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Complete the steps in the proof that show quadrilateral [tex]\operatorname{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:
[tex]KI = \sqrt{(2 - (-2))^2 + (1 - 0)^2} = \sqrt{17},[/tex]
[tex]KE = \square,[/tex]
[tex]IT = \square,[/tex]
and [tex]TE = \square.[/tex]

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

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Answer :

GinnyAnswer
To confirm that quadrilateral KITE 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, we will use the distance formula to calculate the lengths of its sides and then analyze the properties of these sides:

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

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

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

4. Calculating [tex]\( TE \)[/tex]:
[tex]\[ TE = \sqrt{(5 - (-1))^2 + (7 - 4)^2} = \sqrt{(6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.708 \][/tex]

Since we have:
- [tex]\( KI \approx 4.123 \)[/tex]
- [tex]\( KE \approx 4.123 \)[/tex]
- [tex]\( IT \approx 6.708 \)[/tex]
- [tex]\( TE \approx 6.708 \)[/tex]

We notice that [tex]\( KI = KE \)[/tex] and [tex]\( IT = TE \)[/tex].

Therefore, quadrilateral [tex]\( \operatorname{KITE} \)[/tex] is a kite because it has two pairs of adjacent sides that are equal.

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