Transcript Lesson 6.2

Bellringer
ο€ͺ Can a triangle have the sides with the given
lengths? Explain
1. 8mm, 6mm, 3mm
2. 5ft, 20ft, 7ft
3. 3m, 5m, 8m
6-2 Properties of
Parallelograms
Theorems
Theorem 6-1: Opposite sides of a parallelogram
are congruent
Theorem 6-2: Opposite angles of a parallelogram
are congruent
Theorem 6-3: The diagonals of a parallelogram
bisect each other.
3π‘₯ βˆ’ 15 = 2π‘₯ + 3
βˆ’2π‘₯
βˆ’ 2π‘₯
1π‘₯ βˆ’ 15 = 3
+15 + 15
π‘₯ = 18
𝑄𝑅 = 3π‘₯ βˆ’ 15
𝑄𝑅 = 3 18 βˆ’ 15
𝑄𝑅 = 54 βˆ’ 15
𝑄𝑅 = 39
𝑃𝑆 = 2π‘₯ + 3
𝑃𝑆 = 2 18 + 3
𝑃𝑆 = 36 + 3
𝑃𝑆 = 39
*Consecutive angles of a parallelogram are sameside interior angles, so they are supplementary.
π‘šβˆ π‘† + 112 = 180
βˆ’112 βˆ’ 112
π‘šβˆ π‘† = 68
πŸπŸ‘πŸ“ βˆ’ 𝒙 = 𝒙 + πŸπŸ“
____ βˆ’ 𝒙 βˆ’ 𝒙_____
πŸπŸ‘πŸ“ βˆ’ πŸπ’™ = πŸπŸ“
βˆ’πŸπŸ‘πŸ“
βˆ’ πŸπŸ‘πŸ“
βˆ’πŸπ’™ = βˆ’πŸπŸπŸŽ
βˆ’πŸ
βˆ’πŸ
𝒙 = πŸ”πŸŽ
π‘šβˆ π΅ = π‘₯ + 15
π‘šβˆ π΅ = 60 + 15
π‘šβˆ π΅ = 75
π‘šβˆ π΄ = 180 βˆ’ π‘šβˆ π΅
π‘šβˆ π΄ = 180 βˆ’ 75
π‘šβˆ π΄ = 105
2π‘₯ + 5 = 5𝑦
π‘₯ = 7𝑦 βˆ’ 16
2(7𝑦 βˆ’ 16) + 5 = 5𝑦
14𝑦 βˆ’ 32 + 5 = 5𝑦
14𝑦 βˆ’ 27 = 5𝑦
βˆ’27 = βˆ’9𝑦
𝑦=3
π‘₯ = 7𝑦 βˆ’ 16
π‘₯ = 7 3 βˆ’ 16
π‘₯ = 21 βˆ’ 16
π‘₯=5
Theorem 6-4: If three (or more) parallel lines cut
off congruent segments on one transversal, then
they cut off congruent segments on every
transversal.
BD  DF
𝑦 = 11
π‘šβˆ πΈ = π‘šβˆ πΊ = 70
π‘šβˆ πΉ = π‘šβˆ π» = 110
π‘Ž = 16
𝑏 = 14
𝐸𝐻 = 7.5
Practice!!
ο€ͺPg. 297-300 #1-22 and 44-52