Chapter 5 Exam Review Day 2
5.8:
5.8 Practice A
1.
It is easy to find the width and length of the rectangle by placing a vertex at the origin.
2.
It is easy to see the length of each leg by placing the vertex with the right angle at the origin.
3.
It is easy to see the width and length of the rectangle by placing a vertex at the origin.
4.
It is easy to see the length of the base and the
placing one vertex at the origin.
height, and to determine that the triangle is isosceles by
5.
AB 
a
 ,
2
a 2  4, m AB 
2
, midpoint of AB:
a

1

BC  2, m AB  undefined, midpoint of AB:
a, 1
a
AC  a, m AB  0, midpoint of AB:  ,
2

0

! ABC is a right triangle because AC is horizontal
and BC is vertical.
When a  2, ! ABC is also isosceles because then AC  BC  2.
6.
JK  a, mJK  undefined, midpoint of JK :
 a
 0, 
 2
KL 
a
a 2  b 2 , mKL   , midpoint of KL:
b
b a
 , 
2 2
b
JL  b, mJL  0, midpoint of JL:  ,
2

0

! JKL is a right triangle because JL is horizontal and JK is vertical. ! JKL is not isosceles because
7. O0, 0, C h, 0; OB 
h2  k 2
8. G  2h, 0, O0, 0, D2, 4k ; FD  4
DE  2 4k 2  h2  2h  1
4k 2  1,
a  b.
9.
Coordinate proof:
Segments CO and DO have the same length.
CO  0    h  h; DO  h  0  h
Segments AC and BD have the same length.
AC  k  0  k ; BD  k  0  k
Segments AO and BO have the same length.
AO 
  h  0 2
DO 
 h  0 2
  k  0 
2
  k  0 
2
h2  k 2 ;
h2  k 2
! ACO  ! BDO by the SSS Congruence
Theorem (Thm. 5.8).
5.7:
1. The figure shows that DAC  E, B  C, and AB  DC. So,
! ACD  ! EBA
by AAS Congruence
Theorem (Thm. 5.11) and EB  AC because corresponding parts of congruent triangles are congruent.
2. The figure shows that ACB  DCB and AC  DC. Use the Reflexive Property of Congruence (Thm. 2.1)
to show BC  BC. Then ! ABC  ! DBC by SAS Congruence Theorem (Thm. 5.5). So, A  D
because corresponding parts of congruent triangles are congruent.
3. Show that PQRS is a parallelogram by definition, then show QPS  RSP by the Alternate Interior Angles
Theorem (Thm. 3.2). Show PS  SP by the Reflexive Property of Congruence (Thm. 2.1). The figure shows
PQ  SR. So, ! PQS  ! SRP by SAS Congruence Theorem (Thm. 5.5), and PR  SQ because
corresponding parts of congruent triangles are congruent.
4. The figure shows that HI  JI and HK  JK . Use the Reflexive Property of Congruence
(Thm. 2.1) to show that IK  IK . Then ! HIK  ! JIK by SSS Congruence Theorem (Thm. 5.8), and
H  J because corresponding parts of congruent triangles are congruent.
5. Place a stake at L so that LM  MN . Find K, the midpoint of ML. Locate the point J so that ML  JL, and J, K,
and N are collinear. Then find JL; Given: M and L are right angles, K is the midpoint of ML. Paragraph
proof: Because they are right angles, M  L. MK  LK by the definition of the midpoint of a segment.
JKL  NKM by the Vertical Angles Congruence Theorem (Thm. 2.6). ! JKL  ! NKM by the ASA
Congruence Theorem (Thm. 5.10). LJ  MN because corresponding parts of congruent triangles are congruent.
6.
DE  5 ; Sample answer: mC  44 by the Triangle Sum Theorem (Thm. 5.1), so you have
A  D, C  F , and AC  DF .
! ABC  ! DEF by the ASA Congruence Theorem (Thm. 5.10).
Therefore, DE  AB because corresponding parts of congruent triangles are congruent.
5.6:
1.yes; AAS Congruence Theorem (Thm. 5.11)
2.no; Although two pairs of sides and one pair of angles are congruent, they are not the corresponding parts needed
for SAS.
3.yes; ASA Congruence Theorem (Thm. 5.10)
4. Done in class.
5. Done in class.
5.5:
6. not congruent.
5.4:
5.1 to 5.4: