ TE and AT ≠ EK), so the quadrilateral is not a parallelogram. Also, the diagonals are not equal (KT ≠ AE), so it's not a rectangle or a rhombus.
Therefore, the quadrilateral KATE with vertices K(0,4), A(3,6), T(6,2) and E(0,-2) is a general quadrilateral.
Solution 2
To determine the type of the quadrilateral KATE with vertices K(0,4), A(3,6), T(6,2) and E(0,-2), we need to calculate the lengths of the sides and diagonals.
Step 1: Calculate the lengths of the sides
- KA = sqrt[(3-0)² + (6-4)²] = sqrt[9 + 4] = sqrt[13]
- AT = sqrt[(6-3)² + (2-6)²] = sqrt[9 + 16] = sqrt[25] = 5
- TE = sqrt[(0-6)² + (-2-2)²] = sqrt[36 + 16] = sqrt[52]
- EK = sqrt[(0-0)² + (-2-4)²] = sqrt[0 + 36] = sqrt[36] = 6
Step 2: Calculate the lengths of the diagonals
- KT = sqrt[(6-0)² + (2-4)²] = sqrt[36 + 4] = sqrt[40]
- AE = sqrt[(0-3)² + (-2-6)²] = sqrt[9 + 64] = sqrt[73]
Step 3: Compare the lengths We see that KA ≠ AT ≠ TE ≠ EK and KT ≠ AE, so the quadrilateral is not a rectangle, square, parallelogram, or rhombus.
Therefore, the quadrilateral KATE is a general quadrilateral.
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