# TODAY IN GEOMETRYβ¦ Independent Practice

## Presentation on theme: "TODAY IN GEOMETRYβ¦ Independent Practice"β Presentation transcript:

TODAY IN GEOMETRYβ¦ Independent Practice
Learning Target 1: Review Ch. 5 concepts-Triangle Midsegments, Centroids and Inequalities Mini Quiz-TODAY! Learning Target 2: Simplifying Radicals Independent Practice

MIDSEGMENT THEOREM: The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long as that side. π΅ π·πΈ β₯ π΄πΆ π·πΈ= 1 2 π΄πΆ πΈ π· π΄ πΆ

If the perimeter of β³πππ=68 πππβππ , find the perimeter of β³πππ. β³πππ= 1 2 (68) If ππ=2π₯β4 and ππ=3π₯β3, what is ππ? 2 ππ =ππ 2 2π₯β4 =3π₯β3 4π₯β8=3π₯β3 β 3π₯ β 3π₯ π₯β8=β3 π=π ππ=2π₯β4 ππ=2 5 β4=6 π π π π π π 1. If ππ=13, find ππ. ππ=2 ππ =2 13 =26 2. If ππ=20, find ππ. ππ= 1 2 ππ = =10

CONCURRENCY OF MEDIANS: The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side. π΄π= 2 3 π΄π΅ π΄ A different way to think about this: π΄π is twice as long as π΅π OR π΅π is half of π΄π ππ π π π΅ CENTRIOD

PRACTICE: π is the centroid of β³π΄π·πΈ, πΆπΈ=27
π΅ πΆ πΈ πΉ 1. πΉπΈ= 2. πΆπ= 3. ππΈ= 4. πΉπ= 5. πΉπ·= 6. π΄πΈ= π΄πΉ=12 12 1 3 πΆπΈ=9 2 3 πΆπΈ=18 1 2 π·π=3 6 πΉπ+ππ·=9 2π΄πΉ=24

π΄π΅+π΅πΆ>π΄πΆ π΅πΆ+π΄πΆ>π΄π΅ π΄πΆ+π΄π΅>π΅πΆ
TRIANGLE INEQUALITY THEOREM: The sum of the lengths of any two sides of a triangle is greater than the length of the third side. πΆ π΅ π΄ π΄π΅+π΅πΆ>π΄πΆ π΅πΆ+π΄πΆ>π΄π΅ π΄πΆ+π΄π΅>π΅πΆ

π΄π΅ < π΅πΆ < π΄πΆ Triangle inequalities: β πΆ=35Β° β π΄=45Β° β π΅=110Β°
π π πΆ π π΄ π΄π΅ < π΅πΆ < π΄πΆ The longest side of a triangle is opposite the largest angle. The smallest side of a triangle is opposite the smallest angle. ππππ π ππ πππ πππππππ ππππ ππππ π ππ πππ ππππππππ ππππ

YES, IT IS POSSIBLE TO CONSTRUCT THIS TRIANGLE!
PRACTICE: Is it possible to construct a triangle with the sides 3, 7, 9? 1. Smallest possible value can be found by subtracting the two largest numbers. 2. Greatest possible value can be found by adding the two smallest numbers. smallest value:π₯>9β7 π>π largest value:π₯<3+7 π<ππ Check smallest given sideβ¦.3 π>π Check greatest given sideβ¦.9 π<ππ YES, IT IS POSSIBLE TO CONSTRUCT THIS TRIANGLE!

CHAPTER 5 QUIZ You have 10 minutes to complete your test
Are you prepared? Pencil? Calculator? Ch.5 Yellow notes allowed If you finish early: Finish any missing assignments. Work on something quietly as the rest of your fellow classmates complete their test. NO TALKING. STAY IN YOUR SEATS. RETURN ANY BORROWED BOOKS AND CALCULATORS. Turn in any missing or corrected assignments.

Perfect Squares: 1 4 9 16 25 36 49 64 81 100 121 144 SIMPLIFYING RADICALS: =9 =5 πππππππ‘ πππ’πππ πππππππ‘ πππ’πππ = 36Β·2 = 36 Β· 2 =6 2 = 16Β·3 = 16 Β· 3 =4 3 = 25Β·7 = 25 Β· 7 =5 7 = 100Β·3 = 100 Β· 3 =10 3

Perfect Squares: 1 4 9 16 25 36 49 64 81 100 121 144 SIMPLIFYING RADICALS WITH VALUES IN FRONT: β8 25 β9 48 =4Β·9 =β8Β·5 =36 =β40 =β9 16Β·3 =β9( 16 Β· 3 ) =β9 4 3 = β9Β·4 3 =β36 3 =2 36Β·2 =2( 36 Β· 2 ) =2 6 2 =(2Β·6) 2 =12 2

Perfect Squares: 1 4 9 16 25 36 49 64 81 100 121 144 MULTIPLYING AND SIMPLIFYING RADICALS: Β· β 4 Β· 25 Β· β2 48 Β·3 2 = 5Β·10 =β 4Β·25 = 50 = 25Β·2 = 25 Β· 2 =π π =β 100 =βππ =(2Β·2)( 72Β·3 ) =4 216 =4 6Β·36 =4 6 Β· 36 =(4Β·6) 6 =ππ π = β2Β· Β·2 =β6( 96 ) =β6( 16Β·6 ) =β6( 16 Β· 6 ) = β6Β·4 6 =βππ π