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]
[tex]\[ TE = \square \][/tex]

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



Answer :

To show that quadrilateral KITE with vertices [tex]\( K (0, -2), I (1, 2), T (7, 5) \)[/tex], and [tex]\( E (4, -1) \)[/tex] is a kite, we need to use the distance formula to determine the lengths of the sides of KITE and show that it has two pairs of adjacent sides that are equal.

The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

Step 1: Calculate [tex]\( KI \)[/tex]:
[tex]\[ KI = \sqrt{(1 - 0)^2 + (2 - (-2))^2} \][/tex]
[tex]\[ KI = \sqrt{1 + 16} \][/tex]
[tex]\[ KI = \sqrt{17} \][/tex]

Step 2: Calculate [tex]\( KE \)[/tex]:
[tex]\[ KE = \sqrt{(4 - 0)^2 + (-1 - (-2))^2} \][/tex]
[tex]\[ KE = \sqrt{16 + 1} \][/tex]
[tex]\[ KE = \sqrt{17} \][/tex]

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

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

We notice that:
[tex]\[ KI = KE = \sqrt{17} \][/tex]
[tex]\[ IT = ET = \sqrt{45} \][/tex]

Since KITE has two pairs of adjacent sides that are equal ([tex]\( KI = KE \)[/tex] and [tex]\( IT = ET \)[/tex]), we can conclude that KITE is a kite.

Therefore, fill in the drop-down menus as follows:

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