Question

5.2 congruent polygons
geometry practice
write a congruence statement for the polygons. identify all pairs of congruent corresponding parts.
(see example 1)
1.
△____≅△____
2.
____≅____
∠a≅______
overline{ab}≅______
∠b≅______
overline{bc}≅______
∠c≅______
overline{ca}≅______
∠h≅______
overline{st}≅______
∠j≅______
overline{tq}≅______
∠k≅______
overline{rq}≅______
∠g≅______
overline{rs}≅______
find the values of x and y. (see example 2)

1. abcd≅efgh
2. △mnp≅△tus

Explanation:

Step1: Identify congruent triangles for problem 1

By matching the corresponding angles and sides (marked with same tick - marks and arc - marks), we can see that \(\triangle ABC\cong\triangle DFE\).

• For angles: \(\angle A\) corresponds to \(\angle D\), \(\angle B\) corresponds to \(\angle F\), \(\angle C\) corresponds to \(\angle E\).
• For sides: \(\overline{AB}\) corresponds to \(\overline{DF}\), \(\overline{BC}\) corresponds to \(\overline{FE}\), \(\overline{CA}\) corresponds to \(\overline{ED}\).

Step2: Identify congruent polygons for problem 2

By matching the corresponding angles and sides (marked with same tick - marks and arc - marks), we can see that \(GHKJ\cong RQTS\).

• For angles: \(\angle H\) corresponds to \(\angle S\), \(\angle J\) corresponds to \(\angle T\), \(\angle K\) corresponds to \(\angle Q\), \(\angle G\) corresponds to \(\angle R\).
• For sides: \(\overline{ST}\) corresponds to \(\overline{HK}\), \(\overline{TQ}\) corresponds to \(\overline{KJ}\), \(\overline{RQ}\) corresponds to \(\overline{KG}\), \(\overline{RS}\) corresponds to \(\overline{GH}\).

Step3: Solve for \(x\) and \(y\) in problem 3

Since \(ABCD\cong EFGH\), then \(\angle A=\angle E\) and \(\angle C=\angle F\).

• \(\angle A = 28^{\circ}\), so \(4y−4 = 28\).
• Add \(4\) to both sides: \(4y=28 + 4=32\).
• Divide both sides by \(4\): \(y=\frac{32}{4}=8\).
• \(\angle C = 135^{\circ}\), so \(10x+65 = 135\).
• Subtract \(65\) from both sides: \(10x=135 - 65 = 70\).
• Divide both sides by \(10\): \(x=\frac{70}{10}=7\).

Step4: Solve for \(x\) and \(y\) in problem 4

Since \(\triangle MNP\cong\triangle TUS\), then \(\angle P=\angle U\) and \(MN = TU\).

• \(\angle P = 24^{\circ}\), so \(2x−50=24\).
• Add \(50\) to both sides: \(2x=24 + 50=74\).
• Divide both sides by \(2\): \(x=\frac{74}{2}=37\).
• \(MN=(2x - y)\) and \(TU = 13\). Substitute \(x = 37\) into \(2x−y=13\).
• \(2\times37−y=13\).
• \(74−y=13\).
• Subtract \(74\) from both sides: \(-y=13 - 74=-61\).
• Multiply both sides by \(- 1\): \(y = 61\).

Answer:

1. \(\triangle ABC\cong\triangle DFE\); \(\angle A\cong\angle D\), \(\angle B\cong\angle F\), \(\angle C\cong\angle E\), \(\overline{AB}\cong\overline{DF}\), \(\overline{BC}\cong\overline{FE}\), \(\overline{CA}\cong\overline{ED}\)
2. \(GHKJ\cong RQTS\); \(\angle H\cong\angle S\), \(\angle J\cong\angle T\), \(\angle K\cong\angle Q\), \(\angle G\cong\angle R\), \(\overline{ST}\cong\overline{HK}\), \(\overline{TQ}\cong\overline{KJ}\), \(\overline{RQ}\cong\overline{KG}\), \(\overline{RS}\cong\overline{GH}\)
3. \(x = 7\), \(y = 8\)
4. \(x = 37\), \(y = 61\)