Question

name _
unit 4 polygons and quadrilaterals summative (show all work)
date _11 - 2024 block _4b goal _80%_
quadrilateral prst is a parallelogram.

1. (overline{sr}cong) _
2. (angle tprcong) _
3. (overline{tc}cong) _
4. (angle7cong) _
5. if (mangle1 = 27^{circ},mangle2 = 30^{circ}), then (mangle5=) _ & (mangle tsr=) _.
6. if (tc = 2x + 7) & (tr = 30), then (x=) _.

Explanation:

Step1: Properties of parallelogram (for questions 1 - 4)

In a parallelogram \(PRST\):

• Opposite sides are equal. So, \(\overline{SR}\cong\overline{PT}\) (for question 1).
• Opposite angles are equal and alternate - interior angles formed by the diagonals are equal. \(\angle TPR\cong\angle RSP\) (for question 2).
• Diagonals bisect each other. So, \(\overline{TC}\cong\overline{RC}\) (for question 3).
• \(\angle7\cong\angle3\) (alternate - interior angles, for question 4).

Step2: Calculate angles (question 5)

In \(\triangle PRS\), \(m\angle1 = 27^{\circ}\), \(m\angle2=30^{\circ}\).

• \(m\angle5\):

Since \(PR\parallel TS\) (property of parallelogram), \(\angle2\) and \(\angle5\) are alternate - interior angles. So \(m\angle5 = 30^{\circ}\).

• \(m\angle TSR\):

In \(\triangle PRS\), \(m\angle PRS=180-(27 + 30)=123^{\circ}\). In parallelogram \(PRST\), \(m\angle TSR=m\angle PRT\). Also, \(m\angle TSR = 180-(m\angle1 + m\angle5)=180-(27 + 30)=123^{\circ}\)

Step3: Solve for \(x\) (question 6)

Since diagonals of a parallelogram bisect each other, \(TC=\frac{1}{2}TR\).
Given \(TC = 2x + 7\) and \(TR = 30\), then \(2x+7=\frac{30}{2}\).

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Answer:

1. \(\overline{PT}\)
2. \(\angle RSP\)
3. \(\overline{RC}\)
4. \(\angle3\)
5. \(m\angle5 = 30^{\circ}\), \(m\angle TSR=123^{\circ}\)
6. \(x = 4\)