At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Given the vertices [tex]\(A(11,-7), B(9,-4), C(11,-1)\)[/tex], and [tex]\(D(13,-4)\)[/tex], and comparing them to the provided answers, we proceed as follows:
First, we determine the lengths of the sides of the quadrilateral ABCD.
For [tex]\(AB\)[/tex]:
[tex]\[ AB = \sqrt{(9 - 11)^2 + (-4 - (-7))^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
For [tex]\(BC\)[/tex]:
[tex]\[ BC = \sqrt{(11 - 9)^2 + (-1 - (-4))^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
For [tex]\(CD\)[/tex]:
[tex]\[ CD = \sqrt{(13 - 11)^2 + (-4 - (-1))^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
For [tex]\(DA\)[/tex]:
[tex]\[ DA = \sqrt{(13 - 11)^2 + (-4 - (-7))^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
All sides of quadrilateral [tex]\(ABCD\)[/tex] are equal to [tex]\(\sqrt{13} \approx 3.605551275463989\)[/tex].
Next, we need to determine if this is a parallelogram. For it to be a parallelogram, opposite sides must be equal and parallel.
Additionally, we check the diagonals to see if they're equal:
[tex]\[ AC = \sqrt{(11 - 11)^2 + (-1 - (-7))^2} = \sqrt{0 + 36} = 6 \][/tex]
[tex]\[ BD = \sqrt{(13 - 9)^2 + (-4 - (-4))^2} = \sqrt{4^2 + 0} = 4 \][/tex]
Since AC and BD do not equal, ABCD is not a parallelogram; however, as all sides are equal and its diagonals do not bisect each other equally, it is a rhombus.
Next, we consider [tex]\(C^{\prime}(11, 1)\)[/tex] instead of [tex]\(C\)[/tex]. We compute the distances:
For [tex]\(B C'\)[/tex]:
[tex]\[ BC' = \sqrt{(11 - 9)^2 + (1 - (-4))^2} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385164807134504 \][/tex]
For [tex]\(C' D\)[/tex]:
[tex]\[ C'D = \sqrt{(11 - 13)^2 + (1 - (-4))^2} = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385164807134504 \][/tex]
For [tex]\(DA\)[/tex]:
[tex]\[ DA = \sqrt{(13 - 11)^2 + (-4 - (-7))^2} = \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.605551275463989 \][/tex]
Given the difference in side lengths [tex]\( \sqrt{13}\)[/tex] and [tex]\( \approx \sqrt{29}\)[/tex], [tex]\(AB' C' D\)[/tex] cannot be a parallelogram or a rhombus. Hence, it’s a general quadrilateral, not fitting any special categories of quadrilaterals listed here.
Therefore:
Quadrilateral [tex]\(ABCD\)[/tex] is a rhombus. Quadrilateral [tex]\(AB'C'D\)[/tex] would be a quadrilateral.
Final Answer:
Quadrilateral [tex]\(ABCD\)[/tex] is a [tex]\(\boxed{\text{Rhombus}}\)[/tex] and [tex]\(AB'C'D\)[/tex] would be a [tex]\(\boxed{\text{Quadrilateral}}\)[/tex].
First, we determine the lengths of the sides of the quadrilateral ABCD.
For [tex]\(AB\)[/tex]:
[tex]\[ AB = \sqrt{(9 - 11)^2 + (-4 - (-7))^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
For [tex]\(BC\)[/tex]:
[tex]\[ BC = \sqrt{(11 - 9)^2 + (-1 - (-4))^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
For [tex]\(CD\)[/tex]:
[tex]\[ CD = \sqrt{(13 - 11)^2 + (-4 - (-1))^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
For [tex]\(DA\)[/tex]:
[tex]\[ DA = \sqrt{(13 - 11)^2 + (-4 - (-7))^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]
All sides of quadrilateral [tex]\(ABCD\)[/tex] are equal to [tex]\(\sqrt{13} \approx 3.605551275463989\)[/tex].
Next, we need to determine if this is a parallelogram. For it to be a parallelogram, opposite sides must be equal and parallel.
Additionally, we check the diagonals to see if they're equal:
[tex]\[ AC = \sqrt{(11 - 11)^2 + (-1 - (-7))^2} = \sqrt{0 + 36} = 6 \][/tex]
[tex]\[ BD = \sqrt{(13 - 9)^2 + (-4 - (-4))^2} = \sqrt{4^2 + 0} = 4 \][/tex]
Since AC and BD do not equal, ABCD is not a parallelogram; however, as all sides are equal and its diagonals do not bisect each other equally, it is a rhombus.
Next, we consider [tex]\(C^{\prime}(11, 1)\)[/tex] instead of [tex]\(C\)[/tex]. We compute the distances:
For [tex]\(B C'\)[/tex]:
[tex]\[ BC' = \sqrt{(11 - 9)^2 + (1 - (-4))^2} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385164807134504 \][/tex]
For [tex]\(C' D\)[/tex]:
[tex]\[ C'D = \sqrt{(11 - 13)^2 + (1 - (-4))^2} = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385164807134504 \][/tex]
For [tex]\(DA\)[/tex]:
[tex]\[ DA = \sqrt{(13 - 11)^2 + (-4 - (-7))^2} = \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.605551275463989 \][/tex]
Given the difference in side lengths [tex]\( \sqrt{13}\)[/tex] and [tex]\( \approx \sqrt{29}\)[/tex], [tex]\(AB' C' D\)[/tex] cannot be a parallelogram or a rhombus. Hence, it’s a general quadrilateral, not fitting any special categories of quadrilaterals listed here.
Therefore:
Quadrilateral [tex]\(ABCD\)[/tex] is a rhombus. Quadrilateral [tex]\(AB'C'D\)[/tex] would be a quadrilateral.
Final Answer:
Quadrilateral [tex]\(ABCD\)[/tex] is a [tex]\(\boxed{\text{Rhombus}}\)[/tex] and [tex]\(AB'C'D\)[/tex] would be a [tex]\(\boxed{\text{Quadrilateral}}\)[/tex].
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.