Download Geometry – AP Book 8, Part 2: Unit 3

Geometry – AP Book 8, Part 2: Unit 3
IMPORTANT NOTE:
For many questions in this
unit, there are multiple correct
answers, e.g. line segment AB
can be written as BA, RST is
the same as TSR, etc.
AP Book G8-16
page 59
AP Book G8-17
page 61
1.
Teacher to check.
1.
2.
b)
c)
Where appropriate, teachers
should be sure to check for
similar equivalent answers in
their students’ work.
3.
AP Book G8-15
page 57
a)
Acute
Circle: V
UVW, WVU
b)
Obtuse
c)
Acute
Circle: E
FED, DEF
d)
Obtuse
d)
Circle: M
LMN, NML
a)
HIJ
b)
ACD
c)
RMK
3.
1.
Answers will vary –
teacher to check.
2.
a)
BA
b)
DE, ED
 2 obtuse: AED, CEB
c)
ST, TS
 2 straight: AEB, CED
a)
Teacher to check.
4.
b)
Line segment
 2 acute: AEC, DEB
4.
6.9 cm = 69 mm
5.
Teacher to check.
6.
a)
b)
7.
S
iii)
K
iv)
H
ii)
ST
iii)
KJ
iv)
HG
a)
RK
b)
BM, BQ
c)
DA, XD
Answers will vary –
teacher to check.
9.
a)
AB, E
b)
RS, FG, C
c)
MN, JI, I
a)
EF, UT, S
b)
AH, DF, P
c)
IE, IG, I
11.
Circle two:
ATB or BTA,
CTB or BTC
5.
For endpoint, circle:
ii)
8.
10.
BONUS
There are 6 unique angles:
6.
7.
8.
5.
R
6.
7.
a)
Acute; 30
b)
Obtuse; 150
c)
Obtuse; 130
d)
Acute; 40
a)
30
b)
130
c)
78
a)
65, 25;
90
b)
30, 60;
90
Circle: B, C
BAC, ABC, BCA
AP Book G8-18
page 63
b)
Circle: L, K, J
KLJ, LKJ, KJL
1.
a)
QR
b)
AB, BC, CA
c)
WX, XY, YW
b)
KMJL
c)
YVTZX
d)
DEF
a)
Circle three:
ALYXBK, BKALYX,
XYLAKB
8.
Teacher to check.
Circle: right
* not necessarily a
rectangle
b)
Circle: left
* not necessarily a
right angle triangle
B
9.
D
Triangle 1:
4 cm
a)
Triangle 2:
Circle: left
a)
Circle: left
b)
Circle: right
c)
Circle: left
a)
4 cm
3 cm
Circle: left
b)
B
a)
a)
c)
4.
D
15 cm
a)
3.
A rectangular gift
box is 15 cm long
and 10 cm wide.
It has a transparent
lid decorated with a
paper flower 5 cm
in diameter. What is
the area of the
bottom of the box?
10 cm
A
Sample answers:
A
Circle: left
a)
b)
C
BONUS
Answers will vary – teacher
to check.
b)
70
b)
Answers will vary –
teacher to check.
Circle: right
Sample sketches:
2.
Q
a)
Teacher to check.
a)
b)
Answers will vary –
teacher to check.
4.
45
3 cm
3 cm
5 cm
10.
Triangle 1:
2 cm
40
6 cm
b)
C
70
120
70
AB, BC, CD, DA
A
b)
Sample answer:
B
F
X
P
Y
A
C
D
AB, BC, CD, DE, EF, FA
BONUS
A
a)
C
E
B
5.
a)
Circle two: D, C
b)
None
b)
1 cm
A
C
Triangle 2:
B
2 cm
100
D
40
T
11.
B
40
a)
Circle: left
b)
Circle: right
1 cm
V
U
2 cm
V‐14 AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
3.
2.
P
c)
Geometry – AP Book 8, Part 2: Unit 3
12.
Answers will vary –
teacher to check.
AP Book G8-19
page 66
Sample answers:
1.
a)
A
B
D
C
A
B
2.
J
b)
CD
b)
EF  KJ
c)
PQ  DE
d)
CB  AB
K
L
M
P
c)
Q
S
R
P
E
C
D
2
2
2
4.
a)
a)
a)
b)
Teacher to check.
a)
A
b)
c)
No;
B doesn’t show that
the wall and ground
are perpendicular.
7.
Says: wall = vertical,
ground = horizontal
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
ACB;
ACB + BCE;
DCE + BCE;
ACD + DCE
60
b)
115
c)
125
C
d)
112, 72
BC  CD
AD  DC
e)
65, 121
f)
52
Yes;
Since its angles are
all 90, any pair of
adjacent sides will
be perpendicular.
3.
g)
90
h)
65, 65
a)
1, 3; 4, 2;
4, 1; 3, 2
40;
ABC =
ABD + DBC
b)
180, 180 ‒ 3;
180, 180 ‒ 3;
They are equal.
 DBC = (90 ‒ 50)
c)
1, 2
b)
B, C
c)
B, C
 3 = 4
2
d)
1, 2; 3, 4
2
e)
1, 5; 5, 1
f)
3, 2; 3, 2
a)
45
b)
53
c)
50
8.
x + 5 = 13 ;
2
c)
a)
A;
It’s simpler, with only
the essential details.
2
1, 5; 3, 7;
1, 5; 3, 7
Teacher to check.
d)
2
b)
b)
A
2
COB; AOD;
BOD; BOC
As straight angles,
2 + 3 = 180
and 2 + 4 = 180,
which means that:
2 + 3 = 2 + 4
c)
x = 13 – 5
= 169 – 25 = 144
 x = 12 m
f)
a)
AB  BC
AD  AB
2.
a)
a)
D
B
 perpendicular
e)
1.
Teacher to check.
b)
 c = 13 cm
14.
AP Book G8-20
page 69
The two lines
are perpendicular
because the angle
was constructed
to be 90.
2
No; 12.5 m
AnswerKeysforAPBook8.2
9.
55, 115, 115
Answers will vary –
teacher to check.
5.
2
2
f)
a – 20
C, A, B
Teacher to check.
c = 5 + 12 = 169
51, 138
3.
4.
 c = 5 cm
b)
105, 75
e)
2.
B; D; A; C
c =a +b
2
2
= 3 + 4 = 25
d)
Teacher to check line
extensions; the lines will
intersect.
Teacher to check
additional information.
a)
60, 90, 30
1.
3.
6.
50, 105
c)
 a = 55, (a – 20) = 35
B
E
13.
The 4 middle angles
are 90.
Sample explanation:
A
a
b)
The 4 corner and
the 4 middle angles
are all 90.
D
A
 x = 18, 4x = 72
11.
135
c)
R
C
x + 4x = 90
so x = 90 ÷ 5
a)
AP Book G8-21
page 71
Q
S
5.
AOB = DOC;
AOD = BOC
a + (a – 20) = 90
2a = 110
B
d)
x
f)
The 4 corner angles
are 90.
L
J
4x
b)
K
M
10.
Teacher to check.
a)
C
D
a)
(continued)
4.
b)
c)
d)
5.
6.
a)
XY
b)
FG || RS
c)
LM || GH
AD = BE = CF = 1.35 cm;
They are all equal.
AP Book G8-22
page 72
1.
9, 3
4, 10
13, 7
8, 14
3, 9
4, 10
13, 7
8, 14
a)
DOB; COB
b)
APD = CPB;
APC = DPB
INVESTIGATION 1
c)
4, 3; 8, 7
A.
3, 4, 7, 8
d)
2 = 5, 1 = 4;
3 = 6
B.
100, 80
100, 80
80, 100
80, 100
DCE = ACB;
DCA = ECB
C.
e)
equal
V‐15
1.
a)
5, 7; 6, 8
b)
3, 4; 6
c)
11, 13; 12, 2
INVESTIGATION 2
A.
3, 4
B.
45, 135, 75, 105
C.
C.
D.
3.
4.
115
b)
111; 69
DC, DC;
BC, BC
c)
x + c = 180
(supplementary)
c)
Answers may vary –
teacher to check.
d)
a + b + c
= x + c
Answers may vary –
teacher to check.
2.
a)
 x = a + b
Sample explanation:
By extending the
length of the sides,
you can use the
presence of equal
corresponding
angles to prove that
the pairs of opposite
sides are parallel.
 4 = 5.
48
AP Book G8-25
page 78
INVESTIGATION
A.
G
B, C, D
A, E, F
B.
D, F;
C, E, G;
A, B
D, F
C, E, G
A, B
c)
78
b)
72, 68
d)
128; 52
c)
92, 88
a)
 = 125°
AP Book G8-24
page 76
 = 55°
1.
180, 180;
180, 150;
30
2.
a)
70
C.
isosceles; equilateral
b)
30, 30;
120
1.
a)
120; obtuse
b)
90; right
c)
80; acute
a)
no (or 0); scalene
e)
105; 95
f)
60; 45
a)
85; parallel
b)
95; not parallel
c)
63; parallel
d)
99; not parallel
b)
none
c)
none
d)
k || l
e)
none
f)
r || s
g)
x || w
h)
y || z, a || b
3.
 
 
 
 
b)
a)
6; 7
b)
5; 6
c)
Answers may vary –
teacher to check.
 = 95°
 = 122°
 = 85°
 = 58°
 
 


 
 
c)
AP Book G8-23
page 74
1.
b)
Since m || n, we know
4 = 8 (corresponding)
and 8 = 5 (opposite)
No
a)
parallel;
corresponding;
opposite, 7;
6
Sample proof:
(since the rays aren’t parallel)
2.
(continued)






 = 125°
4.
 = 55°

 

 

 

 
NOTE:
Students should give
6 unique angle pairs.
4.
OPL, OPL; OPL;
parallel; MN || KL
8, 5;
5.
a)
a || c
11, 2;
b)
d || g (given), e || f
3; 10
c)
p || k (given)
INVESTIGATION
6.
d)
35; 60; 85
e)
55; 35
b)
2; isosceles
f)
43; 43; 69
c)
3; equilateral
a)
90
b)
90
c)
2.
3.
4 m or 7 m
(see G8-18 #9 for details)
60
4.
Teacher to check.
a)
DCA; equal;
DCA
5.
a)
b)
ECB;
equal, ECB
c)
= ACD + ACB
+ BCE
= 180
right
acute obtuse
scalene



isosceles




equilateral
b)
Neither an obtuse
or a right equilateral
triangle is possible;
b)
55, 125
c)
45
d)
70, 60
Answers may vary –
teacher to check.
e)
120
f)
120, 120
Sample explanation:
g)
54, 50, 83
h)
132, 92, 44
i)
45, 75, 60
a)
a + b
∆DEG:
obtuse, isosceles
b)
a + b + c = 180
(form a triangle)
∆HJI:
right, isosceles
6, 5
NOTE: Angle measures
may vary slightly but each
pair of supplementary
angles must add to 180.
Since a || b, we know that
1 = 5 (corresponding)
and 1 + 3 = 180
(supplementary)
107, 73, 73, 107;
equal
 5 + 3 = 180.
V‐16 180, 90, 40;
50
93
B.
a)
Teacher to check.
5.
c)
a)
A.
7.
3.
A, E, F;
B, C, D;
G
6.
An equilateral
triangle always has
3 equal angles, each
one measuring 60
(which is acute).
6.
∆ABC:
right, scalene
∆DEF, ∆GEF:
right, scalene
AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
Geometry – AP Book 8, Part 2: Unit 3
Geometry – AP Book 8, Part 2: Unit 3
its other two angles
must be acute – this
is why b) has only
one solution.
∆NLM:
right, scalene
∆NLK:
right, isosceles
since KNM = 105
8.
a)
Teacher to check.
b)
i)
Yes;
2 options
ii)
No
iii)
Yes;
Infinite options
as AB’s length
changes
Explanations and
sketches may vary –
teacher to check.
a)
AP Book G8-26
page 80
1.
b)
57 (OAT)
80 (SATT)
c)
55 (CAT)
65 (SATT)
d)
75 (AAT)
45 (AAT)
60 (SAT or SATT)
Sample answer:
Both a and
b are obtuse
angles.
a
In a triangle, the
angles must add to
180 so, if one is
greater than 90
(obtuse), the sum
of the other two
must be less than
90 ‒ meaning
they’re both acute.
b)
e)
f)
g)
(CAT)
(SAT)
(OAT)
(CAT or SAT)
h)
35
35
55
125
(SATT)
(AAT)
(AAT or SATT)
(SAT)
i)
81
39
39
60
(AAT)
(SATT)
(AAT)
(SATT)
Sample answer:
a = b = 90
b
Again, the angles’
sum must be 180
so, if one is 90, the
other two must add
to 90 ‒ so they
must both be acute.
9.
a)
Triangle 1: 30, 120
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
Triangle 2: 75, 75
b)
35, 35
c)
In a), we are given
an acute angle.
2.
55 (OAT)
95 (AAT)
30 (SATT)
115
65
115
65
No
a
119 (SAT)
119 (AAT)
119 (CAT)
Specific theorems used
by students may vary –
teacher to check.
Also teacher to check any
numbered “helper” angles
marked by students.
In b), we are given
an obtuse angle.
a)
1 = 67° (EAT)
2 = 83° (SATT)
They have a
different number of
solutions because,
if a given triangle
has an obtuse angle,
b)
1 = 32° (CAT)
2 = 58° (CAT)
c)
1 = 75° (AAT/SAT)
2 = 130° (SAT)
AnswerKeysforAPBook8.2
Again, specific theorems
used may vary – teacher
to check.
a)
1 = 2 = 37 (ITT)
b)
1 + 2 = 60 (EAT)
1 = 2 = 30 (ITT)
c)
d)
4.
No
b
3.
In a), we’re given
an acute angle,
which might either
be one of the two
equal angles in the
isosceles triangle,
or the unequal angle
– this is why it has
two possible
solutions.
∆KNM:
obtuse, scalene
7.
(continued)
We know that the
other two angles in
the triangle are each
55 (ITT), so:
1 = 2 = 125 (SAT)
1 = 80
2 = 50
g)
by EAT, SATT:
4x – 30 = (x + 5) + 40
3x = 75
x = 25
 angles are 40, 30 and 110.
AP Book G8-27
page 82
1.
(EAT)
(ITT)
NOTE:
In c) and d), the missing
angles can be found
using ITT only – students
don’t necessarily have to
solve for x.
Teacher to check marked
angles and sides.
a)
2.
rectangle
b)
parallelogram
c)
square
d)
(right) trapezoid
a)
rhombus
b)
rectangle
BONUS
b)
trapezoid
by SATT:
70 + (x + 20) + (3x + 10) = 180
4x + 100 = 180
4x = 80
x = 20
3.
 angles are 70, 40 and 70.
4.
Teacher to check.
5.
Similarities:
 Both are quadrilaterals.
c)
by ITT:
6x + 5 = 35
6x = 30
x =5
all
b)
no
c)
no
d)
some
Differences:
 A parallelogram has
two pairs of parallel
sides; a trapezoid has
only one.
 angles are 35, 35 and 110.
d)
by ITT, SATT:
96 + 2x + 2x
96 + 4x
4x
x
a)
 A parallelogram has
equal opposite sides
and angles; this isn’t
true for a trapezoid.
= 180
= 180
= 84
= 21
6.
a)
It’s a parallelogram
with 4 right angles.
b)
Their sides are not
all equal.
c)
A trapezoid has one
pair of parallel sides;
a parallelogram has
two.
d)
It’s a parallelogram
with all sides equal.
 angles are 96, 42 and 42.
e)
by SATT:
90 + (2x + 15) + (4x + 45)
6x + 150
6x
x
= 180
= 180
= 30
=5
 angles are 90, 25 and 65.
f)
by SAT, SATT:
45 + (5x + 15) + (4x + 12)
9x + 72
9x
x
= 180
= 180
= 108
= 12
 angles are 45, 75 and 60.
7.
Teacher to check for
proper sketches.
a)
One square plus
many different
rhombuses
V‐17
b)
c)
8.
One rectangle plus
many different kites,
parallelograms,
and quadrilaterals
with an indentation
B.
2; 2
C.
180°; 180°; BAD; 180°;
ADC
1.
Explanations and
(counter) examples may
vary – teacher to check.
The kites and any
shapes with an
indentation aren’t
parallelograms.
a)
10.
a)
a
Two correct options:
ABC and BCD or
ADC and BAD
b)
Yes; Yes
c)
Two correct options:
ABC and BAD or
ADC and BCD
d)
No (185); No
e)
Trapezoid
a)
Teacher to check.
Property
Shapes
No e/s
F, G, J
1 pair e/s
I
B, C
No p/s
D, F, H
1 pair p/s
G, I, J
c)
No r/a
B, D, E, G, I
1 r/a
F
2 r/a
H, J
4 r/a
A, C
No e/a
F, G
1 pair e/a
D, H, J
In the above
trapezoid, we see
that a + b ≠ 180°.
A, C
AP Book G8-28
page 84
Here, the two sets
of opposite angles
add to 180° but this
quadrilateral is a kite,
not a trapezoid or a
parallelogram.
INVESTIGATION
A.
Teacher to check.
In the top parallelogram,
the opposite angles are
130° and 50°.
In the other, the opposite
angles are 110° and 70°.
V‐18 False;
Sample counterexample:
2 pairs e/a B, E, I
4 e/a
b
2.
a)
It doesn’t talk about
a rectangle being a
parallelogram.
a
b
d
c
2
2
8
2
2
29
2
2
29
2
2
8
a=
2 +2 =
b=
2 +5 =
and the reminder at
the top of page 83:
c=
2 +5 =
d=
2 +2 =
A + D = 180°
 AB || DC
 a = d and b = c
so this is a kite.
a
b)
d
So ABCD must be
a parallelogram.
Since ABCD is a
parallelogram and
it has 4 right angles,
we know ABCD is a
rectangle.
b
c
a=5
b=
2
2
40
2
2
40
2
2
25
2 +6 =
2 +6 =
c=
 Jean is correct.
BONUS
If a quadrilateral has two pairs
of equal angles (let’s call them
pair a and pair b), then only
two possible cases exist:
a
d)
C
Sample counterexample:
False;
a)
B
False;
Sample counterexample:
2 pairs p/s A, B, C, E
Using the following
sketch,
3.
Isosceles trapezoids
have 2 pairs of equal
angles but are not
parallelograms.
2 pairs e/s A, D, E, H
4 e/s
c
So, from the
reminder box on
page 83, we know
that each pair of
adjacent angles
(a/b, b/c, c/d and
d/a) adds to 180°.
b)
c)
4.
A + B = 180°
 AD || BC
b
d
b)
Yes, this will always
be true. (A square
is also a rectangle.)
D
True;
In a parallelogram,
the opposite sides
are always parallel.
Trapezoid, parallelogram
and rhombus
b)
A
Sample explanation:
Teacher to check that
drawings are correct.
9.
(continued)
d= 4 +3 =
=5
A, B, E
 a = d and b = c
so this is a kite.
c)
a
b
1. the pairs of equal angles
are
opposite each other;
c
d
2. the pairs of equal angles
are
adjacent to each
other.
a= 4 +3 =
=5
2
2
25
2
2
41
2
2
45
2
2
13
Case 1: Opposite
Looking clockwise, the angle
order is a, b, a, b.
b=
c=
3 +6 =
As a quadrilateral, we know:
d=
2 +3 =
a + b + a + b = 360°
2(a + b) = 360°
a + b = 180°
From this, as in #2 c) above,
we see that all four adjacent
angle pairs add to 180°, so
opposite sides will be parallel
 it’s a parallelogram.
Case 2: Adjacent
Looking clockwise, the angle
order here is a, a, b, b.
As a quadrilateral, a + b =
180° also applies here. So,
with two adjacent (and distinct)
angle pairs adding to 180°,
this shape will have one set of
parallel sides  it’s a
trapezoid.
4 +5 =
 this is not a kite.
5.
Teacher to check drawn
diagonals.
The kites have diagonals
that intersect at a right
angle.
6.
a)
a = 5 units
c = 2 units
b = d = 3 units
b)
Correct prediction:
A kite’s diagonals
are perpendicular
and one of the
diagonals bisects
the other.
Teacher to check
drawings of the
two other kites.
AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
Geometry – AP Book 8, Part 2: Unit 3
Geometry – AP Book 8, Part 2: Unit 3
Teacher to check.
7.
a), b)
Students should
notice that kites
always have
one pair of equal,
opposite angles:
7.
B
A
C
O
8.
A = 180° ‒ B ‒ C
= 180° ‒ S ‒ T
= R
Rhombus
a), b)
K
7.
1.
A
D
E
F
H
O
G
L
C
8.
Square
b)
Rhombus
c)
Square
d)
Rhombus
2.
3.
9.
10.
No
Yes
d)
Yes
No
a)
4.
Teacher to check.
5.
a)
A
S
i)
C = F
B = D
A = E
4.
Sample:
6.
iii)
Y
O
Sample:
D
2
XY = ZY
M2
2
2
OX + OW = OZ + OW
2
XW = ZW
2
Teacher to check.
AnswerKeysforAPBook8.2
11.
2
 XW = ZW
Teacher to check.
2.
EF, FG, EG
3.
P, R, Q
4.
5.
12.
7.
a)
No
b)
Yes (SAS)
c)
No
d)
Yes (SSS)
V = Y
UW = XZ
W = Z
8.
None of Tom’s statements
are correct;
AP Book G8-31
page 91
D = S, E = L,
F = M
b)
ASA;
∆RST  ∆ZXY
W = R, X = T,
Y = S;
c)
SSS;
∆MNO  ∆FGH
a)
b)
H
Teacher to check triangle
sketches.
∆PQR ∆KLM
VW = YZ
54°
C
Corresponding angles
SAS;
∆DEF
No; she forgot to check if
the corresponding sides
were equal.
36°
4 cm
Teacher to check triangle
sketches.
F
Sample sketch:
C

2.
6.4 cm
5 cm
No, they aren’t congrent
– their equal sides are
opposite different angles.
Corresponding sides
a)
WY = RS, XW = TR,
XY = TS
54°
∆ABC  ∆XVW
1.
36°
One equal side:
BC = GK
Sample:
∆PQR ∆KLM
 XY = ZY
6.
Sample:
∆PQR ∆KLM
K
5 cm
G
Three equal angles:
A = K, C = G,
B = H
∆PQR ∆KLM
BONUS
2
2
2
2
OX + OY = OZ + OY
M1
Teacher to check.
B
A
G = J,
H = K,
I = L
∆GHI  ∆JKL
Kite
2
Teacher to check.
5.
∆IHG  ∆LKJ
AP Book G8-30
page 88
M2
M1, M2
X
b)
1.
C
ii)
a)
2
i)
All six angles
are equal.
Z
M1, M2
M1
b)
10.
B
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
c)
7: ∆ACB  ∆RTS
iii)
Rectangle
W
No, there is
nothing to indicate
that FE = ED.
A, C
5: ∆WXY  ∆RTS
R
c)
b)
d)
Yes
∆CBA  ∆FED
P
C, DB; N, AB
D
Yes
ii)
O
a)
c)
b)
a), b)
Midpoint at 4 cm;
Teacher to check.
A
Yes
These four angles
are equal.
Square
Corresponding? Equal?
No
∆ABC  ∆DEF
6 cm; 3 cm;
Teacher to check that
proper midpoint is
marked.
b)
M
Q
BONUS
No, it can’t. By definition, a line
is infinite and has no endpoints
 it can have no midpoint.
6.
c)
9.
3.
3.7 cm
a)
AP Book G8-29
page 86
1.
N
2.
No;
They’re equal but
are positioned
differently in relation
to the corresponding
angles.
a)
 A = R, even though
they aren’t marked.
8.
B
c)
In this case, B = S
and C = T, so:
D
c)
The third angle in any
triangle will be equal to
180° ‒ 1 ‒ 2.
3 cm
c)
(continued)

D
B

H

G
NOTE: ∆BCD and ∆FGH
have two equal angles so
they are similar by AA.
a)
No, this is just
further evidence
of their similarity.
No;
They aren’t equal.
b)
Yes (ASA)
c)
Yes (AAS)
Yes
d)
No
V‐19
9.
a)
13.
a)
(continued)
No, by SSS and
Pythagorean;
One’s sides that are
c)
b)
10.
b)
AP Book G8-32
page 94
Case 1:
1.
A
D


E
Here, ∆ABC  ∆DEF.
2.
53;
53 + 53 = 106
b)
75;
75 + 75 = 150
c)
20;
20 + 20 = 40
a)
Case 2:
A

B
F
D

E
b)
C
Here, ∆ABC  ∆DEF.
INVESTIGATION
A.
A = D, CB = EF and
AB = DE
B.
No: the corresponding
angle isn’t between the
corresponding sides.
C.
No
D.
No, neither has its
corresponding angle
between its
corresponding sides –
only SAS is a proper
congruence rule.
3.
4.
5.
V‐20 T
F
iii)
T
iv)
T
a)
B.
They will be equal (think
of an isosceles triangle).
C.
∆ACM  ∆BCM by SAS:
AM = BM (given),
AMC = BMC (given),
and side CM is common.
D
C
8.
a)
DC;
BAC = DCA
b)
BC;
BCA = DAC
c)
∆ABC  ∆CDA (ASA)
d)
BC = DA, AB = CD;
Yes
a)
Measurements
may vary slightly but
relationships must
remain to be correct.
AE = CE = 2.8 cm
BE = DE = 1.5 cm
b)
SSS
This congruency means
that corresponding sides
AC and AB are equal.
They are equal.
A
B
INVESTIGATION 1
T
T
7.
AP Book G8-33
page 97
b)
e)
SSS
A
AC = BC
BONUS
Explanations will vary –
teacher to check.
d)
C
B
E
A.
SSS
CB, DC;
∆CBD;
∆CBD, SSS;
CBD;
CBD, 90º;
AC;
AC
D
F
SAS
c)
9.
a)
F
2
Yes, by SSS and
Pythagorean;
Both triangles have sides
that are 9 cm, 12 cm and
15 cm in length.
6.
i)
ii)
From c), we know
that AE = CE and
BE = DE
 the parallolograms’ diagonals
(AC and BD) bisect
each other at E.
This proves that BD
is the perpendicular
bisector of AC.
T
b)
d)
But AEC is a
straight angle, so:
AEB = BEC = 90
and  AC  BD.
f)
BAE = DCE (AAT)
ABE = CDE (AAT)
and AB = CD,
 we know that
∆ABE  ∆CDE (ASA)
Since ∆ABE  ∆CBE
(SAS), we know that:
AEB = CEB.
d)
ASA
c)
Sample explanation:
From d), we know
∆ABE  ∆CBE and
 AE = CE.
Correct prediction:
The diagonals of
a parallelogram will
bisect each other.
Teacher to check
test shapes.
Explanations may
vary – teacher to
check.
c)
a)
triangles is 13 – 5 =
12 cm long.
12.
Yes;
A square’s diagonal
is always at a 45
angle to the vertex.
(Teacher to check
drawing.)
Teacher to check
label for E.
∆ABE  ∆CBE (SAS)
e)
No;
In this rectangle,
the diagonal is at a
20/70 angle to the
vertex.
Yes, by SSS and the
Pythagorean Theorem;
The third side of both
2
3.
a)
F
d)
No, by SATT/ITT;
One has angles that
are 80°, 50° and 50°.
The other’s are 70°,
55° and 55°.
No, it depends where
the equal side and angle
are positioned within the
triangles.
B
11.
2 and
b)
Teacher to check
angles are marked.
1, 2 and 5, and the
other’s sides are
1,
DBA
INVESTIGATION 2
A.
BC
B.
Isosceles
C.
C
A
D.
D
B
No, AB and CD would
have to be perpendicular;
∆ADC  ∆BDC (SSS)
 CDA = CDB, but
this can only happen if
both angles equal 90°,
meaning AB  CD.
AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
Geometry – AP Book 8, Part 2: Unit 3
Geometry – AP Book 8, Part 2: Unit 3
1.
b)
a)
B
A
O
She should choose
this intersection
point, which we’ll
label as O;
radius = OA = OB =
OC ≈ 3.2 cm
4.
b)
A
(continued)
7.
L
B
c)
b)
A
M
NOTE:
By drawing ∆KLM and its
three sides’ perpendicular
bisectors, you identify
the centre of the circle
through K, L and M.
O
B
In all cases, the centre O
lies on the perpendicular
bisector of AB.
This happens because
OA = OB (both are radii)
and all points equidistant
from A and B lie on their
perpendicular bisector.
a)
B
A
O
Teacher to check drawings.
NOTE:
The centre of a right triangle’s
circumcentre is at the midpoint
of the triangle’s hypoteneuse.
a)
AD : EH = 1 : 3
O
C
b)
Yes
c)
By definition, all of a
rectangle’s angles
are 90°.
B
c)
A
2.
B
C
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
In all cases, the three
perpendicular bisectors
go through centre O –
that is, they intersect at O.
This happens because
OA = OB = OC (all radii)
and all points equidistant
from two endpoints lies
on their line segment’s
perpendicular bisector.
All three intersect at
the same point.
AnswerKeysforAPBook8.2
3.
4.
5.
6.
9.
Side ratio = 1 : 2 = 2 : ?
So the length of the
unknown side, indicated
by ? = 2 × 2 cm = 4 cm.
O
a)
AB : EF = 2 : 6 = 3 : 1
BC : FG = 1 : 3
A
a)
6 cm
b)
12 cm
c)
15 cm
No; A square has all sides
equal, while a rectangle
only needs its opposite
sides to be equal.
No; A square always has
two pairs of parallel sides
but, by definition, a
trapezoid has exactly one
pair of parallel sides.
To be similar, triangles
must have proportionate
corresponding sides and
all their corresponding
angles must be equal.
Since the corresponding
sides in a pair of
congruent triangles are
equal, they will have a 1:1
proportion – which means
they also meet the
requirement for similarity.
CD : GH = 2 : 6 = 3 : 1
b)
No;
By AA, equilateral
triangles will always
be similar, but they
are only congrent
if they have at least
one equal,
corresponding side.
To be congruent, all their
corresponding angles
and sides must be equal.
AP Book G8-34
page 99
1.
3.
8.
BONUS
C
Teacher to check
sketch.
13.
a)
Yes, they are
congruent;
The angles in any
equilateral triangle
are all 60° so, since
one corresponding
side is equal, we
know (by ASA) that
∆EFG  ∆QRS.
K
O
2.
a)
3.2 cm
b)
Sides
14.
15.
x = 20, y = 25
b)
x = 16, y = 12
a)
No: they have only
one equal angle.
b)
No: not all of the
corresponding sides
are proportionate.
AP Book G8-35
page 101
A.
11.
A and C are similar:
their corresponding sides
all have a ratio of 1 : 2.
Teacher to check.
In all cases, the resulting
triangles should be similar
with a scale factor of 2.
Sample answer:
NOTE: Students will need
to use the Pythagorean
Theorem in this question.
a)
AB = 3 cm
AC = 5 cm
BC = 5.7 cm
b)
A'B' = 6 cm
A'C' = 10 cm
B'C' = 11.4 cm
c)
A = A' = 86°
B = B' = 62°
Based on the relationship
between corresponding
sides/angles, we find:
Congruent A and H
D and I
Similar
E and J
C and G
AC : DF
= AB : DE
= BC : EF
= 2:3
a)
Yes, 1 : 3 = 2 : 6
Teacher to check drawing;
new figure = 2 units high
2.7 cm
Yes;
Angles A = D
B = E
C = F
Yes, 1 : 4 = 1 : 4
Yes, 2 : 3 = 4 : 6
120.5°
F
E
c)
No, 1 : 2.5 ≠ 2 : 5.5
39°
20.5°
4.8 cm
b)
e)
B
6.6 cm
D
INVESTIGATION
d)
C
4.4 cm
39°
20.5°
120.5° 1.8 cm
A
No, 1 : 2 ≠ 3 : 3
a)
10.
12.
NOTE:
Measurements may
vary slightly but the
2 : 3 proportion must
remain.
C = C' = 32°;
Yes
B.
d)
Scale factor = 2
a)
AB = 4.8 cm
AC = 2.2 cm
BC = 5.4 cm
Teacher to check.
V‐21
b)
Teacher to check
∆A'B'C'.
3.
A'B' = 14.4 cm
A'C' = 6.6 cm
d)
C.
B'C' = 16.2 cm
4.
No, since they meet
the SSS rule in
A above, we know
they are similar.
e)
Teacher to check.
a)
Teacher to check.
b)
Teacher to check.
c)
Yes:
M = M' = 80°
d)
A'B' A'C'
AB = AC = 2
L'K' 12 cm
LK = 3 cm
b)
While ∆K*L*M*
side lengths will vary,
all three ratios will
be equal if correct.
c)
d)
1.
Yes;
Scale factor will vary
– teacher to check.
e)
Teacher to check.
a)
SAS
b)
AA
c)
2.
M = 80°, M* = 80°
(sum of the triangles’
angles must be 180°)
a)
SSS
A'B' 1.65 cm
AB = 3.3 cm
B'C' 2.35 cm
BC = 4.7 cm
Once reduced, both
equal 0.5.
b)
C = C' = 19°
c)
No
d)
No
Both a square and a
rectangle have four right
angles but there is no
proportion between their
sides.
 E
 
2
1 : s = 50 : 25
50
1
2
s = 25
25
1
2
s = 50 = 2
s=
1
2
 PQ : UV = 2 : 1
(enlargement)
3.
a)
M
2 cm
 their scale factor
is 4.
Teacher to check.
Sample answer - AAAA:
 D
 
C 
 
c)
60°
b
A
B 
 
b
120°
2
1 : s = 1 : 100
s = 10
 PQ : UV = 1 : 10
(enlargement)
120°
b
a
but obviously A' ≠ A so
the triangles aren’t similar.
a)
b
60°
a
a
2
1 : s = 125 : 5
125
1
2
s = 5
5
1
2
s = 125 = 25
1
s= 5
 PQ : UV = 5 : 1
(dilation)
b)
NOTE: 2a = b
a
C
Here we can see that:
5.
a)
This square and rhombus
have a side proportion of
1 : 2 but their angles are
different.
3 cm
43°
2 cm
Let s represent the scale
factor of the two triangles.
Sample answer - SSSS:
B
C'
K'M' 10.8 cm
KM = 2.7 cm
a)
NOTE:
Teachers should see an
example used in each
student explanation.
B'
A
M'L' 8 cm
ML = 2 cm
D.
6 cm
2.
Answers may vary –
teacher to check.
Teacher to check.
A'
K'M' 10.8 cm
KM = 2.7 cm
M'L' 8 cm
ML = 2 cm
7.
Sample example:
4 cm
c)
No;
In Question 2, we showed
that ASS isn’t sufficient
so AS certainly isn’t.
(continued)
b)
A is common
L
ABD = ACE
ADB = AEC
6.
c)
They are similar.
a)
K
b)
2 : 1; 2 : 1
c)
They are similar,
by SAS.
d)
KN = 2 × KO
KO 1
 KN = 2
e)
Two correct options:
KLO and M or
KOL and N;
Each corresponding
pair is equal.
f)
2 cm
30°
30°
K
We know:
M = 120° (SATT)
AP Book G8-36
page 104
R
120°
INVESTIGATION 1
Q
A.
Teacher to check drawn
parallelograms; Yes; 3
We know:
P = Q = 30° (ITT)
B.
2, 2, 6, 6;
 ∆KLM and ∆PQR
are similar (AA).
Base of B = Base of A ×
scale factor
C.
D.
1
2:1
4:1
4
6:1
36 : 1
36
2 : 6 = 1: 3
4 : 36 = 1 : 9
If the base ratio of
A (original) : B (image) is
y
image
x : y,  s = original = x ;
MN || LO (CAT);
LMNO is a trapezoid.
Area of A =
1.
a)
b)
b)
Area of B
2
s
Teacher to check.
P
2
1 : s = 1 : 81
s=9
 PR = QR = 9 × 2
= 18 cm
INVESTIGATION 2
A.
∆ABC:
AC = 3.3 cm
BC = 4.3 cm
AB = 6.8 cm
∆DEF:
FD = 5.1 cm
DE = 3.2 cm
FE = 2.5 cm
Correct ratio: 1 : 9
Teacher to check
area calculations.
c)
1
1
2 bh : 2 (3b)(3h)
1
9
2 bh : 2 bh
1:9
V‐22 AnswerKeysforAPBook8.2
COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
Geometry – AP Book 8, Part 2: Unit 3
Geometry – AP Book 8, Part 2: Unit 3
B.
6.8
AB
FD = 5.1 ≈ 1.33
b)
 yes, they are similar –
their corresponding sides
are proportionate.
C.
4
s ≈ 3 = 1.3
D.
14.4 cm, 10.8 cm
E.
P(∆ABC) 14.4
P(∆DEF) = 10.8 = 1.3 = s
5.
P(∆ABC)
s
a)
C = C (common)
ABC = BDC = 90°;
They’re also similar
(AA).
c)
Yes: ∆ABD and
∆BCD are similar,
since they are both
similar to ∆ABC.
Perimeter ratio
= Per A : Per B = 1 : 5
6.
2.88 m = 288 cm
Sample answer:
7.
a)
3
A
b)
B
Perimeter A (original) = 10
Perimeter B (image) = 50
 the ratio between their
perimeters is also 1 : 5.
General rule:
For similar shapes,
the ratio between the
perimeters is the same
as the ratio between the
corresponding sides.
4.
a)
A
2
24(0.75) = 13.5 cm
k)
8.
QR
PQ
PR
QS = PS = PQ
d)
PQ, PQ = 10 cm
e)
10
PQ
PS = 8 = 1.25
f)
24 cm
2
Since ∆KLM and ∆LNM
are similar, we know that
MKL = MLN and
MLK = MNL.
But, from ∆LNM, we see
that MLN + MNL = 90°
since NML = 90° (given).
 KLN = MLN + MLK
= 90°
a)
c)
2
24 + 13.5 = 37.5 cm
PRQ = PQS,
RPQ = QPS
A (original) : B (image)
= 2 : 10 = 3 : 15
= 1: 5
 their scale factor is 5.
2
PQR = PSQ,
 ∆PQS and ∆PRQ
are similar (AA).
Many non-congruent
24 cm , multiply it
by the square of the
scale factor;
NOTE:
In a) and b), there are
many possible answers
and students can’t be
expected to find them all.
The more they can find,
the better.
P = P (shared)
15
10
j)
9.
PQR = PSQ
= 90° (given)
Quadrilaterals
Sample answers:
90°; 90°; BCA;
They’re similar (AA).
b)
b)
QR
RS
QS
PQ = QS = PS
6
QS
s = PS = 8 = 0.75
 ∆ABC and ∆DAC
are similar (AA).
Teacher to check pair of
shapes drawn.
2
i)
Since it’s given that
BCD = 2ADC,
we also know that
ADC = ACD.
4.3
BC
DE = 3.2 ≈ 1.34
F.
BAC = ACD (AAT)
BAC = BCA (ITT)
3.3
AC
FE = 2.5 = 1.32
 P(∆DEF) =
(continued)
Many non-similar
c)
Answers will vary –
teacher to check.
Triangles
8 non-congruent
2
2
2
g)
(1.25) , 37.5 cm
h)
Explanations may
vary – teacher to
check.
PSQ = QSR
= 90° (given) 
PQS + RQS
= 90° (given)
6 non-similar
PQS + QPS
= 90° (SATT)
B

COPYRIGHT©2011JUMPMATH:NOTTOBECOPIED
so QPS = RQS 
D

C
 ∆PQS and ∆QRS
are similar (AA).
From this similarity,
we also know that
SQP = SRQ. 
AnswerKeysforAPBook8.2
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