Download Geometry – AP Book 7, Part 2: Unit 5

Geometry – AP Book 7, Part 2: Unit 5
AP Book G7-13
8.
page 105
1.
Teacher to check.
2.
4.
J = 66°
b)
Teacher to check.
A = E = B
= D
a)
VW = EF
C = F
WX = FG
ii)
∆ABC  ∆EDF
VX = EG
iii)
Answers may
vary – teacher
to check.
JK = ST
b)
JL = SU
a)
A = D
i)
All the angles
are equal.
ii)
Teacher to
check – any
order is correct.
iii)
Teacher to
check – any
order is correct.
B = E
C = F
b)
M = P
N = Q
O = R
6.
A = D = K
12.
C = F = M
a)
B = E
C = F
13.
b)
R = W
c)
COPYRIGHT©2010JUMPMATH:NOTTOBECOPIED
d)
Measurements may
vary slightly:
RS = WY
BC = 2.0 cm
DE = 4.6 cm
None of Tom’s are correct.
There is more than one
correct solution – teacher
to check.
a)
Teacher to check.
Teacher to check.
AP Book G7-15
page 111
INVESTIGATION 1
A.
a)
Only one unique
triangle is possible
(though it may be
rotated, reflected,
etc.)
b)
Yes, there is only
one way to arrange
any combination of
side lengths into a
triangle.
a)
There is only one
way to complete
the triangle (all the
triangles drawn will
be congruent).
b)
Yes, if you know any
two side lengths and
the angle between
those sides, there is
only one way to draw
the triangle.
a)
They are congruent.
b)
Yes, if you know one
side length and the
angles on either side
of it, only one
triangle is possible.
a)
It is impossible
to make more than
one unique triangle.
Sample solution:
∆ABC  ∆XV W
AP Book G7-14
4 squares:
B, C, D, E, F, G
B.
B  F, C  D, E  G, A  I,
H K
C.
Yes
D.
No, for instance A and H
have the same areas but
are not congruent.
E.
No, if two shapes are
the same size and shape
(which is the definition of
congruence), they must
also have the same area.
EF = 2.0 cm
DF = 5.0 cm
AB = KL = BC = LM
HI
= 4.6 cm
AC = KM
IJ
= 2.0 cm
E = P = G = O
HJ = 5.0 cm
F = N
KL = 5.5 cm
EF = PN = FG = NO
LM = 3.9 cm
EG = PO
KM = 2.4 cm
AnswerKeysforAPBook7.2
Teacher to check.
1.
a)
No, if their
corresponding side
lengths are different,
then they won’t have
the same shape.
No, they are not
congruent. ∆ABC
has no right angles.
b)
B.
5 squares:
A, H, I, K
AB = 4.8 cm
B = L
d)
A.
AC = 3.4 cm
A = K = C = M
15.
∆DEF  ∆HIJ
2
BONUS
Teacher to check.
M = 118°
PR = LM
T = X
RT = WX
3.
L = 23°
R = M
S = Y
ST = YX
b)
K = 39°
c)
2
Area of ∆ADC: 6 m
I = 91°
INVESTIGATION
a)
Area of ∆ABC: 6 m
H = 23°
Q = L
R = M
14.
2.
F = 66°
Q = K
BC = EF
AC = DF
E = 91°
page 110
PR = KM
B = E = L
7.
QR = LM
a)
C = 120°
a)
i)
No, if they’re
congruent, their
corresponding side
lengths would be
equal, which means
their perimeters must
also be equal.
B = 38°
11.
Teacher to check.
b)
A = 22°
10.
KL = TU
5.
Measurements may
vary slightly:
D = 23°
a)
b)
b)
The angles of a triangle
always add to 180° so,
if any two corresponding
angles are equal, the third
has to be equal as well.
∆ABC  ∆RST
9.
3.
No, not necessarily.
Kali forgot to check that
the side length of the two
shapes is equal (they
might just be similar,
not congruent).
C.
D.
U‐23
Geometry – AP Book 7, Part 2: Unit 5 (continued)
1.
a)
Yes, the third angle
can always be found
by subtracting the
first two from 180°.
Three angles, along
with one side length,
will always form a
unique triangle.
10.
b)
c)
c)
E
INVESTIGATION 2
d)
A
A.
c)
Uppermost (middle)
point
e)
B
f)
D
BC = DE, AC = EF and
B = D
ASA
No
3.
∆RST  ∆ZXY
D.
No, neither one is a
congruence rule. The
equal angles must be
between the equal sides –
otherwise more than one
triangle can be drawn
according to the specs.
B.
SSS
4.
Q = Y
5.
Need to know if any
corresponding sides are
equal.
Teacher to check.
OA = OB
a)
No
b)
Yes
a)
2
b)
1
c)
0
page 120
5.
Teacher to check.
6.
Teacher to check.
8.
a)
13.
a)
Teacher to check.
b)
Teacher to check.
c)
Teacher to check.
a)
Yes, they are: all
6 angles will be 60°
and, since one pair
of corresponding
sides is equal, all
6 sides will be the
same length.
Sample answer:
b)
Yes, ∆ABC  ∆GHI since
corresponding sides and
angles are all equal.
Teacher to check sketch.
15.
Teacher to check –
there are 2 possible
locations for point D:
B
C
A
D2
2.
b)
E
c)
D
d)
A
e)
C, F
a)
SSS, A = D
b)
SAA, BC = EF
or SAA, AB = DE
3.
A  C (SAS)
4.
a)
5.
ASA
b)
SAS
c)
SSS
d)
SSS
a)
Teacher to check.
b)
Teacher to check.
c)
i)
∆ADP  ∆BDP
by SSS
ii)
∆ADP  ∆BDP
by SSS
iii)
IMPORTANT
The sketch
should include
pre-drawn lines
for AP and BP,
as shown here:
A
B
BD = CD
AP Book G7-17
page 118
Teacher to check
sketch.
1.
Teacher to check.
2.
Teacher to check.
No, not necessarily.
It’s possible that
their side lengths
are different (making
them similar but not
congruent).
3.
If 2 of the 3 corresponding
angles are equal, the third
will also be equal since the
sum in both cases must
be 180°.
1.
D1
b)
BONUS
In order to form a triangle,
the sum of the two shorter sides
must be greater than the length
of the longest side.
AP Book G7-18
Congruent triangles
are also similar since their
corresponding angles are
equal. But similar triangles
aren’t always congruent:
their angles are equal but
their corresponding side
lengths may be different.
Teacher to check.
No, they aren’t always
congruent – it depends
where the equal sides are
in relation to the given
angle, e.g:
c)
12.
14.
9.
Teacher to check.
Teacher to check.
Sample answer:
8.
b)
7.
Possible answers:
As above.
Teacher to check.
Teacher to check.
A ≠ D, C ≠ F,
BA ≠ ED, CA ≠ FD
a)
4.
a)
11.
Teacher to check.
b)
E
C
Left point
C.
Teacher to check.
a)
b)
Top point
SAS
3.
b)
1.
5.
page 115
2.
Teacher to check.
7.
Circle: 7
No, since neither one has
the equal angle between
the two equal sides.
2.
a)
AP Book G7-16
b)
∆MNO  ∆FGH
6.
Check, from left
to right: 2, 3, 6, 7
a)
∆ABC  ∆DEF
b)
a)
a)
scalene right
b)
isosceles obtuse
c)
isosceles obtuse
d)
isosceles obtuse
e)
equilateral acute
Answers will vary, re:
those identified by side
lengths. Teacher to check.
4.
a)
AC = BC
b)
AC = AB
c)
AB = AC = BC
P
∆ADB  ∆APB
by SSS
* radii of equal circles
* radii of the same circle
* radii of equal circles
U‐24 AnswerKeysforAPBook7.2
COPYRIGHT©2010JUMPMATH:NOTTOBECOPIED
b)
Geometry – AP Book 7, Part 2: Unit 5 (continued)
AP Book G7-19
9.
Teacher to check.
page 122
10.
Question 5 shows that
∆ABD  ∆CBD by SSS,
so we know from this that
ABD = CBD.
1.
Teacher to check.
2.
a)
Teacher to check.
b)
Teacher to check.
c)
AD is also a median
in i), but not in ii).
3.
a)
No
b)
No, only the median
through point A is a
perpendicular
bisector.
c)
4.
In an isosceles
triangle, a median
to the unequal side
(or, to say it slightly
differently, through
the vertex between
the two equal sides)
is a perpendicular
bisector.
1.
a)
QR = PR, OP = OQ
b)
Teacher to check.
Teacher to check.
∆ABD  ∆CBD (SSS)
c)
A = C
d)
6.
ADB = CDB
F.
DE  FC
3.
Teacher to check.
page 128
4.
Teacher to check.
5.
Teacher to check.
The following sides will
likely be marked – teacher
to check.
6.
a)
QT = QS
a)
b)
TU = SU
c)
∆QTU  ∆QSU
d)
SSS
a)
The angles opposite
the equal sides are
equal (isosceles).
7.
8.
ABD = CBD
ADB = CDB
∆DFC = ∆EFC so all their
corresponding angles are
equal. We also know from
this that both equal 90°,
since 180° ÷ 2 = 90°.
AP Book G7-22
AC
a)
Yes
Teacher to check.
90°
b)
b)
All the angles are
equal (equilateral).
a)
180° ÷ 3 = 60°
b)
Teacher to check.
c)
CAD = 30°
1.
a)
b)
COPYRIGHT©2010JUMPMATH:NOTTOBECOPIED
* BM bisects ABC
7.
8.
2.
Teacher to check.
3.
a)
b)
c)
Teacher to check.
6.
4 × 90° = 360°
2 × 180° = 360°
c)
360° ÷ 60° = 6
a)
45°
b)
Teacher to check.
c)
Teacher to check.
a)
DBC = 120°
b)
DBC = 30°
c)
DBC = 30°
d)
DAB = 30°
a)
DBC = 90°
b)
DBC = 45°
c)
DBC = 135°
a)
Teacher to check.
b)
Teacher to check.
AP Book G7-24
page 131
1.
2.
No, the two angles
already sum to 180°
– this means that
KFL would be 0°,
which is impossible.
Teacher to check.
a)
Teacher to check.
b)
AD = BD
c)
90°
d)
90°
e)
∆ADC  ∆BDC
Teacher to check
explanation (likely
using ITT or SAS).
AP Book G7-23
page 127
1.
a)
Teacher to check.
page 129
b)
Teacher to check.
1.
Teacher to check.
2.
QOR = 90°
* ∆ABM  ∆CBM
A.
Teacher to check.
DC is an angle bisector.
B.
D = E
Students can check
by measuring to find
that BC = CA.
C.
Yes (SAS)
AnswerKeysforAPBook7.2
90° + 90° + KFL
= 180° + KFL
a)
b)
AP Book G7-21
INVESTIGATION
Teacher to check for
median from A to BC,
which bisects CAB.
Right angles occur
at (and all around)
points K and L.
* FC is a shared side and
we know, from Step 4,
that DF = EF.
f)
AC = BC
g)
AE = BE
h)
No, any point E
along CD will work
since ∆AEB will
always be isosceles.
POR = 180°
3.
BD is an extension
of the median to the
isosceles triangle’s
unequal side, so
we can use the
Isosceles Triangle
Theorem.
* the result will be an
isosceles triangle
8.
Use Yen’s method,
and then bisect the
60° angle to form
two 30° angle.
* BM is a perpendicular
bisector of AC
AC = AB
5.
b)
* BM is a median of ∆ABC
* BM  AC
4.
7.
CDA = 90°
9.
The following statements
are true (the others should
be crossed out as false):
* BM is a line of symmetry
for ∆ABC
e)
2.
ADB = CDB
5.
E.
AP Book G7-20
page 125
180°
DCF + ECF = DCE
which is a straight line and
therefore equal to 180°
Since ABD + CBD =
ABC, BD bisects ABC.
AB = BC, AD = DC
∆ABD  ∆CBD, SSS
D.
a)
Teacher to check.
b)
SSS
i)
No
c)
DBA and BDA
are both equal to
DBC
j)
Median, line of
symmetry
INVESTIGATION
d)
∆ABE  ∆CBE
AC = BC
Teacher to check
explanations.
A.
Isosceles
B.
Teacher to check.
U‐25
Geometry – AP Book 7, Part 2: Unit 5 (continued)
C.
No, since ∆ABC is
isosceles, the angle
bisector of ACB must
also be the perpendicular
bisector of AB.
3.
BD = 3 cm
Teacher to check
explanation.
Sample answer:
COPYRIGHT©2010JUMPMATH:NOTTOBECOPIED
If ∆ABC is isosceles and
A = B = 60°, it follows
that ∆ABC is actually an
equilateral triangle. This
means that AB = 6 cm.
If D is the midpoint of AB,
then BD = 3 cm.
U‐26 AnswerKeysforAPBook7.2