Unit 5 HW 5
p. 258 # 2, 5, 7 – 12, 15
Mrs. McCaleb
2. 2x + 5 = 4x – 17
22 = 2x
11 = x
5. A) 9x(2x – 5)
18x2 – 45x
B) 9x + 9x + 2x – 5 + 2x – 5
22x – 10
C) Area: 18(4.2)2 – 45(4.2)
128.52 sq units
Perimeter: 22(4.2) – 10
82.4 units
3(11) – 2 ≠ 27
BC ≠ AD
So ABCD is not a rhombus
7.
1.
2.
3.
4.
5.
6.
7.
8.
8.
1.
2.
3.
4.
5.
6.
7.
8.
Statements
AB = BC
<BAC = <BCA
AC bisects <BAD
<BAC = <CAD
<CAD = <BCA
BC // AD
AB not // CD
ABCD is a trapezoid
Reasons
1. Given
2.
3. Given
4. Def bisect
5. Transitive prop
6. Alt int <s =  // lines
7. Given
8. Def trap (a quad with exactly 1 pr //
sides is a trapezoid)
Statements
YTWX is a parallelogram
YP perpendicular TW
ZW perpendicular TY
<WZT and <YPT are rt <s
<WZT = <YPT
<T = <T
TP = TZ
ΔTZW = ΔTPY
YT = WT
TWXY is a rhombus
Reasons
1. Given
2.
3.
4.
5.
6.
7.
8.
Def perpendicular
Right angle thm
Reflexive prop]
Given
ASA
CPCTC
A parallelogram with a pair of
consecutive sides = is a rhombus
P
9. Given: <Q is a right <
M is the midpt PR
AM // QR
JM // PQ
Prove: AMJQ is a rectangle
1.
2.
3.
4.
5.
A
Statements
<Q is a right <
PQ perpendicular QR
AM // QR
JM // PQ
PQ perpendicular AM
MJ perpendicular QR
6.
<MAQ is right
<MJQ is right
AMJQ is a rectangle
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Statements
ID bisects RB
BK = KR
BI = IR
IK = IK
ΔIKB = ΔIKR
<IKB = <IKR
<IKB supp <IKR
<IKB and <IKR are rt <s
ID perpendicular BR
BIRD is a kite
10.
Q
M
J
R
Reasons
1. Given
2. Def perpendicular
3. Given
4. In a plane, if a line is perpendicular to
one of 2 // lines, then it is perpendicular
to the other (Thm 43, p. 227)
5. Def perpendicular
6. A quad with 3 right <s is a rectangle
Reasons
1. Given
2. Def bisect
3. Given
4. Reflexive
5. SSS
6. CPCTC
7. Def linear pair
8. If 2 <s are both = and supp, they are rt
9. Def perpendicular
10. If 1 diagonal is the perpendicular bis
of the other, then the quad is a kite.
OR . . . you could use a detour proof; after getting ΔIKB = ΔIKR, use
CPCTC to get <BIK = <RIK and reflexive prop for ID = ID. Now you can say that
ΔBID = ΔRID, so BD = RD by CPCTC. BIRD is a kite b/c it has 2 disjoint pairs of
consecutive sides congruent (BI = IR, and BD = DR)
11.
1.
2.
3.
4.
5.
6.
7.
8.
Statements
ABDE is a parallelogram
ED // AC
ACDE is a trapezoid
AE = DB
BC is base of isos ΔBCD
BD = CD
AE = CD
ACDE is an isosceles trap
12.
1.
2.
3.
4.
5.
6.
7.
8.
Reasons
Given
Opp sides of a parallelogram are //
A quad with 1 pr of // sides is a trap
Opp sides of a parallelogram are =
Given
Def isosceles triangle
Transitive
The non// sides are =
Because ABCD is a parallelogram,
AB = CD and BC = AD. And since we’re
given that NC = AM, segment subtraction
show us that BN = MD.
So, the red is ½ the perimeter of
the parallelogram, and the blue is the
other ½. Half of 52 is 26; since the
perimeter of ABNM is 36, the black line
(NM) must be 36 – 26 = 10.
15.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Statements
parallelogram PQRS
QR // PS
<QAP = <APS
PA bisects <QPS
<QPA = <APS
<QAP = <QPA
QP = QA
QP = RS
A is the midpt of QR
QA = AR
AR = RS
<RAS = <RSA
<RAS = <ASP
<RSA = <ASP
SA bisects <PSR
Phew!
Reasons
Given
Opp sides of a parallelogram are //
// lines  alt int <s =
Given
Def angle bisector
Transitive (steps 3 & 5)
1.
2.
3.
4.
5.
6.
7.
8. Opp sides of a parallelogram are =
9. Given
10. Def midpt
11. Transitive (steps 10, 7, 8)
12.
13. // lines  alt int <s =
14. Transitive (steps 12, 13)
15. Def angle bisector