geometry review august 2011. 5 7 10 12 p q r s t p q r s t

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Geometry Review August 2011

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Page 1: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

Geometry ReviewAugust 2011

Page 2: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 3: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 4: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

5

7

10

π‘₯

12

π‘₯+10 512

=10π‘₯+10

5 (π‘₯+10 )=1205 π‘₯+50=120

5 π‘₯=705π‘₯5

=705

π‘₯=14

Page 5: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

P

Q

RS

TP

Q

RS

T

Page 6: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

7

25

π‘Ž2+𝑏2=𝑐2

72+𝑏2=252

49+𝑏2=625𝑏2=576

βˆšπ‘2=√576𝑏=24

Page 7: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 8: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

𝑑=√ (π‘₯2βˆ’π‘₯1 )2+( 𝑦2βˆ’ 𝑦1 )2

𝑑=√ (9βˆ’1 )2+(2βˆ’βˆ’4 )2

𝑑=√ (8 )2+ (6 )2

𝑑=√64+36𝑑=√100𝑑=10

π‘Ÿ π‘¦βˆ’π‘Žπ‘₯𝑖𝑠 (π‘₯ , 𝑦 )β†’ (βˆ’ π‘₯ , 𝑦 )

Page 9: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

30

80 180βˆ’30βˆ’80=70

Page 10: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 11: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

4 π‘₯+2 𝑦=142 𝑦=βˆ’4 π‘₯+14𝑦=βˆ’2π‘₯+7

π‘š=βˆ’2π‘šβˆ₯=βˆ’2

𝑦=βˆ’2π‘₯+𝑏2=βˆ’2 (2 )+𝑏2=βˆ’4+𝑏6=𝑏

Page 12: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

π‘Ÿ 𝑦=π‘₯ (π‘₯ , 𝑦 )β†’ (𝑦 ,π‘₯ )π‘Ÿ 𝑦=π‘₯ (βˆ’3π‘Ž ,4𝑏 )β†’ (4𝑏 ,βˆ’3π‘Ž )

Page 13: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 14: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

A

M

5

25

2

B

Page 15: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 16: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

𝑦=4

XX

Page 17: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 18: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

P

Q

R2 π‘₯

3 π‘₯

5 π‘₯

2 π‘₯+3 π‘₯+5 π‘₯=18010 π‘₯=180π‘₯=18

2 π‘₯=2 βˆ™18=363 π‘₯=3 βˆ™18=545 π‘₯=5 βˆ™18=90

Page 19: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

π‘š

𝑛

Page 20: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

If the diagonals do not bisect each other, then the quadrilateral can not be a parallelogram!

Page 21: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

π‘₯+2 𝑦=32 𝑦=βˆ’π‘₯+32 𝑦2

=βˆ’π‘₯2

+32

𝑦=βˆ’12π‘₯+32

π‘š=βˆ’12

π‘šβŠ₯=2

Page 22: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

𝑉= h𝐡𝐡=π‘π‘Žπ‘ π‘’π‘Žπ‘Ÿπ‘’π‘Žπ΅=

12h𝑏

𝐡=12βˆ™6 βˆ™4

𝐡=12

𝑉=12 βˆ™10𝑉=120

Page 23: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

6 2 4

π‘₯

π‘₯

8

𝐴𝐸 βˆ™πΈπ΅=𝐢𝐸 βˆ™π·πΈ8 βˆ™ 4=π‘₯ βˆ™π‘₯32=π‘₯2

√32=√π‘₯2√32=π‘₯

√16 βˆ™2=π‘₯4√2=π‘₯

Page 24: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

𝑆=(π‘›βˆ’2 )180𝑆=(6βˆ’2 )180𝑆=(4 )180𝑆=720

hπΈπ‘Žπ‘ πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ=7206

hπΈπ‘Žπ‘ πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ=120

𝑂𝑅hπΈπ‘Žπ‘ 𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ=

3606

hπΈπ‘Žπ‘ 𝐸π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ=60

hπΈπ‘Žπ‘ πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ=180βˆ’60hπΈπ‘Žπ‘ πΌπ‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ=120

Page 25: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

π‘šπ΄π΅=𝑦2βˆ’ 𝑦1π‘₯2βˆ’π‘₯1

π‘šπ΄π΅=6βˆ’20βˆ’8

π‘šπ΄π΅=4βˆ’8

π‘šπ΄π΅=βˆ’12

π‘šβŠ₯=2

𝑀𝑖𝑑π‘₯=π‘₯2+π‘₯12

𝑀𝑖𝑑π‘₯=8+02

𝑀𝑖𝑑π‘₯=82

𝑀𝑖𝑑π‘₯=4

𝑀𝑖𝑑𝑦=𝑦2+𝑦 12

𝑀𝑖𝑑π‘₯=2+62

𝑀𝑖𝑑π‘₯=82

𝑀𝑖𝑑π‘₯=4

𝑀𝑖𝑑𝐴𝐡=(4,4 )

𝑦=π‘šπ‘₯+𝑏4=2 βˆ™4+𝑏4=8+π‘βˆ’4=𝑏

𝑦=2 π‘₯βˆ’4

Page 26: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

π‘Ž2+𝑏2=𝑐2

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‘₯+1βˆ’π‘₯2βˆ’π‘₯2

49=2π‘₯+148=2π‘₯24=π‘₯π‘₯+1=24+1=25

Page 27: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 28: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

In a trapezoid, the bases are parallel.

Parallel lines intercept congruent arcs between them.

180βˆ’80=1001002

=50

5050

𝐡𝐢=50

Page 29: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T
Page 30: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

𝑉=43πœ‹ π‘Ÿ3

𝑉=43πœ‹ βˆ™93

𝑉=43πœ‹ βˆ™729

𝑉=972πœ‹

Page 31: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

πΆπ‘’π‘›π‘‘π‘’π‘Ÿβˆ’ (5 ,βˆ’4 )π‘Ÿ=6

(π‘₯βˆ’5 )2+(𝑦+4 )2=62

(π‘₯βˆ’5 )2+(𝑦+4 )2=36

Page 32: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

Statements Reasons

1.βˆ π΄πΆπ΅β‰…βˆ π΄πΈπ· 1. Given

2 .βˆ π΄β‰…βˆ π΄ 2. Reflexive Postulate

3. Ξ” 𝐴𝐡𝐢 Ξ” 𝐴𝐷𝐸 3. AA AA

Page 33: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

A

B

C

6

4

3

2 6

4

3

2 πΆπ‘’π‘›π‘‘π‘Ÿπ‘œπ‘–π‘‘ : (7,5 )

Page 34: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

180βˆ’70=1101102

=55

55

55

180βˆ’55=125

125

180βˆ’125βˆ’28=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.

Page 35: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

𝐷2 𝑇 βˆ’3,1

G’’

S’’

H’’

Page 36: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

π‘₯+2π‘₯

=π‘₯+64

4 (π‘₯+2 )=π‘₯ (π‘₯+6 )4 π‘₯+8=π‘₯2+6 π‘₯βˆ’4 π‘₯βˆ’4 π‘₯

8=π‘₯2+2π‘₯βˆ’8βˆ’80=π‘₯2+2π‘₯βˆ’80=(π‘₯+4 ) (π‘₯βˆ’2 )π‘₯+4=0π‘₯=βˆ’4

π‘₯βˆ’2=0π‘₯=2Reject

Page 37: Geometry Review August 2011. 5 7 10 12 P Q R S T P Q R S T

x

y

A

B

C

10

8

5

410

4

5

2D E

F

π‘š=π‘Ÿπ‘–π‘ π‘’π‘Ÿπ‘’π‘›

π‘šπ΄π·=54

π‘šπ·πΈ=06=0

π‘šπΉπΈ=54

π‘šπ΄πΉ=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.

𝑑𝐴𝐷 :π‘Ž2+𝑏2=𝑐2

𝑑𝐴𝐷 :52+42=𝑐2

25+16=𝑐241=𝑐2

√ 41=𝑐6.4=𝑐𝑑𝐷𝐸=6

Since two consecutive sides of parallelogram ADEF are not congruent, then ADEF is not a rhombus.

6.4

6