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.

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?

100

200

300

400

500

100

200

300

400

500

100

200

300

400

500

100

200

300

400

500

Prove

It!

100

200

300

400

500

Be Reasonable

Go To Final Jeopardy!

1000

Given: CAT DOG

mC = 72, mG = 45

AT = 12, DG = 15

Identify whether each of the following are true or false:1.mO = 632.mA = 453.CA = 2

100

1. TRUE since 180 - (72 + 45) = 63

2. FALSE since mA = mO !!

3. FALSE since 2 + 12 < 15 – it couldn’t be a !

100

Given: CAT DOG

mC = 72, mG = 45

AT = 12, DG = 15Question: True or False?1.mO = 632.mA = 453.CA = 2

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

200

1 2 3

200

1 2 3

1. 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. Given2.AE bisects BED 2. Given3.BEM DEM 3. Definition of angle bisector4.AE AE 4. Reflexive property of 5.ABE ADE 5. AAS Theorem

300

Given: ABE ADE,

AE bisects BEDProve: ABE ADE

A E

B

D

M

Identify the 3 missing reasons in the proof above.

400

Step Reason .1.c || d 1. Given2.1 3 2. Given3.1 2 3. 4.2 3 4. 5.a || b 5.

12

3

a

b

c d

400

Step Reason .1.c || d 1. Given2.1 3 2. Given3.1 2 3. Corresponding ’s Postulate4.2 3 4. Substitution Property of 5. a || b 5. Alternate Exterior ’s CONVERSE Theorem

12

3

a

b

c d

Given: AE bisects BD,

AE bisects BAD

Prove: BAM DAM

500

A E

B

D

M

Step Reason .1.AE bisects BD 1. Given2.AE BD 2. Given3.BM DM 3. Definition of segment bisector4.AMB, AMD are 4. Definition of right angles5.AMB AMD 5. Definition of right ’s6.AM AM 6. Reflexive property of 7.BAM DAM 7. SAS Theorem

500

Given: AE bisects BD,

AE BDProve: BAM DAM

A E

B

D

M

100

F

O

X

B

Given: OX bisects FOB mBOX = 4x + 14, mFOB

= 84Find: x

4x + 14 = 42 (half of 84!!)

x = 7

100OX bisects FOB MBOX = 4x + 14, mBOX = 84 Find mBOX.

F

O

XB

200

F

O

X

B

Given: OX bisects FOB mFOX = 2x + 21, mBOX

= 5x – 3Find: mFOB.

2x + 21 = 5x – 3

24 = 3x

x = 8

Each half angle = 37, so…

mFOB = 74

200OX bisects FOB MFOX = 2x + 21, mBOX = 5x – 3 Find mFOB.

F

O

XB

Given: 1 || 2, 3 || 4Find: ma, mb

300

a

b 31 110

1 2

3

4

300

a

b 31 110

ma = 31, mb = 39

since (mb + 31 + 110 = 180)

1 2

3

4

Solve for x.

400

x

26

145

x = 61

400x

26

145 35

2626

35

Based on the following, find mDAC.

500

A

B

E

49

97

D

C

mDAC = 180 – (41 + 83)

mDAC = 56

500A

B

E

49

97

D

C

83

41

100

x y

860

Find x and y:

100

x y

860

x = 83

y = 16

30

200

Find x and y:

x

y

8

45

200

x

y

8

4524

24 2

28

2

2

2

8

y

x

300

The diagonal of a square is 7 inches.

How long is a side of the square?

300

2

27

2

2

2

7side

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 30-60-90 .

Altitude = 53 feet

400

60

30

10 feet

5 feet5 feet

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

500

Perimeter = (123) 3 =

363 inches

500

60

30

18 inches

63 in

123 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

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

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

300

1.Non-perpendicular, intersecting lines in the same plane

2.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 Theorem1 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 DACNotice 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.

1000

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 = x

2.Rotated 90 clockwise

3.Reflected about the origin

4.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)

2.Rotated 90 clockwise … (–9, 3)

3.Reflected about the origin … (9, –3)

4.Translated 3 down and 2 right. … (11, –6)