Select the correct answers from each drop-down menu.

Complete the steps in the proof that show quadrilateral KITE with vertices [tex]$K (0,-2), I (1,2), T (7,5)$[/tex], and [tex]$E (4,-1)$[/tex] is a kite.

Using the distance formula,
[tex]KI = \sqrt{(2 - (-2))^2 + (1 - 0)^2} = \sqrt{17}[/tex]

[tex]KE = \square[/tex]

[tex]IT = \square[/tex]

Therefore, KITE is a kite because [tex]\square[/tex]



Answer :

Certainly! Let's complete the steps to prove that the quadrilateral [tex]\( KITE \)[/tex] with vertices [tex]\( K (0,-2) \)[/tex], [tex]\( I (1,2) \)[/tex], [tex]\( T (7,5) \)[/tex], and [tex]\( E (4,-1) \)[/tex] is a kite.

Using the distance formula, the distances between the points are calculated as follows:

1. Calculate [tex]\( KI \)[/tex]:
[tex]\[ KI = \sqrt{(I_x - K_x)^2 + (I_y - K_y)^2} = \sqrt{(1 - 0)^2 + (2 - (-2))^2} = \sqrt{1 + 16} = \sqrt{17} \][/tex]

2. Calculate [tex]\( KE \)[/tex]:
[tex]\[ KE = \sqrt{(E_x - K_x)^2 + (E_y - K_y)^2} = \sqrt{(4 - 0)^2 + (-1 - (-2))^2} = \sqrt{16 + 1} = \sqrt{17} \][/tex]

3. Calculate [tex]\( IT \)[/tex]:
[tex]\[ IT = \sqrt{(T_x - I_x)^2 + (T_y - I_y)^2} = \sqrt{(7 - 1)^2 + (5 - 2))^2 = \sqrt{36 + 9} = \sqrt{45} \][/tex]

4. Calculate [tex]\( ET \)[/tex]:
[tex]\[ ET = \sqrt{(T_x - E_x)^2 + (T_y - E_y)^2} = \sqrt{(7 - 4)^2 + (5 + 1))^2} = \sqrt{9 + 36} = \sqrt{45} \][/tex]

Summarizing the results, we have the lengths:
[tex]\[ KI = \sqrt{17}, \quad KE = \sqrt{17}, \quad IT = \sqrt{45}, \quad ET = \sqrt{45} \][/tex]

Both pairs of adjacent sides [tex]\( KI \)[/tex] and [tex]\( KE \)[/tex] as well as [tex]\( IT \)[/tex] and [tex]\( ET \)[/tex] are equal in length.

Therefore, [tex]\( KITE \)[/tex] is a kite because it has two pairs of adjacent sides of equal length.

Now, let's fill in the drop-down menu selections:

- [tex]\(\sqrt{17}\)[/tex]
- [tex]\(KE = \sqrt{17}\)[/tex]
- [tex]\(IT = \sqrt{45}\)[/tex]
- [tex]\(ET = \sqrt{45}\)[/tex]
- [tex]\(Therefore, \ KITE \ is \ a \ kite \ because \ it \ has \ two \ pairs \ of \ adjacent \ sides \ of \ equal \ length.\)[/tex]