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Three vertices of parallelogram PQRS are show: Q(8, 5), R(5, 1), S(2, 5)

Place statements and reasons in the table to complete the proof that shows that parallelogram PQRS is a rhombus.

User Sany
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1 Answer

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Final answer:

Upon computing the lengths of sides PR and QS of parallelogram PQRS using the distance formula, it is found that PR \(\\eq\) QS, indicating PQRS is not a rhombus because not all sides are equal.

Step-by-step explanation:

To prove that parallelogram PQRS with vertices Q(8, 5), R(5, 1), and S(2, 5) is a rhombus, we need to demonstrate that all four sides are of equal length. In a parallelogram, opposite sides are parallel and equal in length by definition. The next step is to compute the distances between the points Q, R, and S to find PR and QS.

Using the distance formula \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), the length of PR is:

\(d_{PR} = \sqrt{(5-2)^2+(1-5)^2} = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = \sqrt{25} = 5\)

The length of QS can be found similarly:

\(d_{QS} = \sqrt{(8-2)^2+(5-5)^2} = \sqrt{6^2+0^2} = \sqrt{36} = 6\)

Since PR \(\\eq\) QS, based on different lengths, the parallelogram PQRS is not a rhombus because the sides PR and QS are not equal. In a rhombus, all sides must have equal length. Therefore, without the need for further proof, we can say that PQRS is not a rhombus due to the unequal lengths of its sides.

User Glen Balliet
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