geometry review august 2011. 5 7 10 12 p q r s t p q r s t
TRANSCRIPT
Geometry ReviewAugust 2011
5
7
10
π₯
12
π₯+10 512
=10π₯+10
5 (π₯+10 )=1205 π₯+50=120
5 π₯=705π₯5
=705
π₯=14
P
Q
RS
TP
Q
RS
T
7
25
π2+π2=π2
72+π2=252
49+π2=625π2=576
βπ2=β576π=24
π=β (π₯2βπ₯1 )2+( π¦2β π¦1 )2
π=β (9β1 )2+(2ββ4 )2
π=β (8 )2+ (6 )2
π=β64+36π=β100π=10
π π¦βππ₯ππ (π₯ , π¦ )β (β π₯ , π¦ )
30
80 180β30β80=70
4 π₯+2 π¦=142 π¦=β4 π₯+14π¦=β2π₯+7
π=β2πβ₯=β2
π¦=β2π₯+π2=β2 (2 )+π2=β4+π6=π
π π¦=π₯ (π₯ , π¦ )β (π¦ ,π₯ )π π¦=π₯ (β3π ,4π )β (4π ,β3π )
A
M
5
25
2
B
π¦=4
XX
P
Q
R2 π₯
3 π₯
5 π₯
2 π₯+3 π₯+5 π₯=18010 π₯=180π₯=18
2 π₯=2 β18=363 π₯=3 β18=545 π₯=5 β18=90
π
π
If the diagonals do not bisect each other, then the quadrilateral can not be a parallelogram!
π₯+2 π¦=32 π¦=βπ₯+32 π¦2
=βπ₯2
+32
π¦=β12π₯+32
π=β12
πβ₯=2
π= hπ΅π΅=πππ ππππππ΅=
12hπ
π΅=12β6 β4
π΅=12
π=12 β10π=120
6 2 4
π₯
π₯
8
π΄πΈ βπΈπ΅=πΆπΈ βπ·πΈ8 β 4=π₯ βπ₯32=π₯2
β32=βπ₯2β32=π₯
β16 β2=π₯4β2=π₯
π=(πβ2 )180π=(6β2 )180π=(4 )180π=720
hπΈππ πΌππ‘πππππ=7206
hπΈππ πΌππ‘πππππ=120
ππ hπΈππ πΈπ₯π‘πππππ=
3606
hπΈππ πΈπ₯π‘πππππ=60
hπΈππ πΌππ‘πππππ=180β60hπΈππ πΌππ‘πππππ=120
ππ΄π΅=π¦2β π¦1π₯2βπ₯1
ππ΄π΅=6β20β8
ππ΄π΅=4β8
ππ΄π΅=β12
πβ₯=2
ππππ₯=π₯2+π₯12
ππππ₯=8+02
ππππ₯=82
ππππ₯=4
ππππ¦=π¦2+π¦ 12
ππππ₯=2+62
ππππ₯=82
ππππ₯=4
ππππ΄π΅=(4,4 )
π¦=ππ₯+π4=2 β4+π4=8+πβ4=π
π¦=2 π₯β4
π2+π2=π2
72+π₯2= (π₯+1 )2
π₯2+49=(π₯+1 ) (π₯+1 )π₯2+49=π₯2+π₯+π₯+1π₯2+49=π₯2+2π₯+1βπ₯2βπ₯2
49=2π₯+148=2π₯24=π₯π₯+1=24+1=25
In a trapezoid, the bases are parallel.
Parallel lines intercept congruent arcs between them.
180β80=1001002
=50
5050
π΅πΆ=50
π=43π π3
π=43π β93
π=43π β729
π=972π
πΆπππ‘ππβ (5 ,β4 )π=6
(π₯β5 )2+(π¦+4 )2=62
(π₯β5 )2+(π¦+4 )2=36
Statements Reasons
1.β π΄πΆπ΅β β π΄πΈπ· 1. Given
2 .β π΄β β π΄ 2. Reflexive Postulate
3. Ξ π΄π΅πΆ Ξ π΄π·πΈ 3. AA AA
A
B
C
6
4
3
2 6
4
3
2 πΆπππ‘ππππ : (7,5 )
180β70=1101102
=55
55
55
180β55=125
125
180β125β28=27
27
is not isosceles because all the angles have different measures, which means all the sides have different lengths, making the triangle scalene, not isosceles.
π·2 π β3,1
Gββ
Sββ
Hββ
π₯+2π₯
=π₯+64
4 (π₯+2 )=π₯ (π₯+6 )4 π₯+8=π₯2+6 π₯β4 π₯β4 π₯
8=π₯2+2π₯β8β80=π₯2+2π₯β80=(π₯+4 ) (π₯β2 )π₯+4=0π₯=β4
π₯β2=0π₯=2Reject
x
y
A
B
C
10
8
5
410
4
5
2D E
F
π=πππ πππ’π
ππ΄π·=54
ππ·πΈ=06=0
ππΉπΈ=54
ππ΄πΉ=06=0
Since the opposite sides of ADEF have equal slopes, they are parallel. Since the opposites sides of ADEF are parallel, ADEF is a parallelogram.
ππ΄π· :π2+π2=π2
ππ΄π· :52+42=π2
25+16=π241=π2
β 41=π6.4=πππ·πΈ=6
Since two consecutive sides of parallelogram ADEF are not congruent, then ADEF is not a rhombus.
6.4
6