1.Team members may consult with each other, but all team members must participate and solve problems to earn any credit. 2.All teams will participate.

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Presentation transcript:

1.Team members may consult with each other, but all team members must participate and solve problems to earn any credit. 2.All teams will participate during all rounds – answers shown simultaneously on white boards. JEOPARDY! Geometry – Bench Mark 1 Review

Angle Madhouse Special Triangles Where Did I Go? Prove It! Be Reasonable Go To Final Jeopardy! 1000

Given:  CAT   DOG m  C = 72 , m  G = 45  AT = 12, DG = 15 Identify whether each of the following are true or false: 1.m  O = 63 m  O = 63  2.m  A = 45 m  A = 45  3.CA = 2CA = 2 100

1.TRUE since 180  - (72  + 45  ) = 63 TRUE since 180  - (72  + 45  ) = 63  2.FALSE since m  A = m  O !!FALSE since m  A = m  O !! 3.FALSE since < 15 – it couldn’t be a  !FALSE since < 15 – it couldn’t be a  ! 100 Given:  CAT   DOG m  C = 72 , m  G = 45  AT = 12, DG = 15 Question: True or False? 1.m  O = 63  2.m  A = 45  3.CA = 2

Identify the Triangle Congruence Theorem which applies for each of the figures above

AAS 2. HL 3. AAS or ASA depending on which two angle pairs you use. All 3 pairs are congruent.

Given:  ABE   ADE, AE bisects  BED Prove:  ABE   ADE 300 A E B D M

Step Reason. 1.  ABE   ADE 1. Given  ABE   ADE 1. Given 2.AE bisects  BED 2. GivenAE bisects  BED 2. Given 3.  BEM   DEM 3. Definition of angle bisector  BEM   DEM 3. Definition of angle bisector 4.AE  AE 4. Reflexive property of AE  AE 4. Reflexive property of  5.  ABE   ADE 5. AAS   Theorem  ABE   ADE 5. AAS   Theorem 300 Given:  ABE   ADE, AE bisects  BED Prove:  ABE   ADE A E B D M

Identify the 3 missing reasons in the proof above. 400 Step Reason. 1.c || d 1. Givenc || d 1. Given 2.  1   3 2. Given  1   3 2. Given 3.  1   2 3.  1    2   3 4.  2   a || b 5.a || b a b cd

400 Step Reason. 1.c || d 1. Givenc || d 1. Given 2.  1   3 2. Given  1   3 2. Given 3.  1   2 3. Corresponding  ’s Postulate  1   2 3. Corresponding  ’s Postulate 4.  2   3 4. Substitution Property of   2   3 4. Substitution Property of  5. a || b 5. Alternate Exterior  ’s CONVERSE Theorem 12 3 a b cd

Given: AE bisects BD, AE bisects  BAD Prove:  BAM   DAM 500 A E B D M

Step Reason. 1.AE bisects BD 1. GivenAE bisects BD 1. Given 2.AE  BD 2. GivenAE  BD 2. Given 3.BM  DM 3. Definition of segment bisectorBM  DM 3. Definition of segment bisector 4.  AMB,  AMD are 4. Definition of   AMB,  AMD are 4. Definition of  right angles 5.  AMB   AMD 5. Definition of right  ’s  AMB   AMD 5. Definition of right  ’s 6.AM  AM 6. Reflexive property of AM  AM 6. Reflexive property of  7.  BAM   DAM 7. SAS   Theorem  BAM   DAM 7. SAS   Theorem 500 Given: AE bisects BD, AE  BD Prove:  BAM   DAM A E B D M

100 F O X B Given: OX bisects  FOB m  BOX = 4x + 14, m  FOB = 84  Find: x

4x + 14 = 42 (half of 84!!) x = OX bisects  FOB M  BOX = 4x + 14, m  BOX = 84  Find m  BOX. F O X B

200 F O X B Given: OX bisects  FOB m  FOX = 2x + 21, m  BOX = 5x – 3 Find: m  FOB.

2x + 21 = 5x – 3 24 = 3x x = 8 Each half angle = 37 , so… m  FOB = 74  200 OX bisects  FOB M  FOX = 2x + 21, m  BOX = 5x – 3 Find m  FOB. F O X B

Given: 1 || 2, 3 || 4 Find: m  a, m  b 300 a b 31  110 

300 a b 31  110  m  a = 31 , m  b = 39  since (m  b + 31   = 180  )

Solve for x. 400 x 26  145 

x = 61  400 x 26  145  35  26  35 

Based on the following, find m  DAC. 500 A B E 49  97  D C

m  DAC = 180  – ( )  m  DAC = 56  500 A B E 49  97  D C 83  41 

100 xy 8 60  Find x and y:

100 xy 8 60  x = 8  3 y = 

200 Find x and y: x y 8 45 

200 x y 8 45 

300 The diagonal of a square is 7 inches. How long is a side of the square?

300

The side of an equilateral triangle equals 10 feet. Find the length of the altitude. 400

The “side” is the long 90  side of a . Altitude = 5  3 feet  30  10 feet 5 feet

The altitude of an equilateral triangle is 18 inches. Find the length of the perimeter of the triangle. 500

Perimeter = (12  3)  3 = 36  3 inches  30  18 inches 6  3 in 12  3 in

Point A is at (2, 5) and Point B is at (–3, 7). Find the new location of each point when they are translated according to the motion rule: (x, y)  (x – 6, y + 1) 100

A’ is at (–4, 6) B’ is at (–9, 8) 100 Point A is at (2, 5) and Point B is at (–3, 7). Find the new location of each point when they are translated according to the motion rule: (x, y)  (x – 6, y + 1)

Point A is at (3, 4) and Point B is at (–1, –5). Find the new location of each point when they are translated 2 units up and 4 units left. 200

Be careful…read the question… (up 2 = y + 2, left 4 = x – 4) A’ is at (–1, 6) B’ is at (–5, –3) 200 Point A is at (3, 4) and Point B is at (–1, –5). Find the new location of each point when they are translated 2 units up and 4 units left.

When A (–5, 2) and B (–3, 6) are reflected about the y-axis, the new coordinates of B’ are: 300

(3, 6) 300 When A (–5, 2) and B (–3, 6) are reflected about the y-axis, the new coordinates of B’ are:

Point A (6, –4) is first reflected about the origin, then reflected about the x-axis. The new coordinates of A’ are: 400

It first moves to (–6, 4), then it moves to (–6, –4). 400 Point A (6, –4) is first reflected about the origin, then reflected about the x-axis. The new coordinates of A’ are:

Daily Double 500

A (–7, 2) is rotated 90  counterclockwise. Find the location of A’. 500

The x-dimension and y- dimension switch every 90  and one sign changes. Since we rotated “left”, both the x and y became negative. (–2, –7) 500

Define inductive and deductive reasoning. Identify key phrases to help identify each type. 100

Inductive = Making a generalization based upon SPECIFIC EXAMPLES or a PATTERN Deductive = USING LOGIC to DRAW CONCLUSIONS based upon ACCEPTED STATEMENTS.

“If Kristina studies well, then Kristina scores at least 95% on the test.” Write the converse and the contrapositive statements. 200

Converse: (Switch the If and then parts) “If Kristina scores at least 95% on the test, then Kristina studied well.” Contrapositive (switch parts AND negate it) “If Kristina does NOT score at least 95% on the test, then Kristina did NOT study well.” “If Kristina studies well, then Kristina scores at least 95% on the test.”

“Two lines in a plane always intersect to form right angles.” Find one or more counterexamples. 300

1.Non-perpendicular, intersecting lines in the same planeNon-perpendicular, intersecting lines in the same plane 2.Parallel lines in the same plane.Parallel lines in the same plane. They have to be lines that LIE IN THE SAME PLANE. “Two lines in a plane always intersect to form right angles.” Find one or more counterexamples.

400 Which 8 pairs of congruent angles could be used to prove p || r? Why?

400  1   5,  2   6,  3   7,  4   8 Corresponding  Converse Theorem  3   6,  4   5 Alternate Interior  ’s Converse Theorem  1   8,  2   7 Alternate Exterior  ’s Converse Theorem

500 Explain how this construction can be used to prove  DAB   DAC by two possible methods.

Prove:  DAB   DAC Notice AB = AC from the first step of the construction. Notice BD = CD from the second step of the construction. Notice AD = AD (reflexive property!). This gives us SSS! Also, remember,  BAD   CAD by definition of “bisects”. This gives us SAS! 500

Write a proof by contradiction for the following Given:  A,  B, and  C are part of  ABC Prove:  A and  B are not both obtuse angles. 1000

Assume:  A and  B are both obtuse angles. This implies the measurements of both  A and  B are both more than 90 . BUT, this contradicts our given statement that the angles are part of  ABC since the angles of a triangle add to 180  ! Therefore, we may conclude:  A and  B are not both obtuse angles Write a proof by contradiction for the following Given:  A,  B, and  C are part of  ABC Prove:  A and  B are not both obtuse angles.

Final

Where’s Waldo??? Determine your final wagers now.

Waldo is hiding at (–9, –7). If Waldo goes through the following transformations, where is his new hideout? 1.Reflected about y = xReflected about y = x 2.Rotated 90  clockwiseRotated 90  clockwise 3.Reflected about the originReflected about the origin 4.Translated 3 down and 2 right.Translated 3 down and 2 right.

Waldo is hiding at (–9, –3). If Waldo goes through the following transformations, where is his new hideout? 1.Reflected about y = x… (–3, –9)Reflected about y = x… (–3, –9) 2.Rotated 90  clockwise … (–9, 3)Rotated 90  clockwise … (–9, 3) 3.Reflected about the origin … (9, –3)Reflected about the origin … (9, –3) 4.Translated 3 down and 2 right. … (11, –6)Translated 3 down and 2 right. … (11, –6)