TODAY IN GEOMETRY… Independent Practice

Slides:



Advertisements
Similar presentations
Date: Sec 5-4 Concept: Medians and Altitudes of a Triangle
Advertisements

Today – Monday, March 11, 2013 Review for QUIZ QUIZ
Median ~ Hinge Theorem.
4.6 Medians of a Triangle.
Geometry Chapter 4 Cipollone.
Today – Monday, February 11, 2013  Warm Up: Simplifying Radicals  Learning Target : Review for Ch. 6 Quiz  CHAPTER 6 QUIZ TODAY!  All Chapter 6 Assignments.
CHAPTER 6: Inequalities in Geometry
Triangle Inequality Theorem:
TODAY IN GEOMETRY…  Learning Target: 5.5 You will find possible lengths for a triangle  Independent Practice  ALL HW due Today!
TODAY IN GEOMETRY…  Review for CH. 4 QUIZ  CH. 4 QUIZ – TODAY!  Learning Goal: 4.3 You will use the side lengths to prove triangles are congruent. 
Today – Wednesday, May 22, 2013  EOC Practice Test #1 – Second half  Learning Target : Assessment of Ch. 10 concepts  CH. 10 TEST TODAY!
OBJECTIVE: 1) BE ABLE TO IDENTIFY THE MEDIAN AND ALTITUDE OF A TRIANGLE 2) BE ABLE TO APPLY THE MID-SEGMENT THEOREM 3) BE ABLE TO USE TRIANGLE MEASUREMENTS.
Geometry Chapter 5 Benedict. Vocabulary Perpendicular Bisector- Segment, ray, line or plane that is perpendicular to a segment at its midpoint. Equidistant-
TODAY IN GEOMETRY…  Algebra Review Quiz today!  Review Segment Addition Postulate and Congruency  Learning Goal: 1.3 You will find lengths of segments.
TODAY IN GEOMETRY…  Warm Up: Advanced Angles Practice  Review for CH. 4 QUIZ  CH. 4 QUIZ – TODAY!
Relationships within triangles
TODAY IN ALGEBRA…  Last minute questions  Mid. Ch. 6 Test Today!  Learning Goal: 6.5 You will solve Absolute Value Equations  Independent Practice.
TODAY IN ALGEBRA…  Warm Up: Adding and Subtracting Integers with cards Activity  Review for Mid Chapter 2 test  MID CHAPTER 2 TEST TODAY!!!  Learning.
Today – Monday, March 18, 2013  Warm up: Simplified Radical Form  Learning Target : Review concepts for Chapter 7 test  CHAPTER 7 TEST  ALL CH. 7 ASSIGNMENTS.
TODAY IN ALGEBRA…  CH.1 TEST TODAY!  ALL HW DUE TODAY and Notebook check today!  Learning Goal: 2.1 You will graph and compare positive and negative.
Honors Geometry Section 4.6 Special Segments in Triangles
TODAY IN GEOMETRY…  CH.8 Review  CHAPTER 8 TEST TODAY!  All Chapter 8 HW due today!
TODAY IN ALGEBRA…  Learning Goal: Review Ch. 2 concepts for the Ch. 2 test  CH. 2 TEST TODAY!  All Ch. 2 HW due TODAY!
MORE TRIANGLES Chapter 5 Guess What we will learn about Geometry Unit Properties of Triangles 1.
Points of Concurrency Line Segments Triangle Inequalities.
Unit 5.
Chapter 5 Review Perpendicular Bisector, Angle Bisector, Median, Altitude, Exterior Angles and Inequality.
Ticket In the Door Write out each of the following: 1.SSS Postulate 2.SAS Postulate 3.ASA Postulate 4.AAS Postulate.
Lesson 3-3: Triangle Inequalities 1 Lesson 3-3 Triangle Inequalities.
Triangle Sum Properties & Inequalities in a Triangle Sections 4.1, 5.1, & 5.5.
Today – Monday, May 20, 2013  EOC Practice Test # - First half  Learning Target : Review Ch. 10 concepts by practicing problems on a practice test 
Warm up: Calculate midsegments
November. Get a worksheet from the front, complete the crossword puzzle!
TODAY IN GEOMETRY…  Warm Up: Simplifying Radicals by squaring  Learning Target : Review for Ch. 6 Quiz  CHAPTER 6 QUIZ TODAY!  Independent Practice.
Relationships Within Triangles Chapter5. Triangle Midsegment Theorem If a segment joins the midpoints of two sides of a triangle, then the segment is.
Triangle Sum Theorem In a triangle, the three angles always add to 180°: A + B + C = 180° 38° + 85° + C = 180° C = 180° C = 57°
4.7 Triangle Inequalities. In any triangle…  The LARGEST SIDE lies opposite the LARGEST ANGLE.  The SMALLEST SIDE lies opposite the SMALLEST ANGLE.
TODAY IN ALGEBRA 2.0…  Learning Target 1: Review Ch.7 Concepts  CH. 7 TEST – TODAY!  ALL HW – DUE TODAY!
TODAY IN GEOMETRY…  Review for Ch.3 Quiz  CH.3 QUIZ - TODAY!
TODAY IN GEOMETRY…  Quick Concept Review  CH. 10 QUIZ TODAY!
Geometry Sections 5.1 and 5.2 Midsegment Theorem Use Perpendicular Bisectors.
Chapter 10 Section 3 Concurrent Lines. If the lines are Concurrent then they all intersect at the same point. The point of intersection is called the.
TODAY IN ALGEBRA…  Review:  Mid Ch. 5 Test TODAY!  Independent Practice.
1 Triangle Inequalities. 2 Triangle Inequality The smallest side is across from the smallest angle. The largest angle is across from the largest side.
Geometry Section 5.5 Use Inequalities in a Triangle.
4.7 Triangle Inequalities
5.5 Inequalities in Triangles Learning Target I can use inequalities involving angles and sides in triangles.
TODAY IN GEOMETRY…  Group POP QUIZ  Learning Target 1: 5.1 Use properties of mid segments of triangles to calculate lengths of sides  Learning Target.
TODAY IN GEOMETRY…  Warm Up: Simplifying Radicals  Learning Target : Review for Ch. 6 Quiz  CHAPTER 6 QUIZ TODAY!  All Chapter 6 Assignments Due TODAY!
TODAY IN GEOMETRY…  Learning Target: 6.6 You will calculate missing sides of triangles using proportionality with parallel lines  Independent Practice.
Triangle Theorems. Warm-Ups 1.What do you think is going well with this class? 2.What is your favorite part of the class? 3.What do you wish was different.
5-2 Median & Altitudes of Triangles
WARM UP March 11, Solve for x 2. Solve for y (40 + y)° 28° 3x º xºxºxºxº.
Chapter 5, Section 1 Perpendiculars & Bisectors. Perpendicular Bisector A segment, ray, line or plane which is perpendicular to a segment at it’s midpoint.
5.3 Trade Routes and Pasta Anyone?
Medians, and Altitudes.
Section 5.4 Theorem – MIDSEGMENT THEOREM The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long.
Medians and Altitudes of a Triangle
Lines Associated with Triangles 4-3D
Triangle Inequalities
Medians and Altitudes Median – A segment whose endpoints are a vertex of a triangle and the midpoint of the side opposite the vertex. Centroid – The point.
5-3 & 5-4: Midsegments and Medians of Triangles
Relationships Within Triangles
Triangle Inequalities
TRIANGLE INEQUALITY THEOREM
Triangle Midsegment Theorem – The segment joining the midpoints of any two sides will be parallel to the third side and half its length. If E and D are.
TRIANGLE INEQUALITY THEOREM
TRIANGLE INEQUALITY THEOREM
Triangle Inequalities
9-8: Transversals to Many Parallel Lines
Presentation transcript:

TODAY IN GEOMETRY… 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. 𝐵 𝐷𝐸 ∥ 𝐴𝐶 𝐷𝐸= 1 2 𝐴𝐶 𝐸 𝐷 𝐴 𝐶

If the perimeter of △𝑅𝑆𝑇=68 𝑖𝑛𝑐ℎ𝑒𝑠, find the perimeter of △𝑈𝑉𝑊. △𝑈𝑉𝑊= 1 2 (68) If 𝑉𝑊=2𝑥−4 and 𝑅𝑆=3𝑥−3, what is 𝑉𝑊? 2 𝑉𝑊 =𝑅𝑆 2 2𝑥−4 =3𝑥−3 4𝑥−8=3𝑥−3 − 3𝑥 − 3𝑥 𝑥−8=−3 + 8 + 8 𝒙=𝟓 𝑉𝑊=2𝑥−4 𝑉𝑊=2 5 −4=6 𝑆 𝑈 𝑉 𝑅 𝑊 𝑇 1. If 𝑈𝑉=13, find 𝑅𝑇. 𝑅𝑇=2 𝑈𝑉 =2 13 =26 2. If 𝑆𝑇=20, find 𝑈𝑊. 𝑈𝑊= 1 2 𝑆𝑇 = 1 2 20 =10

A different way to think about this: 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. 𝐴𝑃= 2 3 𝐴𝐵 𝐴 A different way to think about this: 𝐴𝑃 is twice as long as 𝐵𝑃 OR 𝐵𝑃 is half of 𝐴𝑃 𝟐𝒙 𝑃 𝒙 𝐵 CENTRIOD

PRACTICE: 𝑃 is the centroid of △𝐴𝐷𝐸, 𝐶𝐸=27 𝐵 𝐶 𝐸 𝐹 1. 𝐹𝐸= 2. 𝐶𝑃= 3. 𝑃𝐸= 4. 𝐹𝑃= 5. 𝐹𝐷= 6. 𝐴𝐸= 𝐴𝐹=12 12 1 3 𝐶𝐸=9 2 3 𝐶𝐸=18 1 2 𝐷𝑃=3 6 𝐹𝑃+𝑃𝐷=9 2𝐴𝐹=24

𝐴𝐵+𝐵𝐶>𝐴𝐶 𝐵𝐶+𝐴𝐶>𝐴𝐵 𝐴𝐶+𝐴𝐵>𝐵𝐶 TRIANGLE INEQUALITY THEOREM: The sum of the lengths of any two sides of a triangle is greater than the length of the third side. 𝐶 𝐵 𝐴 𝐴𝐵+𝐵𝐶>𝐴𝐶 𝐵𝐶+𝐴𝐶>𝐴𝐵 𝐴𝐶+𝐴𝐵>𝐵𝐶

𝐴𝐵 < 𝐵𝐶 < 𝐴𝐶 Triangle inequalities: ∠𝐶=35° ∠𝐴=45° ∠𝐵=110° 𝑎 𝑐 𝐶 𝑏 𝐴 𝐴𝐵 < 𝐵𝐶 < 𝐴𝐶 The longest side of a triangle is opposite the largest angle. The smallest side of a triangle is opposite the smallest angle. 𝒔𝒊𝒅𝒆 𝒃 𝒊𝒔 𝒕𝒉𝒆 𝒍𝒐𝒏𝒈𝒆𝒔𝒕 𝒔𝒊𝒅𝒆 𝒔𝒊𝒅𝒆 𝒄 𝒊𝒔 𝒕𝒉𝒆 𝒔𝒉𝒐𝒓𝒕𝒆𝒔𝒕 𝒔𝒊𝒅𝒆

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:𝑥>9−7 𝒙>𝟐 largest value:𝑥<3+7 𝒙<𝟏𝟎 Check smallest given side….3 𝟑>𝟐 Check greatest given side….9 𝟗<𝟏𝟎 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: 1. 81 2. 25 3. 72 4. 48 5. 175 6. 300 =9 =5 𝑃𝑒𝑟𝑓𝑒𝑐𝑡 𝑆𝑞𝑢𝑎𝑟𝑒 𝑃𝑒𝑟𝑓𝑒𝑐𝑡 𝑆𝑞𝑢𝑎𝑟𝑒 = 36·2 = 36 · 2 =6 2 = 16·3 = 16 · 3 =4 3 = 25·7 = 25 · 7 =5 7 = 100·3 = 100 · 3 =10 3

Perfect Squares: 1 4 9 16 25 36 49 64 81 100 121 144 SIMPLIFYING RADICALS WITH VALUES IN FRONT: 1. 4 81 2. −8 25 3. 2 72 4. −9 48 =4·9 =−8·5 =36 =−40 =−9 16·3 =−9( 16 · 3 ) =−9 4 3 = −9·4 3 =−36 3 =2 36·2 =2( 36 · 2 ) =2 6 2 =(2·6) 2 =12 2

Perfect Squares: 1 4 9 16 25 36 49 64 81 100 121 144 MULTIPLYING AND SIMPLIFYING RADICALS: 1. 5 · 10 2. − 4 · 25 3. 2 72 ·2 3 4. −2 48 ·3 2 = 5·10 =− 4·25 = 50 = 25·2 = 25 · 2 =𝟓 𝟐 =− 100 =−𝟏𝟎 =(2·2)( 72·3 ) =4 216 =4 6·36 =4 6 · 36 =(4·6) 6 =𝟐𝟒 𝟔 = −2·3 48·2 =−6( 96 ) =−6( 16·6 ) =−6( 16 · 6 ) = −6·4 6 =−𝟐𝟒 𝟔

Simplified Radical Form (SRF) HOMEWORK #1: Simplified Radical Form (SRF) Half sheet