Download Honors Geometry Chapter 7 Review Name

Honors Geometry Chapter 7 Review
Name ___________________
Solve each proportion. SHOW ALL WORK!
1)
y
48
=
3
y
5)
Verify that
A
2)
x
x3
=
4
x
3)
2x
x 2
=
3
3x+1
x3
20
=
5
x 3
4)
BCD Show all work.
ACE
5
B
4
E
2.5
D
2
C
Explain why the triangles are similar (state the theorem used to prove similarity) Solve for x.
6)
7)
8)
8
84
14
50
x
10
x
5
46
6
Find the value of x.
9)
15
Find the value of x. Then find the lengths of the segments.
10)
12
6
x +2
x
9
3x - 3
21
10
11) Find YW and WZ
YXW  WXZ
12) Find x and lengths of segments
a b c
X
12.8
x+ 1
8
Y
t
t +2
W
Z
2x+ 2
x+4
19
13) A triangle with side lengths 5, 10 and 15 is similar to another triangle with longest side of
length 24. What is the perimeter of the larger triangle?
14) A base angle and vertex angle of an Isosceles triangle are in a ratio of 7:10. Find the measure of
the angles.
15) A rectangle with sides 10 cm and 8 cm. Another rectangle has sides 15 cm and 12cm. Find the
ratio of their perimeters and their areas.
16) The ratio of the perimeters of 2 similar polygons is 16 : 49. What is the similarity ratio? What is
the ratio of their areas?
17) Are the polygons similar? If so, 1st state the similarity statement and 2nd give the similarity ratio of
left polygon to right polygon.
A
M
20
15
B
N
24
18
12
16
D
26
C L
19.5
P
18) Prove that the triangles are similar. Find all the lengths to support your proof.
Given: A( 4, 3), B(0, 4), C(4, 5), D(1, 1), and E( 2, 3)
Prove: ACE BCD
y
C
B
A
D
x
E
Answers
1) y = 12
2) x = 6 & x = −2
3
2
3) x =
&x=
2
3
4) x = −7 & x = 13
5) SAS~
4
2

sides proportional
9 4.5
NLM  OLP Vertical angles  & 3 angles in triangle = 180 so LOP is 84
x=9
EFI  EGH & EIF  EHG If lines  corr angles 
T  T reflexive RQT  SUT given x = 10
C  C reflexive &
6) AA~
7) AA~
8) AA~
9) x = 14
10) x = 6 x + 2 = 8 3x − 3 = 15
can not keep this part x = −7 x + 2 = −5 3x − 3 = −24
10
11) t =
3
11
12) x =
x + 1 = 6.5 2x + 2 = 13 x + 4 = 9.5 can not keep this part x = −1 x + 1 =0
2
13) x = 48
14) x = 7.5 52.5 : 75
2
4
15) perimeters
areas
3
9
16) similarity ratio
16
16
256
Areas ( ) 2 =
49
49
2401
17) ABCD~MNPL
similarity ratio
18) AC = 2 17
4
3
BC = 17
EC = 10
CD = 5
AE = 2 10
BD = 10
2 17 10 2 10
sides proportional SSS~
 
5
17
10
Or use
AC EC

and state C  C reflexive SAS~
BC DC