1.team members may consult with each other, but all team members must participate and solve problems...
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?
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)