Geometry 201 unit 5.7
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Transcript of Geometry 201 unit 5.7
Warm Up
1. Write the angles in order from smallest to largest.
2. The lengths of two sides of a triangle are 12 cm and 9 cm. Find the range of possible lengths for the third side.
∠X, ∠Z, ∠Y
3 cm < s < 21 cm
Example 1A: Using the Hinge Theorem and Its Converse
Compare m∠BAC and m∠DAC.
Compare the side lengths in ∆ABC and ∆ADC.
By the Converse of the Hinge Theorem, m∠BAC > m∠DAC.
AB = AD AC = AC BC > DC
Example 1B: Using the Hinge Theorem and Its Converse
Compare EF and FG.
By the Hinge Theorem, EF < GF.
Compare the sides and angles in ∆EFH angles in ∆GFH.
EH = GH FH = FH m∠EHF > m∠GHF
m∠GHF = 180° – 82° = 98°
Example 1C: Using the Hinge Theorem and Its Converse
Find the range of values for k.
Step 1 Compare the side lengths in ∆MLN and ∆PLN.
By the Converse of the Hinge Theorem, m∠MLN > m∠PLN.
LN = LN LM = LP MN > PN
5k – 12 < 38
k < 10
Substitute the given values.
Add 12 to both sides and divide by 5.
Example 1C Continued
Step 2 Since ∠PLN is in a triangle, m∠PLN > 0°.
Step 3 Combine the two inequalities.
The range of values for k is 2.4 < k < 10.
5k – 12 > 0
k < 2.4
Substitute the given values.
Add 12 to both sides and divide by 5.
Check It Out! Example 1a
Compare m∠EGH and m∠EGF.
Compare the side lengths in ∆EGH and ∆EGF.
FG = HG EG = EG EF > EH
By the Converse of the Hinge Theorem, m∠EGH < m∠EGF.
Check It Out! Example 1b
Compare BC and AB.
Compare the side lengths in ∆ABD and ∆CBD.
By the Hinge Theorem, BC > AB.
AD = DC BD = BD m∠ADB > m∠BDC.
Example 2: Travel Application
John and Luke leave school at the same time. John rides his bike 3 blocks west and then 4 blocks north. Luke rides 4 blocks east and then 3 blocks at a bearing of N 10º E. Who is farther from school? Explain.
Example 2 Continued
The distances of 3 blocks and 4 blocks are the same in both triangles.
The angle formed by John’s route (90º) is smaller than the angle formed by Luke’s route (100º). So Luke is farther from school than John by the Hinge Theorem.
Check It Out! Example 2
When the swing ride is at full speed, the chairs are farthest from the base of the swing tower. What can you conclude about the angles of the swings at full speed versus low speed? Explain.
The ∠ of the swing at full speed is greater than the ∠ at low speed because the length of the triangle on the opposite side is the greatest at full swing.
Example 3: Proving Triangle Relationships
Write a two-column proof.
Given:Prove: AB > CB Proof:
Statements Reasons
1. Given
2. Reflex. Prop. of ≅
3. Hinge Thm.
Check It Out! Example 3a
Write a two-column proof.
Given: C is the midpoint of BD.
Prove: AB > ED
m∠1 = m∠2 m∠3 > m∠4
1. Given
2. Def. of Midpoint
3. Def. of ≅ ∠s
4. Conv. of Isoc. ∆ Thm.
5. Hinge Thm.
1. C is the mdpt. of BDm∠3 > m∠4, m∠1 = m∠2
3. ∠1 ≅ ∠2
5. AB > ED
Statements Reasons
Proof:
Write a two-column proof.
Given:
Prove: m∠TSU > m∠RSU
Statements Reasons
1. Given
3. Reflex. Prop. of ≅
4. Conv. of Hinge Thm.
2. Conv. of Isoc. Δ Thm.
1. ∠SRT ≅ ∠STRTU > RU
∠SRT ≅ ∠STRTU > RU
Check It Out! Example 3b
4. m∠TSU > m∠RSU
Statements Reasons
1. Given
2. Reflex. Prop. of ≅
3. Conv. of Hinge Thm.3. m∠XYW < m∠ZWY
Given:
Prove: m∠XYW < m∠ZWY
4. Write a two-column proof.
Lesson Quiz: Part III
Proof:
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