Preparing for the Regents Examination Geometry, AK

15 x 2 5 180
x 155
16 2x x 180
x 60
17 x 85
18 (x 10) x 90
x 40
19 60 x x x 180
x 40
20 6x 2 0 4x 10
x 15
21 40 mz 180
mz 140
22 a mDOE 75
b mAOB 15
c mDOC 105
23 2x 1 5 4x 1
x 7
mPRT 2(7) 15 29
mRTQ 180 mPRT
180 29
151
24 3x 2 5 10x 4
x 3
mWVY 3(3) 25 34
mWVZ 180 mWVY
180 34
146
25 (5x 2y) (5x 2 y) 180
x 18
m AEC mBED
11y 5x 2y
1 1y 5(18) 2y
9 y 90
y 10
mAED mCEB 5x 2y
5(18) 2(10) 70
4-3 Congruent Polygons
(pages 64–66)
1 (3) IHG MNL
2 (4) a side of one is congruent to a side of the
other
3 (4) IV
4 (1) I
5 (2) II
6 a −−
AC
b BAC
c −−
BC
d CBA
7 a ACD BCD
b ADC CBA
8 a mE 44
b mB 110
c mC 26
9 Answers will vary. For instance, a 3-4-5
triangle has the same angle measure as a
6-8-10 triangle. They are not congruent.
10 Answers will vary.
11 Answers will vary.
12 a Vertical angles are congruent; the triangles
are congruent by SAS.
b AC BD
2 x x 5
x 5
AC BD 10
AE BE 11
CE DE 12
13 The triangles are congruent by SAS.
ACE BDE by corresponding parts of
congruent triangles are congruent. Therefore,
ACE BDE. So 5x 7x 18, and x 9.
mACE 45; mBDE 45; mBED 45.
mDBE 180 45 45 90. Therefore,
DBE is a right triangle.
14 x 2 4x x 14
x 2 5x 14 0
(x 7)(x 2) 0
x 7 and x 2
When x 7, AB AB 21
When x 2, AB AB 12
15 No, the angle must be included between the
pairs of sides.
(page 67)
Note: Since there are many variations of proofs,
the following is simply one set of acceptable
statements to complete each proof. Depending
on the textbook used, the wording and format
of reasons may differ, so they have not been
supplied for the method of congruence applied
in each problem. (These solutions are intended
to be used as a guide—other possible solutions
may vary.)
4-3 Congruent Polygons 15