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Quadrilateral ABCD has vertices A(−2,1), B(4,4), C(5,0), and D(−1,−2).

Which side is the longest
A.BC
B.CD
C.DA
D.AB


Sagot :

Step-by-step explanation:

Refer to the attachment ♪♪

View image VirαtKσhli

Using the distance formula,  [tex]\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}[/tex] to find the length of each sides, the longest side is about 6.71 units which is: d. AB

To find the longest side, we would apply the distance formula to find the length of each side.

Given:

  • A(−2,1)
  • B(4,4)
  • C(5,0)
  • D(−1,−2)

Distance formula = [tex]\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}[/tex]

Length of AB:

  • Let,

[tex]A(-2,1) = (x_1, y_1)\\\\B(4,4) = (x_2, y_2)[/tex]

  • Substitute

[tex]AB = \sqrt{(4 - 1)^2 + (4 - (-2))^2}\\\\AB = \sqrt{9 + 36}\\\\AB = \sqrt{45} \\\\AB = 6.71[/tex]

Length of BC:

  • Let,

[tex]B(4,4) = (x_1, y_1)\\\\C(5,0) = (x_2, y_2)[/tex]

  • Substitute

[tex]BC= \sqrt{(0-4)^2 + (5-4)^2}\\\\BC = \sqrt{16 + 1}\\\\BC = \sqrt{17} \\\\BC =4.12[/tex]

Length of CD:

  • Let,

[tex]C(5,0) = (x_1, y_1)\\\\ D(-1,-2) = (x_2, y_2)[/tex]

  • Substitute

[tex]CD = \sqrt{(-2 - 0)^2 + (-1 - 5)^2}\\\\CD = \sqrt{4 + 36}\\\\CD = \sqrt{40} \\\\CD =6.32[/tex]

Length of DA:

  • Let,

[tex]D(-1,-2) = (x_1, y_1)\\\\ A(-2,1) = (x_2, y_2)[/tex]

  • Substitute

[tex]DA = \sqrt{(-2 - 1)^2 + (-1 - (-2))^2}\\\\DA= \sqrt{9 + 1}\\\\DA = \sqrt{10} \\\\DA = 3.16[/tex]

Therefore, the using the distance formula,  [tex]\sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}[/tex] to find the length of each sides, the longest side is about 6.71 units which is: d. AB

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