Asked by izuku midoriya
Problem #9 Parallelogram "P" "Q" "R" "T"PQRT is shown. Which statement proves that parallelogram "P" "Q" "R" "T"PQRT is a rectangle? 15 squared negative 8 squared equals "T" "Q" squared 8 positive 15 equals "T" "Q" 15 negative 8 equals "T" "Q" 8 squared positive 15 squared equals "T" "Q" squared
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Answered by
GPT-4o mini
To prove that parallelogram \(PQRT\) is a rectangle, we need to show that either the angles are right angles or the diagonals are equal, given the lengths. In this case, we are looking for a statement relating to the sides and/or the diagonals based on the provided information.
The relationship you provided contains expressions involving squares, possibly deriving from the Pythagorean theorem. Let's break down the equality and apply it, assuming \(TQ\) is the side we're evaluating. The expression \(15^2 + (-8)^2 = TQ^2\) or \(8^2 + 15^2 = TQ^2\) suggests that \(TQ\) could be a diagonal in the triangle formed by a rectangle.
- **Calculation**:
\(15^2 + (-8)^2 = 225 + 64 = 289\)
Thus, \(TQ^2 = 289\) or \(TQ = \sqrt{289} = 17\).
Since both expressions for \(TQ\) lead to the same squared length \(TQ^2\), it indicates right triangles are formed confirming the diagonals \(PQ\) and \(RT\) are equal in length, which is a property of rectangles.
Thus, if you establish that the lengths of diagonals or the rectangles formed (in accordance with 15 and 8) conclude with \(TQ^2\), you have sufficient proof.
Therefore, the statement that provides the necessary evidence for parallelogram \(PQRT\) being a rectangle is:
**\(15^2 + (-8)^2 = TQ^2\)**.
This is the correct option for proving that \(PQRT\) is a rectangle.
The relationship you provided contains expressions involving squares, possibly deriving from the Pythagorean theorem. Let's break down the equality and apply it, assuming \(TQ\) is the side we're evaluating. The expression \(15^2 + (-8)^2 = TQ^2\) or \(8^2 + 15^2 = TQ^2\) suggests that \(TQ\) could be a diagonal in the triangle formed by a rectangle.
- **Calculation**:
\(15^2 + (-8)^2 = 225 + 64 = 289\)
Thus, \(TQ^2 = 289\) or \(TQ = \sqrt{289} = 17\).
Since both expressions for \(TQ\) lead to the same squared length \(TQ^2\), it indicates right triangles are formed confirming the diagonals \(PQ\) and \(RT\) are equal in length, which is a property of rectangles.
Thus, if you establish that the lengths of diagonals or the rectangles formed (in accordance with 15 and 8) conclude with \(TQ^2\), you have sufficient proof.
Therefore, the statement that provides the necessary evidence for parallelogram \(PQRT\) being a rectangle is:
**\(15^2 + (-8)^2 = TQ^2\)**.
This is the correct option for proving that \(PQRT\) is a rectangle.
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