Download Similarity - Frost Middle School

Similarity
Lesson 12-7
4. a. yes
b. SAS Similarity Theorem
(pp. 756–763)
c. NPR ∼ NOQ
Mental Math
a. true
5. a. yes
b. false
b. SSS Similarity Theorem
c. false
c. JKL ∼ XWV
Activity
6. a. no
Steps 1–6.
A
3.55 cm
F
E
5.34 cm
6.40 cm
6.00 cm
5.01 cm
b. BZY cannot be similar to either CZY or
BCY because it shares only one congruent
angle with each, and CZY cannot be similar
to BCY because they only share one
congruent angle.
7. 25 ft
B
4.26 cm
C
D
XY
Step 7.
YZ
BC
AB = ___
to XYZ, we have
a size change of k = ___
Yes; ABC ∼ DEF
Guided Example 1
x = _____
x
100 cm
1 m _______
____
= ______
;
; 80; 150,000
15 m
80 cm 1500 cm
80 cm
Guided Example 2
∠VXW; ∠UXT; ∠WVX; ∠TUX; WVX; AA
XY
YZ
X Y = k · XY = AB and Y Z = k · YZ = BC.
Also, because size transformations preserve
angle measure, ∠Y ∠Y = ∠B by transitivity.
Thus, ABC X Y Z by the SAS Congruence
Theorem. Thus, ABC can be mapped onto
XYZ by a composite of size changes and
reflections, so ABC ∼ XYZ.
9. 15 ft
Questions
1. Apply a size change to a triangle to show its
image is congruent to its preimage.
2. The SSS Similarity Theorem, the AA Similarity
Theorem, and the SAS Similarity Theorem: The
SSS Similarity Theorem states that 2 triangles
are similar if the 3 sides of the first triangle are
proportional to 3 sides of the second triangle.
The AA Similarity Theorem states that 2 triangles
are similar if 2 angles of the first triangle are
congruent to 2 angles of the other. The SAS
Similarity Theorem states that 2 triangles are
similar if the ratios of 2 pairs of corresponding
sides are equal and the included angles are
congruent.
3. a. yes
b. AA Similarity Theorem
c. IJM ∼ FLM
A228
BC
AB = ___
8. We are given ∠B ∠Y and ___
. Applying
Geometry
10. a. AEC and BDC are right triangles with
∠C as a common acute angle.Therefore, by
the AA Similarity Theorem, AEC ∼ BDC.
b. BD; DC
c. 20 ft 7.5 in.
11. 75 ft 2 in.
12. a. ∠LTR ∠DTC, because they are vertical
angles,
and ∠RLT ∠CDT because
___ ___
ABCD and these are alternate interior
angles. Therefore, by the AA Similarity
Theorem LRT ∼ DCT.
b. 12 ft
13. a. 2.4
b. SAS Congruence Theorem
14. a. ABE ∼ DFE, ABE ∼ CFB,
DFE ∼ CFB
16
b. __
3
c. 14
28
d. __
3
15. If two isosceles triangles have congruent vertex
angles with measure x, then all of their remaining
180 - x
. By the AA
angles have measure ______
2
Similarity Theorem, the two triangles are similar.
Conclusions
Justifications
____
1. A, BD bisects ∠CBE. 1. Given
16.
2. ∠CBD ∠DBE
3. ∠BEC ∠CDB
4. BFE ∼ BCD
Conclusions
1. m∠STU = 90,
m∠SVT = 90
2. ∠STU ∠SVT
17.
3. ∠S ∠S
4. STU ∼ SVT
2. definition of
bisection
3. The angles
intercept the
same arc.
4. AA Similarity
Theorem
Justifications
1. Given
2. Reflexive Property of
Equality, Angle
Congruence Theorem
3. Reflexive Property of
Congruence
4. AA Similarity Theorem
15 √
3
15
18. __
and _____
2
2
19. a. 125,000,000 times
b. No; the area of a cross section of its muscles
would only be 250,000 times larger, so the
1
larger ant would be able to lift __
of its own
10
weight.
20. a. true
b. false
___
___
; CD
; AB CD; therefore, there is no size
21. AB
transformation S such that S(A) = B and
S(C) = D.
A229
Geometry
22. By the convenient location for a rectangle, L =
(–a, b), M = (a, b), N = (a, –b), and O = (–a, –b).
Thus, E = (0, b), F = (a, 0), G =
(0, –b), and H =
___
(–a,___
0). Therefore, the slope of GE is
undefined,
___
so GE___
is vertical while the slope of___
FH ___
is 0 and
thus FH is horizontal. Therefore, GE⊥FH.
23. 3h
24. Yes. It is given that LMNK
is a parallelogram,____
___
P is the
midpoint
of
LM
,
O
is
___
___
___ the midpoint of MN,
and KO and KP intersect LN at Q and R. By the
Vertical Angles Theorem and the Parallel Lines
Theorem, ∠LQP ∠NQK and ∠NLM ∠LNK.
Thus, by the AA Similarity Theorem, QLP
∼
___
QNK. Because P is the midpoint of LM, we
know the ratio of similitude is 2, so 2LQ = QN.
By the same reasoning, ∠ORN ∠KRL and
∠KLN ∠LNM. Again by the AA Similarity
Theorem, NRO
____ ∼ LRK. Because O is the
midpoint of MN, the ratio of similitude is 2, so
LR = 2RN. So LN = LQ + QN = 3LQ, and LN
= RN + LR = 3RN. Thus, LQ = RN, and by
substitution, LQ = QR = RN.