Complete the steps in the proof that show quadrilateral [tex]\(KITE\)[/tex] 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]\[ K I = \sqrt{(2 - (-2))^2 + (1 - 0)^2} = \sqrt{17} \][/tex]
[tex]\[ K E = \square \][/tex]
[tex]\[ I T = \square \][/tex]
[tex]\[ T E = \square \][/tex]

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



Answer :

To determine if 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, we will use the distance formula to find the lengths of its sides.

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

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

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

Thus, we obtain:
[tex]\[ KI \approx 4.123, \quad KE \approx 4.123, \quad IT \approx 6.708, \quad TE \approx 6.708 \][/tex]

5. Determining if [tex]\(KITE\)[/tex] is a kite:
A quadrilateral is a kite if it has two distinct pairs of adjacent sides that are equal. From the calculated distances, we see:
[tex]\[ KI = KE \quad \text{and} \quad IT = TE \][/tex]

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

Here are the completed steps:
- [tex]\(KE = \sqrt{17}\)[/tex]
- [tex]\(IT = \sqrt{45}\)[/tex]
- [tex]\(TE = \sqrt{45}\)[/tex]
- Therefore, [tex]\(KITE\)[/tex] is a kite because [tex]\(KI = KE\)[/tex] and [tex]\(IT = TE\)[/tex]

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