U8D5 Have Out: HW, packet, GP NB, pencil, highlighter, ruler, calculator, & red pen Bellwork: Determine the length of x. C 1) C 2) D 9 6 5 4 x E D B B.

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U8D5 Have Out: HW, packet, GP NB, pencil, highlighter, ruler, calculator, & red pen Bellwork: Determine the length of x. C 1) C 2) D 9 6 5 4 x E D B B 12 7 x E A A

C  C (Reflexive Property) C  C (Reflexive Property) 1) C 2) C D 6 9 4 5 E B x D B 7 12 x A E A A  CBD (Given) E  CDB (Given) C  C (Reflexive Property) C  C (Reflexive Property) +1 ACE~BCD (AA Thm.) +1 ACE~BCD (AA Thm.) x 5 = 9 + 12 9 x 4 = 7 + 6 6 +1 +1 9x = 5(9 + 12) 6x = 4(7+ 6) 9x = 105 6x = 52 35 3 = u x = u 26 3 x = 105 9 x  11.67 u +1 x  8.67 u +1

S-44 In unit 5, you studied transformations and congruence. The figure at right shows a pool table and seven balls. A Suppose you wanted to shoot the cue ball (C) and hit the ball labeled A. Since you cannot hit it directly (it is blocked by several other balls), you decide you will hit the cue ball into the right side of the table so that it rebounds and continues on to hit A. How can you determine where to hit the side of the table so that ball C will hit ball A? C

S-44 D A’ A B C a) Reflect (flip) ball A across the right side of the table and label the point A’. (An easy way to reflect A is to fold the paper along the right side of the pool table.) Label the point where AA’ crosses the side of the pool table D. b) You have a clear line of sight from C to A’. Draw the segment CA’, label the point where it crossed the side of the pool table B, then draw BA.

S-44 A C A’ D B Statements Reasons E Statements Reasons c) Use the properties of a reflection to prove that ADB  A’DB. A C A’ D B Statements Reasons 1) AD = A’D & BD  AA’ 1) Properties of Reflection 2) ADB  A’DB 2) Definition of Perpendicular 3) 3) Reflexive Property 4) ADB  A’DB 4) SAS E d) Label the lower right corner of the table E. Prove that mEBC = mDBA. Statements Reasons 5) mDBA = mDBA’ 5)  s   parts 6) mEBC = mDBA’ 6) Vertical  Thm 7) mEBC = mDBA 7) Substitution

Angles of Incidence & Reflection Add to your notes. a b In the figure at right, the Angle of Incidence (a) is the angle of approach. The Angle of Reflection (b) is the angle of rebound. A ma = mb b This means that the angle of incidence = angle of reflection!!! a C

S-45 Latoya looks down into a mirror that is lying on the ground between her and a flagpole. As she looks into the mirror, she can see the top of the pole. Her eye level is five feet from the ground and she is standing 4 feet from the mirror, which is 30 feet away from the base of the pole. a) Draw a picture of the situation, filling in any measurements you know, and prove that the triangles are similar. b) How tall is the pole? x 5ft 4 ft 30 ft

S-45 a) Draw a picture of the situation, filling in any measurements you know, and prove that the triangles are similar. b) How tall is the pole? E B x 5 ft A C 30 ft D 4 ft The pole is 37.5 feet tall. Statements Reasons 1) AB and DE are  to AD 1) Given (heights) 2) A  D 2) Def. Of Perpendicular / Rt. angles are  3) mACB = mDCE 3) Angle of incidence = angle of reflection 4) ABC  DEC 4) AA ~ Theorem

Work on S-46 & 48

The height of the pole is about 37.46 feet. Latoya forgot her clinometer, so she devised another method to find the height of the flagpole. Her eye is at A, 5 feet above the ground. A friend holds an 8 foot stick vertically 32.5 feet from the pole (so FD = 32.5’). If D’A = 3.31 feet, how tall would Latoya measure the flagpole to be if she can just see the top of the flagpole over the stick? Show how she would do this based on the given data. S-46 A C E F’ A  A (reflexive Prop.) AD’C  AF’E (Definition of Perpendicular / Rt. s ) 3’ 3.31’ 32.5 x ACD’  AEF’ by AA  Thm A A’ D F F’ D’ C E x x 3 = 32.5 + 3.31 3.31 8’ 3.31x = 3(35.81) 3’ 3.31x = 107.43 3.31’ x  32.46 ft 5’ height  32.46 + 5 height  37.46 ft The height of the pole is about 37.46 feet.

Use the information in each diagram to solve for x. Show that they are similar 1st. a) ABC  DEF b) Careful! A D 6 A A  A by reflexive property. 4 12 9 6 B C 7 7 4 E F x x The s are  by AA  Thm. x 7 = 6 4 x 7 = 9 6 or x 7 = 12 + 4 12 4x = 7(6) 12x = 7(16) 4x = 42 12x = 112 x = 10.5 u x = 9 u 1 3

Homework S 49 - 54