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Complete the steps in the proof that show quadrilateral KITE with vertices [tex]\(K (0,-2), I (1,2), T (7,5), E(x,y)\)[/tex].

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].

Select the correct answers from each drop-down menu.


Sagot :

To solve this problem step-by-step, we'll first calculate the lengths of the sides of the quadrilateral KITE.

We'll use the distance formula:
[tex]\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

Given the vertices of quadrilateral KITE:
- [tex]\( K(0, -2) \)[/tex]
- [tex]\( I(1, 2) \)[/tex]
- [tex]\( T(7, 5) \)[/tex]
- [tex]\( E(6, -1) \)[/tex]

Let's calculate the lengths of the sides one by one:

1. Calculate [tex]\( KI \)[/tex]
[tex]\[ KI = \sqrt{(1 - 0)^2 + (2 - (-2))^2} = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.123 \][/tex]

2. Calculate [tex]\( KE \)[/tex]
[tex]\[ KE = \sqrt{(6 - 0)^2 + (-1 - (-2))^2} = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37} \approx 6.083 \][/tex]

3. Calculate [tex]\( IT \)[/tex]
[tex]\[ IT = \sqrt{(7 - 1)^2 + (5 - 2)^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.708 \][/tex]

4. Calculate [tex]\( TE \)[/tex]
[tex]\[ TE = \sqrt{(6 - 7)^2 + (-1 - 5))^2} = \sqrt{(-1)^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37} \approx 6.083 \][/tex]

So after calculating each side, let's fill in the drop-down values:

[tex]\[ K E = \sqrt{37} \][/tex]
[tex]\[ I T = \sqrt{45} \][/tex]
[tex]\[ T E = \sqrt{37} \][/tex]

Finally, KITE is indeed a kite because the lengths corresponding to [tex]\( KE \)[/tex] and [tex]\( TE \)[/tex] are equal:
[tex]\[ KE = TE = \sqrt{37} \][/tex]

A kite in geometry is defined as having two distinct pairs of adjacent sides that are equal in length.

Therefore, our final filled proof is:

Using the distance formula,
[tex]\[ K I = \sqrt{(2 - (-2))^2 + (1 - 0)^2} = \sqrt{17} \][/tex]
[tex]\[ K E = \sqrt{37} \][/tex]
[tex]\[ I T = \sqrt{45} \][/tex]
[tex]\[ T E = \sqrt{37} \][/tex]

Therefore, KITE is a kite because [tex]\( KE = TE \)[/tex].
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