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Geometry Semester 1 Final
Review packet continued
Geometry Review Semester 1
1. A proof is given
(3)
I have split the statements into a statement and
then a reason in a separate column, as we
have been doing.
Notice we have switched to congruent triangles
in the following problems.
1. ABCD is a parallelogram
Prove: AB = DC and AD =Reason
BC
Statement:
1. AB||DC and AD||BC
1. Definition of a parallelogram
2. Angle b is congruent to angle d
and angle a is congruent to angles
c
2. For parallel lines cut by a
transversal, alternate interior
angles are congruent
3. DB is congruent to DB
3. Reflexive property
4. Triangle DAB is congruent to
Triangle BCD
4. ASA congruence postulate
5. AB congruent to DC and AD
congruent to BC
5. Corresponding parts of congruent
triangles are congruent
The reason for step 5 of the proof
was
That corresponding parts of congruent triangles
are congruent.
2. What is the flaw in the proof?
Angle Angle Angle is a triangle similarity
postulate - not a triangle congruence postulate.
Remember we noticed that for similar triangles
corresponding angles were congruent, but corresponding
sides were proportional, but could be different size
triangles. For congruent triangles, corresponding sides
have to be congruent. Knowing all angles are congruent, is
not enough to prove triangles are congruent.
3.
Draw out the information given.
N
E
D
F
M
O
3.
We can see that if the angles at E and N are
congruent then we will have SAS congruence
N
E
D
F
M
O
Answer D) Angle E is congruent to angle N.
4. Which triangle congruence
theorem should be used...
Label the diagram with the given information.
A
C
B
D
4. Which triangle congruence
theorem should be used...
BD is congruent to itself (reflexive property)
A
C
B
D
SSS congruence postulate applies
5. Complete the 4th line of the proof
(Diagram needs an F where angle 2 is.)
1. Given
2. Perpendicular lines meet to form right
angles.
3. Right angles are congruent.
4.
5. Corresponding parts of congruent triangles
are congruent.
Statement: 4. Triangle ABC is
congruent to triangle EFD
Reason:
4. Triangle ACB is congruent to triangle
CFD ASA Congruence postulate
Geometry: Semester 1 Review (4)
1) If triangle BDF is similar to triangle CEG,
what is the ratio from triangle BDF to triangle
CEG?
BD = 8
CE = 10 corresponding sides
DF = 4
EG = 5 corresponding sides
Ratio 4 : 5
Geometry: Semester 1 Review (4)
1) If triangle BDF is similar to triangle CEG,
what is the ratio from triangle BDF to triangle
CEG?
BD = 8
CE = 10 corresponding sides
DF = 4
EG = 5 corresponding sides
Ratio 4 : 5
(Scale factor 5/4, check 6 x 5/4 = 7.5, yes).
Rectangle - 90 degree vertices,
opposite sides are congruent
2)
100°
40°
40°
AAS congruence
3. Mark the information on the
diagram.
E
D *
C
W
X*
Y
3. If we know XY congruent to DC,
then we have SAS congruence
E
D *
C
W
X*
Or if we know
that the angle at
E is congruent to
the angle at W,
then we have
ASA congruence
Y
4. Diagonals of a quadrilateral are
congruent and bisect each other.
Review page 62 in your Unit 2 workbook
4. Diagonals of a quadrilateral are
congruent and bisect each other.
P 62 Unit 2 workbook
4. Diagonals of a quadrilateral are
congruent and bisect each other.
P 62 Unit 2 workbook
4. Diagonals of a quadrilateral are
congruent and bisect each other.
Best description of the quadrilateral is a
rectangle.
Question 5 is the same as qu. 3 on
sheet (3)
Geometry Review Semester 1 (5)
1. Find geometric mean.
1a
5=x
x 15
x2 = 75 = 3(25)
x = √3√25 = 5√3
Geometry Review Semester 1 (5) geometric mean
1b
5=x
x 45
x2 = 225
x = √225 = 15
Geometry Review Semester 1 (5)
1c
9=x
x 4
x2 = 36
x = √36 = 6
Geometry Review Semester 1 (5)
1d
8=x
x 10
x2 = 80 = 16(5)
x = √16√5= 4√5
2. Determine x, y and z. Exact
answers.
y is the geometric mean of 4 and 9 (altitude
of the large right triangle)
4cm = y
y
9cm
y2 = 36cm2
y = √36cm2 = 6cm
2. Determine x, y and z. Exact
answers.
Using Pythagorean theorem
x2 = y2 + (4cm)2
x2 = (6cm)2 + (4cm)2
x2 = 36cm2 +16cm2 = 52cm2
x = √52cm2 = √4√13cm2 = 2√13cm
2. Determine x, y and z. Exact
answers.
Using Pythagorean theorem
132 = 2√132 + z2
169 = 52 + z2
z2 = 169 - 52 =117= 9(13)
z = √9(13) = 3√13
3. Determine x, y and z. Exact
answers.
y is the geometric mean of 3 and 8 (altitude
of the large right triangle)
3cm = y
y
8cm
y2 = 32cm2
y = √2(16)cm2 = 4√2cm
3. Determine x, y and z. Exact
answers.
Using Pythagorean theorem
x2 = (4√2)2 + (3cm)2
x2 = (32 + 9)cm2
x = √41cm
3. Determine x, y and z. Exact
answers.
Using Pythagorean theorem
z2 = (4√2)2 + (8cm)2
z2 = (32 + 64)cm2
z = √98cm2 = √2(49)cm2 = 7√2cm
4. Determine x, y and z. Exact
answers.
Using Pythagorean theorem
152 = 92 + x2
225 - 81 = x2
x = √144 = 12
4. Determine x, y and z. Exact
answers.
12 is the geometric mean of 9 and y
(altitude of the large right triangle)
9 = 12
12
y
y = 12(12) = 144 = 16
9
9
4. Determine x, y and z. Exact
answers.
Using Pythagorean theorem
z2 = (12)2 + (16)2
z2 = (144 + 256)cm2
z = √400 = 20
Extra work on Dilations
Unit 3A Workbook, page 17
Dilations (pg 17)
1. On Graph A plot triangle ABC with
A(2,1), B(3, 2) and C (1,3)
2. On the SAME graph, Plot D-3 as triangle
A’B’C’
A(2,1)(-3)
maps to A’(-6, -3)
B(3, 2)(-3) maps to B’
C’(1, 3)(-3) maps to C’
Dilations (pg 17)
1. On Graph A plot triangle ABC with
A(2,1), B(3, 2) and C (1,3)
2. On the SAME graph, Plot D(-3, O)as
triangle A’B’C’
A(2,1)(-3) maps to A’(-6, -3)
B(3, 2)(-3) maps to B’ (-9, -6)
C’(1, 3)(-3) maps to C’ (-3, -9)
Dilations cont.
3.Looking at Graph A and your two
triangles we can see that
angle ABC is congruent to angle A’B’C’
angle BCA is congruent to angle B’C’A’
angle CAB is congruent to angle C’A’B’
Dilations cont.
4.Find the length of line segment AB
2
2
2
AB =(1 +1 ) = 2
So length of AB = √2
Count squares - it’s easier!
5.Find points A’B’. Find the length of line
segment A’B’
2
2
2
A’B’ = {(-9 - -6) +(-6 - -3) }
= (32 + 32) = (9 + 9) = 18
A’B’ = √ (2)(9) = 3√ 2
A’B’= √ (2)(9) = 3√ 2
6. How do the lengths of AB and A’B’
relate to our original dilation?
The original dilation was scale factor
-3, and we see the length of AB
multiplied by 3 gives us the length of
A’B’ ;
3AB = A’B’