Honors Solutions - Math-U-See

HONORS SOLUTIONS

Honors Solutions Lesson1 Lesson 1 Honors 1.

Begin by putting x's to show that tyle er and madison do not like tacos. that leaves Jeff as the one who has tacos as his favorite. since you know Jeff's favorite, you can also put x's in Jeff's row, under ice cream and stea ak. We are told that madison is allergic to any ything made with milk, so we can put an x acros ss from her name, under ice cream. now we can see that tyler is the only one who can have ice e cream as his favorite, leaving madison with steak. Jeff

2.

tyler

yes

X

X

X

X

yes

X

X

yes

Lisa

X

yes

X

s∩r = 8 s ∪ r = 30 1 (the twisted ring you started with is called a mobius strip.)) 1st time : one long loop is created 2nd time : two interlocked loops are created

(52 + 5) ÷ 6

+ 10 =

5 + 10 = 15 5.

( 42 ÷ 7) + 6 − 1 = (6) + 6 − 1 = 12 − 1 = 11

Lesson 3Lesson 3 Honors

X

X

yes

X

Celia

X

yes

X

X

Donna

yes

X

X

X

Adam

X

X

X

yes

spring summer autumn winter David

X

X

yes

X

Linda

X

X

X

es ye

shauna

yes

X

X

X

April

X

yes

X

Geometry Solutions

3.

12

(25 + 5) ÷ 6 + 10 = (30 ÷ 6) + 10 =

reading tennis cooking eating

4.

2.

8

black brown blonde yes X X

Ca aitlyn

George

18 + 20 = 38 38 − 30 = 8 days had both sun rain 10 0

4.

We use similar reasoning for the rest of the problems. remember that once you hav ve a "yes" in any row or column, the rest of th he possibilities in that row and in that column n can be eliminated.

3.

1.

ice cream tacos steak X yes X

mad dison

mike


1. 2 2. 4 3. L ∩ F Lesson 4. L 3 ∪F 1. 2 2. 4 3. L ∩ F 4. 5. L 8∪ F

=3 = 12

=3 = 12 3

leaves

2 2

flowers 6. 3

seeds

1 7 K ∪ m −4P = 12 K ∩m∩P = 4 X 8x4 = 4 x 3 3 3 commutative property K m −fPor=multiplication 12 is ∪ true K− ∩6m≠∩6P−=94 9 3 x 4 = 4 x 3 property commutative commutative property is false subtraction H o n o r s L e s so n 1 -for H ONORS L ESSON 3 207 is true f or multiplication ( ) 11. 2 + 1 + 5 = 2 + (1+ 5) 10. 9 6 ≠ 6 − 9property as−sociative commutative property is true for addition is false for subtraction 12. 2 ÷ 8 ≠ 8 ÷ 2 7. 8. 5. 9. 6. 7. 8. 10. 9.

11.

(2 + 1) + 5 = 2 +property (1+ 5) commutative ociative property assfalse is for division

3. 4.

L ∩F = 3 L ∪ F = 12

7.

Lesson 3

Hono rs Lesso n 3 - H ONORS LESSON 5

1. 5. 2. 6. 3. 7. 4. 8. 9.

2 8 4 3 L ∩F = 3 K ∪ m − P = 12 L ∪ F = 12 K ∩m∩P = 4 3x4 = 4 x 3 mustard commutative3property is true for multiplication 1 0 5. 8 10. 9 −6 ≠ 6−9 4 ketchup 6. 3 commutative property pickles 7. K ∪ m −8P = 122 1 is false for subtraction 8. K ∩ m ∩ P = 4 ( ) ( ) 11. 9. 32x+41= +45x=32 + 1+ 5 sociative property as commutative property is true ffor addition is true or multiplication 12. 9 2÷ 10. −8 6≠ ≠8 6÷ −2 9 commutative commutative property property is false for division is false for subtraction 11.

(2 + 1) + 5 = 2 + (1+ 5)

12.

associative property is true for addition 2÷8 ≠ 8 ÷2 commutative property is false for division

8.

9.

10.

($1.75 + D) + D = $3.25 2D + $1.75 = $3 3.25 2D = $1.50 D = $.75 $.75 + $1.75 = $2.50 Drink is $.75 5 sandwich is $2.50 let X = number of isaac's customers 2X = number of Aaron's customers X+2X = 105 3X = 105 X = 35 2X = 70 isaac has 35 customerrs Aaron has 70 customers X + 2X = 18 3X = 18 X = 6 feet; 2X = 12 feet A + ( A + 20) = 144 2A + 20 = 144 2A = 124 A = 62 apples in one box 62 + 20 = 82 apples in the other box

Lesson 5 1. Lesson 5 1.

Honors Lesson 5 Lesson 5 1.

Honors Lesson 4Lesson 4 1. 2. 3. 4. 5.

6.

7.

45º nnW nne no, he should have corrrected 67.5º 5X − 6 = 2X + 18 5X − 2X = 18 + 6 3X = 24 X=8 2C + 10 = 43 − C 3C = 33 C = 11

2. 2.

3. 2. 3.

4 small 2 medium 4 small 1 large 2 medium total 17 large total 47 small 2 medium 1 large 7 total

3.

($1.75 + D) + D = $3.25

2D + $1.75 = $3 3.25 2D = $1.50 D = $.75 $.75 + $1.75 = $2.50 Drink is $.75 5 sandwich is $2.50 8. let X = number of isaac's customers 2X = number of Aaron's customers X+2X = 105 luti ons 208 so 3X = 105 X = 35 2X = 70 isaac has 35 customerrs Aaron has 70 customers 9. X + 2X = 18

4. 4.

4.

1 started with geometry 2 that are half of first triang g le 1 started with 6 small 2 that are half of first triang gle 7 (you may need to 6 overlapping small 16 tottal with drawmay these 7 started overlapping (you need to 1 separately 16 tot t al draw these 2 that are half of first triang gleto be able to count each separately to be 6 small

2.

H o n o r s L e s so n 5 - H ONORS LESSON 7

4. 4.

1 started with 2 started that arewith half of first triang gle 1

3.

6 that smallare half of first triang 2 gle 7 overlapping (you may need to 6 small 16 tottal drawmay these 7 overlapping (you need to separately 16 tottal draw theseto be able to count each separately to be one. see Above.) able to count each one. see Above.)

5.

triangles, squares, trapezoids, pen ntagons

4.

answers will vary

5.

P = 6X + .5 ( 6) X P = 6X + 3X P = 9X P = 9X

5. 6.

P = 9 (8 ) P = $72 6.

Lesson Honors7 Lesson 7

6.

1.

extend all segments AD  Xy  BC AB  rs  DC corresponding angles are congruent

2.

Honors Lesson 6 Lesson 6

cut by transversal AB corresponding ∠ ' s ADF and ABe are both 90º

1. 3.

2.

4.

3.

triangles, squares, trapezoids, pen ntagons

geometry

yes; extend DF and BC these 2 line segements are

G

extend DC to include point G m∠A = 100º sin ce AB and DC are parallel, m∠GDA is 100º . m∠eDF is 80º , since it is supplementary to ∠GDA. m∠DeF = 90º - definition of perpendicular A CAB = 90ºº ( given)

BAD = 45º −definition of bisector ADB = 90º - definition E of perpendicular D F C ABD = 45º - from information given DBe = 135º - supplementary angles all other corners work out the same way.. solutions

4.

answers will vary

5.

P = 6X + .5 ( 6) X P = 6X + 3X

B

209

see how labeled must addthe upangles to 180are º . We know the for easy reference. measure of a and that of ABC. Add a and together, d are 25º and subtract the these definition bisector result fromofthe total 180º that are in a triangle: p and 20) º= 180 − 115 = 65º 180 − (o25are + 90 definition of bisector i = 65º

m∠A = 100º sin ce AB and DC are parallel, m∠GDA is 100º . m∠eDF is 80º , since it is supplementary toLESSON ∠GDA. 8 Hono rs Lesso n 7 - H ONORS m∠DeF = 90º - definition of perpendicular 4.

CAB = 90ºº

(given)

us sing similar reasoning, and looking iatand j arees45 º BFC, ABF, DBC and triangle AeC, definitio o n of bisector ADC, we can find the following:

BAD = 45º −definition of bisector ADB = 90º - definition of perpendicular ABD = 45º - from information given DBe = 135º - supplementary angles all other corners work out the same way.. A

C

B

m = 115º now gºles r = 95look º at triangle AeB. its f =ang 85 must add up to 180 º . We know b = 110º . g = 70the º measure f a and of ABC. now we konow twothat angles from Add each these andgsles. ubtract thewith of the together, smaller triang Armed result from the total 180 º that are this knowledge, and the fact that in a triangle: there are 180º in a triangle, we can

E

(25remaining 180 + 90 ) = 180 − 115 = 65º find −the angles: ic==65 45ºº e = 70º

D

A 2.

Lesson 8Lesson 8 Honors 1.

Look at the drawing below to see how the angles are labeled for easy reference. a and d are 25º definition of bisector

a

D

b g

p and o are 20º definition of bisector

i B

i and j are 45º definitio on of bisector

2.

now look at triangle AeB. its ang gles must add up to 180º . We know the measure of a and that of ABC. Add these together, and subtract the result from the total 180º that are in a triangle:

i = 180º − (K + m)

h = 180º − (30 + 90) ( ) i=180 h = 60ºº− 90 + 30 = 60º º º if ++ih= =90 180

180 − (25 + 90 ) = 180 − 115 = 65º i = 65º us sing similar reasoning, and looking at triangle es AeC, BFC, ABF, DBC and ADC, we can find the following: m = 115º r = 95º f = 85º 210 so b luti = 110ons º. g = 70º now we know two angles from each of the smaller triang gles. Armed with this knowledge, and the fact that there are 180º in a triangle, we can find the remaining angles:

us q s=ing 65ºsimilar reasoning, and n = looking 45º at triangle e s AeC, BFC, ABF, k = 70ºº hDBC = 65and º ADC, we can youfind can the alsofollowing: use what you m = 115º know about vertiical angles r = 95º and complementary f = 85 º angles b = 110º . to find some g =angles. 70º e of the now know two each of b, d, we j and k are allangles 90º from defiinition of the smaller triang gles. Armed with perpendicular knowledge, and the fact t hat ( ) dcthis 180º in a = 180º − a + b there are 180º in a triangle, we can triangle theº remaining F 90) angles: cfind = 180 −f( 60 + = 30º c = 45 º e = 70 º a ( ) i = 180 K m º − + 180 º in r e q =c 65º n = 45º triangle q kh = 70ºº ( h = 65º i=180º −n90 + 30) = 60º k you can also use what you i + i = 90º Angle eGC p know about vertiical angles is 90º m complementary j l and o angles because E find some to e of the angles.C of the b, d, j and k are all 90º defiinition of definition of perpendicular perpendicular. 180 180º in a c − ( aº + b ) 60= + i =º 90 triangle i = 30º ( ) c= 180ºº − h = 180 − (i60 + j+) 90 = 30º 180ϒ in a

f + 60 = 180º f = 120º c + g = 90º 60 + i = 90º i = 30º h = 180º − (i + j)

3.

180 º in a triangle triangle

eGC is Angle BeC Angle i1s80 90 º º because of the definition Angle AGeofis perpendicul 90 º becausear. of the geometryof definition 180 ϒ in a perpendicular. triangle

30 + g = 90º h g = 180 60º º − (30 + 90) h = 60º f + hthe Angle BeC is = 180 º process for this use same one. 180º

2.

2.

b, j ºand k are all 90º c =d,45 q = 65º c= k = 180 70ºº º − ( a + b )

defi nition e = i70 º of perpendicular n =h45 º º − (30 + 90) = 180 180 º= in h =h65 º60aº riangle you can also usetwhat f + hyou = 180º (60 + about c = 180º −know 90) = 30 º vertiical angles ) i = 180º − (and K+m in a= 180º 180 º+ 60 complementary f angles triangle to find some e of the fangles. = 120º ) = 90 A i=180 − (90 30all 60ºº defi b, d, jºand k +are ofº cinition + g = 90 i + i = 90º

triiangle

Angle BeC is H o n1o rºs L e s so n 8 - H ONORS LESSON 9 80

H Angle AGe is 90º because of the definition of perpendicular.

Angle eGC perpendicular is 90 180 º ºin a c = 180º − ( a + b ) because triangle c = 180º − ( 60 + 90) = 30º of the definition 30 90º i = 180º − (K + m) 180 º +inga=of perpendicul ar. g = 60 º triangle 60 + iº=− 90 (90º + 30) = 60º i=180 ii +=i 30 º 3. Angle use eGC the same process for this one. = 90º h = 180º − (i + j)

h = 180º − (30 + 90) h = 60º f + h = 180º 60 + i = 90º + 60 if = 30º= 180º f == 120 180ºº − (i + j) h c + g = 90º h = 180º − (30 + 90) h = 60º f + h = 180º 30 g == 180 90º º f ++ 60 g = 60 º f = 120º

A

180 ϒºineamber that you can also use reme is 90 what i angle tr becauseyou know aboutt vertical angles and suppleme entary of or thecomplementary B angles as definition of a shortcut. = 22.5º Angle BeC perpendiculisar. = 45º 180º = 67.5º = 90º 180ϒ in a = 112.5º Angle AGe is iangle tr = 135º 90º because of the definition Angle BeCof is perpendicular. 180º

c + g = 90º Angle AGe is use the same process for this 90º one. because reme ember that you can also use of the b G c what you know aboutt vertical anglesof D definition g d l or complementary and suppleme entary perpendicular. i angles as a shortcut. 30 + g = 90º m = 22.5º e f h j k g = 60º B E F = 45Cº = 67one. .5º 3. use the same process for this = 9 0 º reme ember that you can also use = 112 .5º what you know aboutt vertical angles = 135eº ntary or complementary and suppleme

geometry

F

C

large rectangle: 15' 6" = 15.5 ft

15.5 x 13 = 201.5 ft 2 small rectangle: 3 x 5 = 15 ft 2

3.a

l arg e trapezoid: 10 + 4 ) = (9) ( 14 ) = (9) (7) = 63 ft2 2 2 small trapezoid:

(9 ) (

(2) (

4 + 8 ) = (2) ( 12 ) = (2) ( 6) = 12 ft2 2 2

total: 2.

= 22.5º = 45º = 67.5º = 90º = 112.5º = 135º

G


angles as a shortcut.

D

201.5 + 15 + 63 + 12 = 291.5 ft 2 it is necessary sometimes to add lines to the drawing to make it clearrer. in figure 1a, dotted lines have been added d to show how one end of the figure has been brroken up. since we know that the long measureme ent is 6.40 in and the space between the dotted d lined is .80 in, we can see that the heights s o l u t i o n s 211 of the trapezoids add up to 5.60 in. since we have been told that the top and bottom are the e same, each trapezoid must have a height of 2..80 in. Area of each trapezoid:

10 + 4 ) = (9) ( 14 ) = (9) (7) = 63 ft2 2 2 small trapezoid:

(9 ) (

4 + 8 ) = (2) ( 12 ) = (2) ( 6) = 12 ft2 2 Hono rs Lesso n 9 - 2 H ONORS LESSON 9 total: 201.5 + 15 + 63 + 12 = 291.5 ft 2

(2) (

2.

it is necessary sometimes to add lines to the drawing to make it clearrer. in figure 1a, dotted lines have been added d to show how one end of the figure has been brroken up. since we know that the long measureme ent is 6.40 in and the space between the dotted d lined is .80 in, we can see that the heights of the trapezoids add up to 5.60 in. since we have been told that the top and bottom are the e same, each trapezoid must have a height of 2..80 in.

figure figure 1a 1a

2.80"

}

trapezoid

}

}

trapezoid

6.40"

shows shows a a different different way way ) figure figure 1b 1b (( of fin ) n ding the area of fin nding the area

Area of each trapezoid: 1.27 + .80 ) = (2.8) ( 2.07 ) = 2 2 2 2.898 in since there are four trapezoids in all, we multiply by 4: 2.898 x 4 = 11.592 in2

(2.8) (

2 Area Area of of large large rectangle rectangle 1 15 5x x 6.4 6.4 = = 96 96 in in2 one one trapezoid trapezoid

long long base base 2 ( ) = 13 15 − 15 − (2 2x x ..8 8) = in2 13..4 4 in short base short base

re c tan gular center portion: .80 in x 15 in = 12 in2

(2 xx 1.27 )= 15 1.27 ) = 15 − − (2 2 12 in2 12..46 46 in

total :

(6.4 -- .8 )÷ he e he eight ight ( 6.4 .8) ÷ 2 2 in2 2= =2 2..8 8 in Area Area of of one one trapezoi trapezoid d

12 + 11.592 = 23.592 in2 3.

area = ( a )(b) or ab ( see figure 2)

4.

area = (2a )(2b ) or 4ab ( see figure 3)

5.

area = (na )(nb ) or n2ab ( see figure 4)

6.

area = n2ab = 52 ( 4) (5) = (25) (20) = 500 ft2

7.

first triangle: a = 1 xy 2 sec ond triangle: a = 1 (2x ) (2y ) = 2xy 2 1 4 times = 2, so new area is 2 four times as great. first square: ( x )( x ) = x2

8.

( )

( )( )

sec ond square: x2 x2 = x4

212

so luti ons

2 = = 36.204 36.204 in in2 2 2x x 36.204 36.204 = = 2 2 72.4 0 8 in 72.408 in

Both Both trapezoids trapezoids

Area Area of of figure figure

96 96 − − 72.408 72.408 = = 2 2 23.592 in 23.592 in

figure 3

figure 2 2a

a b

2b

geometry

H o n o r s L e s so n 9 - H ONORS LESSON 12

figure 5 figure 4

Honors Lesson11Lesson 11 1. green,

na 2x

x y

nb

2y

figure 6

buttons zipper zipper buttons Chris

yes

x

x

x

Douglas

x

yes

x

x

Ashley

x

x

x

yes

naomi

x

x

yes

x

planning refresh −

x

games

Honors Lesson10 Lesson10Lesson 10 E

blue,

red,

2.

x2

1- 4. 1- 4.

green,

F

G

place

birthday

ments

for party

guest

sam

x

x

yes

x

Jason

x

x

x

yes

e shane

yes

x

x

x

troy

x

yes

x

x

3. train boat airplane car Janelle

yes

x

x

x

Walter

x

x

x

yes

Julie

x

yes

x

x

Jared

x

x

yes

x

4.

A

D 5. 5. 6. 6.

7 - 8. 7 - 8.

B

C

your answer should be close your answer should be close to 0.6 61803. to 0.6 61803. see illustration above. see illustration above. the ratio should be close to what the ratio should be close to what you got in #5. you got in #5.

hot dog pizza

chicken tossed soup salad

molly

yes

x

x

x

a tina

x

x

x

yes

Logan

x

x

yes

x

sam

x

yes

x

x

5.

Answers will vary.

Honors Lesson 12 Lesson12 1.

2. 3.

geometry

4 - 8.

60º since the sections are all equ ual, the center angles are all the same. 360º ÷ 8 = 45º solutions

213

Lesson12 1. 1. Lesson12 1. Hono rs Lesso n 12 - H ONORS LESSON 13

2. 2. 3. 3. 2. 3. 4 -- 8. 8. 4

60º 60º since the the sections sections are are all all equ equ ual, al, since u 60º the center angles are all the the center angles are since the sections areall allthe equ ual, same. 360º ÷ ºº all the same. 360ºangles ÷8 8= = 45 45 the center are same. 360º ÷ 8 = 45º

4 - 8.

9. 9. 9.

in in #3, #3, you you divided divided 360 360ºº by by 8 8 to #3, find dyou thatdivided each small small triangle to find d that each in 360ºtriangle by 8 has a dcentral central angle of 45 45 since has a angle of ºº .. since to find that each small triangle a hexagon has six sides, you wa a hexagon has six sides, wan ntt has a central angle of 45ºyou . since tohexagon constructhas sixsix triangles inside the to construct six triangles inside a sides, you wathe nt circle e . 360 º ÷6 = 60 º circle e. 360º ÷6 60º to construct six= triangles inside the #1, in. 360 #1, ºyou you learned how to to circle ein ÷6 =learned 60º how constru c t an equilateral triangle with constru c t an equilateral triangle in #1, you learned how to with each angle equa a l to 60 º . After drawing each angle equa al to 60º .triangle After drawing constru ct an equilateral with a circle andequa onealdiam diam eter, ter, use the the same a circle and one e use same each angle to 60 º . After drawing procedure to construct equil l ateral procedure to construct equil l ateral a circle and one diameter, use the same triangles your triangles inside inside your circle, circle, using a rradius adius procedure to construct equilusing laterala of the circle circle as your your starting pointaeac eac ch h of the as point c triangles inside yourstarting circle, using radius time. After you have constructed four time. have starting constructed of theAfter circleyou as your pointfour eac ch triang les,connect their points, you triang gAfter les,connect their points, and and you time.g you have constructed four will have have an n inscribed inscribed regular hexagon. will an n triang gles,connect theirregular points,hexagon. and you

10 10 -12. -12. will have an n inscribed regular hexagon. 10 -12.

214

so luti ons

Lesson13Lesson 13 Honors 1.

see illustration.

3 x 9 = 27 units2 1 (2x2) = 2 units2 3. 2 1 (3x1) = 1.5 units2 2 1 (1x6) = 3 units2 2 1 (3x2) = 3 units2 Lesson13 2 1. see see illustration. illustration. 1. 2 1 3 + 3 = 92 4. + .5 +27 .5 units2 2. 2. 3 3x x9 9= = 27 units units2 2 27 9.5) = 17.5 units 1 1 (-2 2 3 =2 units2 x2 2x 2 = 2 units 3.. 2 5. see 2 illustration for 5 & 6. 1 1 (3 2 6. see for 2 5 & 6. x 3illustration x1 1) = 1..5 5 units =1 units 2 2 7. 41 = 32 units2 2 1x8 1 ( 1x x6 6) = 3u =3 units nits2 2x2 = 2 units2 8. 12 Lesson13 1 3x2 = 3 units2 2 1 1 (3x x3 2)) = 3.5 units =1 units2 2 (1illustration. 1. see 2 2 2 2 3+ 3= 92 5 units units2 4. +x1 +3 =9 2 3 4. ..5 2. 31+ (129..x55=3++)27 units2 = 3units 2 2 1 (--29.5 27 9.5 = 17.5 units 2 2 17.5 units x2) = 3. 27 =2 units 1 2 2 (2illustration x2) = 2 units 5. for 5. see see illustration for 5 5& & 6. 6. 2 1 2 ( ) 6. see see illustration for 5 & 6. 3 x 1 1 5 = . units 6. illustration for 5 & 1 (1x7 ) = 3.5 units2 6. 2 2 2 2 2 7. 7. 4 41x x8 8 32 units (1x== 632 ) =units 3 units 1 (1 ) 2 2 x2 units = 1 unit2 8. 8. 1 x2 =2 12x 2= 2 units2 1 (3x2) = 3 units2 1 1x3 = 1.5 units2 2 1 9. 22 +(11.x53+) 3=+12 3.5 + 1 = 13 units2 .5+ units 2 2 21+ 1 5 += 3 +19 3= 9.522units2 4. 32 .13 units 1 (-2 unitts s2 2x x3 3) = 3 uni =3 2 10. 27 see 2 - illustration. 9.5 = 17.5 units2 1 1 (2 2 5 & 6. 22 x = 5. 10 see x2 2=) 50 2uunits units =2 x2illustration 5 nitsfor 2 2 6. see illustration for & 6. 25 1 2 1 ((1 2 5x 2)) = 3 5 1 x7 7 3..units 5 units units =5 2 2x8 = 32 units 7. 42 22 1x 15x= 11 2 2 1 = (1=x222)5units 1 unit =units unit 8. 12x2 2 1 1x5) = 2.5 units2 1 ((1 2 2 .5+ units 9. 2 22+ + 1.x53+) 3=+12 3.5 + 1 = 13 units2 9. 2 1.5 + 3 + 2 + 3.5 + 1 = 13 units 1 x1) = 2 units22 2 1 (-413 32 u 32 ==19 19 units nits x3)= 3 uni ts2 2 (-213 2 10. see illustration. 10. see illustration. 11x 1 = 1 unit2 2 (x25x2=) 50 = 2uunits 2 10 nits2 10 units 2 1x (65x1=) 50 = 3 units2 2 1 (1 2 22 5 2 3.units 5 units x7 5x 2) = 5 units =5 2 1 (3x4) = 6 units2 22 1x(15x= 2 12 5=units units 2)5 1 unit 1 x5= 2 51 + 2).5 + 2 + 1+ 3 +2 5 units2 1+(51 26 = 24.5 x 5 ..5 units = 2 1 x 5 units = 2 5 9. 22 .5 + 3 + 2 + 3.5 + 1 = 13 units2 2+ 1 50 - 24.5 = 25.5 units2 1 2 1 (-4 22 x 2 32 units x1 units 413 1) == 2 units = 19 11. see 2 illustration. 2 geometry 10. see illustration. 2 2 2 4x x1 8= =1 32unit units 1 1 = 1 unit 1 10 x 5 = 50 units22 1 ) = .335 units 1 ((1 2 units2 x 6 1 x1 6x 1) == units 1 2 ( ) 2 5x2 = 5 units2 2 1 2 1 ((3 22 units 1)) == 3x x4 4 = 16 6.5units units 2 12x 5 = 5 units2 2 2 1x1=1 51+ +(5 51+ + 2unit +3 3+ +26 6= = 24 24.5 .5 5 units units2 5 2 + 5 )..55=++222.5++ 11units x5 2.

1 (2x3) = 3 unitts2 2 1 (2x2) = 2 units2 2 1 (1x7 ) = 3.5 units2 2 1 (1x2) = 1 unit2 2 9. 10.

2 + 1.5 + 3 + 2 + 3.5 + 1 = 13 units2 32 - 13 = 19 units2

1 x 1 = 1 unit 1 ( 6x1) = 3 units2 2 1 (3x4) = 6 units2 2

H o n o r s L e s so n 1 3 - H ONORS LESSON 14 2

5 + 5 + 2.5 + 2 + 1+ 3 + 6 = 24.5 5 units 50 - 24.5 = 25.5 units2

11.

4 x 8 = 32 units2 1 (1x1) = .5 units2 2 1 (3x1) = 1.5 units2 2 1x1=1 unit2

see illustration. 10 x 5 = 50 units2 1 (5x2) = 5 units2 2 1 x 5 = 5 units2

1 (1x2) = 1 unit2 2 1 (1x2) = 1 unit2 2 1 x 1 = 1 unit2

1 (1x5) = 2.5 units2 2 1 ( 4x1) = 2 units2 2 1 x 1 = 1 unit2 1 ( 6x1) = 3 units2 2 1 (3x4) = 6 units2 2

1 (5x1) = 2.5 units2 2 1 (2x1) = 1 unit2 2 .5 + 1.5 + 1 + 1 +1 + 1 + 2.5 + 1 = 9.5 units2

5 + 5 + 2.5 + 2 + 1+ 3 + 6 = 24.5 5 units2 50 - 24.5 = 25.5 units2 11.

see illustration.

32 - 9.5 = 22.5 units2

see illustration. 4 x 8 = 32 units2 1 (1x1) = .5 units2 2 1 (3x1) = 1.5 units2 2 1x1=1 unit2 1 (1x2) = 1 unit2 2 1 (1x2) = unit2 1 2 1 x 1 = 1 unit2 1 (5x1) = 2.5 units2 2 1 (2x1) = 1 unit2 2 .5 + 1.5 + 1 + 1 +1 + 1 + 2.5 + 1 = 9.5 units2 32 - 9.5 = 22.5 units2


1 (3x4) = 1 (12) = 6 units2 2 2

2.

A=

s (s − a ) (s − b) (s − c )

A=

6 ( 6 − 3) ( 6 − 4) ( 6 − 5)

A=

6 (3) (2)(1)

A=

36

A = 6 un nits2 yes 3.

A=

16 (16 − 7) (16 − 10) (16 − 15)

A=

16 (9) ( 6) (1)

A=

864

A = 29.39 units2 geometry

4.

5.

A=

52 (52 − 36) (52 − 28) (52 − 40)

A=

52 (16) (24)(12)

A=

239, 616

solutions

A = 489.51 units2 V = πrr2h 2 V = 3.14 (2) (10)

215

2.

A=

s (s − a ) (s − b) (s − c )

A=

6 ( 6 − 3) ( 6 − 4) ( 6 − 5)

A=

6 (3) (2)(1)

9.

A n= 1436- H ONORS LESSON 14 Hono rs Lesso nits2 A = 6 un yes

3.

A=

16 (16 − 7) (16 − 10) (16 − 15)

A=

16 (9) ( 6) (1)

A=

864

10.

A = 29.39 units2 4.

5.

A=

52 (52 − 36) (52 − 28) (52 − 40)

A=

52 (16) (24)(12)

A=

239, 616

A = 489.51 units2 V = πrr2h 2 V = 3.14 (2) (10) V = 3.14 ( 4) (10) V = 125.6 in3

6.

V = πrr2h 2 V = 3.14 (1) (10) V = 3.14 (1) (10) V = 31.4 in3

7.

it is 1 the first one 4 V = πr2h 2 V = 3.14 (2) (5) V = 3.14 ( 4) (5) V = 62.8 in3 it is half the first one.

8.

V = πr2h 2 V = 3.14 ( 4) (10) V = 3.14 (16) (10) V = 502.4 in3

9.

it is fo our times the first one. V = πr2h

11. 12.

it is fo our times the first one. V = πr2h 2 V = 3.14 (2) (20)

V = 3.14 ( 4) (20) V = 251.2 cu in3 it is two times the e first one When the height is doubled, the volume is doubled. When the height is halved, the volume is halved. When the radius is double ed, the volume increases by a factor of 4. When the radius is halved, the volume decreases by a factor of 4. the student may use his own words s to express this. Answers will vary. take the formula, and multiply both sides by 2: 2 V = πr h 2V = 2πr2h now rearrange the factors: V = πr2h 2V = πrr22h take the formula, and multiply both sides by 4: 2 V = πr h 4V = 4πr2h re write the 4 on the right side as 22: 4V = 22πr2h re arrange the factors: 4V = π22r2h 2 4V = π (2r ) h there is more than one way to set this up. As long as you show the same resultts as by experimentation, the answer is correct..

2 V = 3.14 (2) (20) V = 3.14 ( 4) (20) V = 251.2 cu in3 10.

216

11. 12.

it is two times the e first one When the height is doubled, the volume is doubled. When the height is halved, the volume is halved. When the radius is double ed, the volume increases by a factor of 4. When he radius is halved, the volume so lutitons decreases by a factor of 4. the student may use his own words s to express this. Answers will vary. take the formula, and multiply both sides by 2:

geometry

V = πr h 2 V = 3.14 (.25) (12) V = 2.355 in3 then find the diffe erence: 9.42 − 2.355 = 7.065 in3

H o n o r s L e s so n 1 5 - H ONORS LESSON 15


3 x 3 x 3 = 27 ft

2.

12 x 12 x 12 = 1,728 in3 8 x 4 x 2 = 64 in3

3. 4.

5.

6. 7.

64 x .3 = 19.2 lb 64 in3 ÷ 1,728 = .037 ft3 .037 x 1200 = 44.4 lbs you could probably lift it, but it would be much heavier tha an expected. First find what the volume would be if it were solid: V = πr2h 2 V = 3.14 (.5) (12) V = 9.42 in3 now find the volume inside the pipe: V = πr2h

8.

2 V = 3.14 (.25) (12) V = 2.355 in3 then find the diffe erence: 9.42 − 2.355 = 7.065 in3 6. 7.

geometry 8.

7.065 × .26 = 1.8369 lb V = 4 πr3 3 3 V = 4 (3.14) (.25) 3 V = .07 in3 (rounded) .07 × .3 ≈ .02 pounds for one bearing 25 ÷ .02 = 1,250 bearings Because we rounded some numbers, the actual number of bearings in the box may be sliightly different. Keep in mind that the startin ng weight was rounded to a whole number. our an nswer is close enough to be helpful in a real life situation, where someone wants to know approximately how many bearings are available witthout counting. the side view is a trapezoid, and the volume of the water is the area of the trapezoid times the width of the pool: A = 3+10 ( 40 ) 2 A = 6.5 ( 40)

7.065 × .26 = 1.8369 lb V = 4 πr3 3 3 V = 4 (3.14) (.25) 3 V = .07 in3 (rounded) .07 × .3 ≈ .02 pounds for one bearing 25 ÷ .02 = 1,250 bearings Because we rounded some numbers, the actual number of bearings in the box may be sliightly different. Keep in mind that the startin ng weight was rounded to a whole number. our an nswer is close enough to be helpful in a real life situation, where someone wants to know approximately how many bearings are available witthout counting. the side view is a trapezoid, and the volume of the water is the area of the trapezoid times the width of the pool: A = 3+10 ( 40) 2 A = 6.5 ( 40) A = 260 ft2 V = 260 (20) V = 5,200 ft3

9.

10.

Volume of the sphere: 3 V = 4 (3.14) (1) units3 3 V = 4.19 Volume of the cube: V = 2 x 2 x 2 = 8 units3 8 − 4.19 = 3.81 units3 Volume of the cylinder: 2 V = 3.14 (1) (2) V = 6.28 units3 Volume of the sphere from #9:: 4.19 units3 6.28 − 4.19 = 2.09 units3 note: you may use the s o l u t i o n s 217 fractional value of π if it seems more convenient.

3 V = 4 (3.14) (1) units3 3 V = 4.19 Volume of the cube: Hono rs Lesso n 15 - H ONORS LESSON 18 V = 2 x 2 x 2 = 8 units3 10.

8 − 4.19 = 3.81 units3 Volume of the cylinder: 2 V = 3.14 (1) (2) V = 6.28 units3 Volume of the sphere from #9:: 4.19 units3 6.28 − 4.19 = 2.09 units3 note: you may use the fractional value of π if it seems more convenient.

Lesson17 Lesson17 1. 1.

2 2 ) V V= 3..14 14 ((2 2)) (( 4 4) =3 3 V = 50 V= 50..24 24 ft ft3

2. 2.

3. 3.

4. 4.

5. 5.

Lesson16Lesson 16 Honors 1. 2.

6. 6.

(r ) πr = πr2

A = LW + LW + LH + LH + WH + WH = 2LW + 2L LH + 2WH

7. 7.

= 2(LW + LH + WH) 3. 4.

(

) ( )

2 s2 + s2 + s2 = 2 3s2 = 6s2 V = 3(11) (3) = 99 ft3 sA = 2 (3x11) + 2 (3x3) + 2 (11x3) = 2 (33) + 2 (9) + 2 (33) = 66 + 18 + 66 = 150 ft2

5.

150 ft2 ÷ 6 faces = 25 ft2 per face

6.

25 = 5 ft the new bin is 5 x 5 x 5. the cube-shaped one holds more. 125 − 99 = 26 ft3 difference.

8. 8. 9. 9.

10. 10. 11. 1. 1 12. 12.


2h V= πrr2 =π V h

V = πr2h

4 4 V = 3 V= 3 4 V= = 4 V 3 3

3 πrr3 π

3 ((33..14 14)) ((2 2))3

3 rounded) V= 33..49 49 ft ) ft3 ((rounded = 33 V 2 2 V= 3..14 14 ((3 3)) (( 6 6)) =3 V 3 V = 169 V= 169..56 56 units units3

3 4 3 4 ((3 V = 3 V= 3..14 14)) ((3 3)) 3 3 V V= = 113 113..04 04 u units nits3 ((rounded rounded)) 2 2 ) V V= 3..14 14 ((1 1)) ((2 2) =3 3 V V= =6 6..28 28 units units3

3 4 3 4 ((3 V V= = 3 3..14 14)) ((1 1)) 3 3 V = 4 V = 4..19 19 units units3 ((rounded rounded)) 113..04 04 ≈ .67 33 33..49 49 ≈ 113 67 ≈ ..67 ≈ .67 . . 24 169 56 50 50.24 169.56 4..19 19 ≈ .67 4 6..28 28 ≈ .67 6 2 2 3 3 2 + 2πrh A= =2 2π πrr2 A + 2πrh 2 A 2 ((3 3..14 14)) ((3 3))2 + 2 ((3 3..14 14)) ((3 3)) (( 6 6)) =2 A= +2

2 units2 A= = 56 + 113 4= = 169 169..56 56 units 56..52 52 + 113..0 04 A 2 A= =4 4 ((3 3..14 14)) ((3 3))2 A 2 A A= units2 = 113 113..04 04 units

113..04 04 ≈ 2 2 113 ≈ 3 169 . 56 169.56 3 the surface surface area area and and volum volum me e of of a a m the 2 sphere appear appear to to be be 2 of of the the sphere 3 3 surface a area rea and and volume volume of of a a surface cylinder with with the the same same dim dim mensions. ensions. cylinder m (Archimedes proved that this is (Archimedes proved that this is the c ase.) the case.)

2 V = 3.14 (2) ( 4)

2.

V = 50.24 ft3 V = 4 πr3 3 3 V = 4 (3.14) (2) 3 V = 33.49 ft3 (rounded)

218 (3)2 (6) 3. Vso = luti 3.14ons

4.

V = 169.56 units3 3 V = 4 (3.14) (3) 3 V = 113.04 units3 (rounded) 2

() ( )

Lesson18Lesson 18 Honors 1. 2. 3.

4, 003 mi 90º; a tangent to a circle is perpendicular to the diameter L2 + 4, 0002 = 4,0 0032

geometry L2 = 4, 0032 − 4, 0002 L2 = 16, 024, 009 − 16, 000, 000

L2 = 24, 009 L= 4.

24, 009 ≈ 155 mi

29, 035 ÷ 5,280 ≈ 5 mi

Lesson18 1. 2. 3.

4, 003 mi 90º; a tangent to a circle is perpendicular to the diameter L2 + 4, 0002 = 4,0 0032 L2 = 4, 0032 − 4, 0002 L2 = 16, 024, 009 − 16, 000, 000 L2 = 24, 009 L=

4. 5.



V = ( 4 ⋅ 4) ( 8 ) V = 128 in3 2.

29, 035 ÷ 5,280 ≈ 5 mi L2 + 4, 0002 = 4, 0052 3.

V = area of base x altitude V= 1 (3x4) x10 2 y = 60 ft3

4.

sA = (2) 1 (3x4) + (3x10) + 2 ( 4x10) + (5x10) sA = 12 + 30 + 40 + 50 sA = 132 ft2

5.

think of the wire as a long, skinny cylinder. 1 ft3 = 12x12x12 = 1,728 in3

L2 = 40, 025 L= 6.

40, 025 ≈ 200 mi

555 ÷ 5,280 ≈ .1 L2 + 4, 0002 = 4, 000.12 L2 + 16, 000, 000 = 16,000,800.01 L2 = 16, 000,800.01− 16, 000, 000 L2 = 800.01 L=

7. 8. 9.

800.01 ≈ 28.3 mi

2 1502 + 4, 0002 = ( X + 4, 000)

10.

Volume of wire = area of base x length

X2 + 8, 000X + 16, 000, 000 22,500 + 16, 000, 000 = X2 + 8, 000X + 16, 000, 000 22,500 = X2 + 8, 000X 0 = X2 + 8, 000X − 22,500

sA = 2( 4x4) + 2 ( 4x8) + 2 ( 4x8) sA = 2 (16) + 2 (32) + 2 (32) sA = 32 + 64 + 64 sA = 160 in2

24, 009 ≈ 155 mi

L2 + 16, 000, 000 = 16, 040, 025 L2 = 16, 040, 025 − 16, 000, 000

V = area of base x altitude

(

)

1,728 = 3.14 x .12 x L 1,728 = .0314L 55, 031.8 in ≈ L 55, 031.8 ÷ 12 ≈ 4,586 ft 6.

or X2 + 8, 000X − 22,500 = 0 8, 000X = 22,500 X = 22,500 ÷ 8, 000 X ≈ 2.8 mi

A = LW

Let L = the circumference and W = the height of the cylinder.

Diameter = 9, so L = 3.14 (9) 28.26 in ≈ L this is one dimension of the rectangle an nd the circumference of the cylinder. 625 = 28.26W 22.12 in = W this is the other dimension of the rectangle and the height of the cylinder. V = area of base x height 2 V = 3.14 (9 ÷ 2) x 22.12 2 V = 3.14 ( 4.5) x 22.12 V ≈ 1, 406.5 in3 7.

geometry

Cylinder will be 4 in high and 4 in3 in diameter. Area of one s o l u2t i o n s 219 circular end = 3.14 (2) = 12.56 in2

area of side = 3.14 ( 4) x 4 = 50.24 in2 50.24 + 12.56 + 12.56 = 75.36 in2 you also could have used what you learned in lesson 17 to find the surface area of the cylinder. Fiirst

dimension of the rectangle and the height of the cylinder. V = area of base x height 2 V = 3.14 (9 ÷ 2) x 22.12

Hono rs Lesso n 19 - H ONORS LESSON 21

2 V = 3.14 ( 4.5) x 22.12 V ≈ 1, 406.5 in3

7.

Cylinder will be 4 in high and 4 in3 in diameter. Area of one

north

2 circular end = 3.14 (2) = 12.56 in2

7.

you also could have used what you learned in lesson 17 to find the surface area of the cylinder. Fiirst find the surface area of the sphere, and then multiply by 3 . (see 2 below for an alternative solution.) alternative solution 2 sA of sphere = 4 (3.14) (2) = 50.24 in2

8.

east 75º

W

south


πy2

3 or 1.5 (50.24) = 75.36 in2 2 A = 2( 4x4) + 2( 4x4) + 2 ( 4x4) A = 32 + 32 + 32 = 96 in2

2.

A = πx2 − πy2 y2 + z2 = x2

the cylinder uses less cardboard. (However, there will be odd-shaped, possibly unuseable pieces left over.)

4.

A = π x2 − y2

5.

A = π z2

3.

z2 = x2 − y2

6.

3. 4. 5. 6.

220

300 ÷ 150 = 2 hours answers may vary the wind blew him off course. 30 ÷ 2 = 15 mm 150 ÷ 2 = 75 mm ∠oWP = 80º ∠oWP = 75º see drawing. your answers to #5 and #6 may vary slightly depending how carefully you drew and measured.

so luti ons

(

( ) A = π ( z2)

A = π( 10 ) 2 2 A = π (5) A = 3.14 x 25 A = 78.5 in2

Honors Lesson 20Lesson 20 1. 2.

P

75 mm

o 20 mm

area of side = 3.14 ( 4) x 4 = 50.24 in2 50.24 + 12.56 + 12.56 = 75.36 in2

7. 8.

9.

)

2

2 A = 3.14 ( 4) A = 3.14 × 16 = 50.24 in2 A = Lx W 50.24 = L × .007 50.24 ÷ .007 ≈ 7,177 in 7,177 ÷ 2 ≈ 3,589 tickets (rounded to the nearest whole number)

geometry

4.

80º; 180º − (50º +50º )

5. 6. 7.

160º; 360º − (100º +100º ) 80º; vertical angles 85º; 180º −95º

(80 ) ONORS LESSON 25 n oº; r s180 L e s−so n º2+2 8.H o 15 85- ºH

Lesson 22Lesson 22 Honors 1. 2. 3. 4.

2.

3. 4.

5.

11. 12.

30º; 180º − (70º +80º ) 30º; alternate interior angles

9.

this bird is red. ∠A is congruent to ∠B. i get 100% on my math test. this triangle has two congruent sides.


10.

checking results with remote interior angles:: 80º +15º = 95º 80º; angle 1 and the 70º angle next to it put together form an angle that is the e alternate interior angle to the 150º angle at the top left. 150º −70º = 80º 70º; alternate interior angles

if i get burned, i touched the hot sto ove. not necessarily true. if two line segments are congruent, they have equal length. true. if a bird is red, it is a cardinal. not necess sarily true. if the leg squared plus the leg squared equals the hypotenuse squared, the triang gle is a right triangle. true. if my plants wilt, i stop watering them. not true if i am sensiblle!


2. 3.

50º; the measure of an inscribed angle e is half the measure of the intercepted arc. 130º; 180º −50º 50º; same reason as #1

4.

80º; 180º − (50º +50º )

5. 6. 7.

160º; 360º − (100º +100º ) 80º; vertical angles 85º; 180º −95º

8.

15º; 180 − (80º +85º )

10.

checking results with remote interior angles:: 80º +15º = 95º 80º; angle 1 and the 70º angle next to it put together form an angle that is the e alternate interior angle to the 150º angle at the top left. 150º −70º = 80º 70º; alternate interior angles

11. 12.

30º; 180º − (70º +80º ) 30º; alternate interior angles

9.

geometry

re asons

AF ≅ eF ∠1 ≅ ∠2

Given Given

CF ≅ CF

re flexive sAs Corresponding parts of congruent triangles

CeF ≅CAF Ce ≅ CA

CDe ≅CBA

sss Corresponding parts ∠CDe ≅ ∠CBA of congruent triangles

2.

Honors Lesson 24Lesson 24 1.

statements

D

e

statements

re asons

tu ≅ rQ ∠tuV ≅ ∠rQV

tuV ≅rQV

Given Given 1 Given sAs 2

tV ≅ rV

CPCtrC

CuV ≅ QV

st ≅ sr

A

sV ≅ sV

tsV ≅rsV ∠tsV ≅ ∠rsV

Given

F

B

re flexive sss CPCtrC

3. statements

re asons

Fe ≅ GH

Given

FH ≅ Ge

Given

eH ≅ eH re flexive sss FeH ≅ GHe





solutions

221

CF ≅ CF

CeF ≅CAF Ce ≅ CA

Lesson 25



re flexive sAs Corresponding parts of congruent triangles

CBA sss27 Hono rs LessoCDe n 25≅ - H ONORS LESSON 1.

∠CDe ≅ ∠CBA statements

2.

AF ≅ eF atements st∠ 1 ≅ ∠2 tu ≅ rQ CF ≅ CF ∠tuV rQV CeF ≅ ∠CAF

Given re asonsGiven Given re flexive Given sAs Corresponding parts uV QV Given Ce ≅≅ CA triangles tuV ≅ rQV of congruent sAs CDe ≅ rV CBA CPCtrC sss tV ≅ parts st ≅≅ ∠ srCBA Corresponding Given ∠CDe of congruent triangles sV ≅ sV re flexive





 

 

tsV ≅rsV

2.

∠stsV ≅ ∠rsV tatements

3.

Corresponding parts of congruent triangles re asons

sss CPCtrC re asons







sV ≅ sV tsV ≅r rsV ∠tsV ≅ ∠rsV





1. 1.

statements re statements asons re asons AB ≅ AC Given AB ≅ AC Given ∠ ∠ArB ArB ≅ ≅∠ ∠AQC AQC P Perpendicular erpendicular ∠ BAr ≅ ∠ CAQ re ∠BAr ≅ ∠CAQ re flexive flexive BAr ≅ CAQ AAs BAr ≅ CAQ AAs or or HA HA

  2. 2.

 

re asons Given

FH ≅ Ge

Given

eH ≅ eH

re flexive sss

FeH ≅GHe

CPCtrC CPCtrC

 

CPCtrC CPCtrC

statements statements eF eF ≅ ≅ GF GF

re re asons asons From From proof proof above above De e finition of Bisector De efinition of Bisector

eX eX ≅ ≅ GX GX FX ≅ FX FX re flexive flexive FX ≅ re eFX ≅ GFX ss s or HL HL eFX ≅ GFX sss or ∠ ≅ ∠ Definition of eXH GXH ∠eXH ≅ ∠GXH Definition of Perpendicular Perpendicular HX X reflexive HX ≅H HX ≅ reflexive eHX GHX sAs or LL LL ≅ eHX ≅ GHX sAs or

3. Fe ≅ GH

CQ CQ ≅ Br ≅ Br

XA XA ≅ ≅ yA yA

3 3..

re flexive sss Q CPCtrC

statements

 

statements re statements re asons asons XB ≅ yB Definiti XB ≅ yB Definitio on n of of bisector bisector ∠ XBA ≅ ∠ yBA Definiton of Perpendicu ar ∠XBA ≅ ∠yBA Definiton of Perpendicullar BA ≅ BA re flexive BA ≅ BA re flexive XBA sAs XBA ≅ ≅ yBA yBA sAs or or LL LL

tu ≅ rQ Given u t statements re asons ∠tuV ≅ ∠rQV Given Fe ≅≅ GH Given uV QV Given FH ≅ Ge Given tuV ≅ rQV sAs s eH re flexiveV tV ≅≅ eH rV CPCtrC sss FeH GHe st ≅≅ sr Given



Honors Lesson Lesson 26 26Lesson 26

 

 

 

 

eH ≅ GH eH ≅ GH

CPCtrC CPCtrC

F H G

e

Honors Lesson 26Lesson 27

e

H

1. statements

re asons

XC ≅ yC

radius of a cirrcle A tangent of a circle ∠PyC ≅ ∠PXC is perpendicular to the radius at that point.

PyC ≅PXC

PC ≅ PC

re flexive HL

PX ≅ Py

CPCtrC

2. statements

222

so luti ons

De ⊥ AB

re asons Given geometry

AC ≅ BC

radius of a circle

FC ≅ FC FCA ≅ FCB ∠ACe ≅ ∠BCe

re flexive HL CPCtrC Property of centra al angle





 ≅ Be  Ae


L

Lesson 26

X

1. statements

re asons

XC ≅ yC

P

radius of a cirrcle C of a circle A tangent ∠PyC ≅ ∠PXC is perpendicular to the radius at that point. PC ≅ PC PyC ≅ PXC





PX ≅ Py

y

1.

De ⊥ AB statements AC ≅ BC

3.

CPCtrC re asons Given re a asons radius of circle


2.



PyC ≅PXCD

HL

PX ≅ Py statements

CPCtrC re asons

oP ≅ Lm statements oC ≅ LC

Given re asons radi us of a circle

De AB PC ⊥ ≅ mC

Given radius of a circle

2.

C CPo AC ≅≅ BCCmL radius of sass circle

3.

oX≅≅FC Ly Definition of Bisector FC re flexive FLCy oCX A ≅≅ FCB HLHL FC A ∠ACe BCe CPCtrC CBPCtrC XC ≅≅ ∠ yC Property of  ≅ Be  Ae e centra al angle

4.





3. statements



Given

oC ≅ LC

radius of a circle



5.

radius of a circle sss

1.

CPCtrC

2. geometry

m∠AeC = 40º +30º = 70º = 35º 2 2 m∠BeD = 35º m∠KPL = 116º −36º = 80º = 40º 2 2

Honors Lesson 29Lesson 29



XC ≅ yC

180 − (50 + 67) = ( 1 )Z 2 1 63 = ( )Z 2 126º = Z B = 180 − 77 = 103º A = 180 − 84 = 96º C = 2x77 = 154º  = 2 ( 63º ) = 126º mQr  = 126º m∠QCr = mQr

oX ≅ Ly Definition of Bisector oCX ≅ LCy HL



67 = ( 1 )X 2 134º = X 50( 1 ) = y 2 100º = y

re asons

oP ≅ Lm PC ≅ mC CPo ≅ CmL

m P

XC yC radius of a cirrcle FC ≅ FC re flexive A tangent HL of a circle FCA ≅ FCB PyC ≅≅ ∠ ∠ACe ∠PXC BCe is perpendic CPCtrCular to the Property of point. radius at that   Ae ≅ Be centra a l angle PC ≅ PC re flexive



y

X

re flexive HL

2.

Lesson 26 statements

C

o

angle

tan

10º 15º 30º 45º 60º

.18 .268 .58 1 1.73

tan10º = a 20 a .18 = s o l u t i o n s 223 20 a = .18 (20) a = 3.6 mi or 19,00 08 ft

1.

angle tan angle tan 10º .18 10 18 15ºº ..268 15 º . 268 30º .58 ºº - H .58 Hono rs Lesso n30 ONORS LESSON 30 4529 1 45 º 60º 1.1 73 60º 1.73 2. tan10º = a a 2. tan10º = 20 20 a .18 = a .18 = 20 (20) a = .1820 a = .18 (20) a = 3.6 mi or 19,00 08 ft a = 3.6 mi or 19,00 08 ft 10º

3. 3.

3. 4. 4.

4. 5. 5.

a

20 mi 3. 3.

tan 45º = a a tan 45º = 150 150 1= a a 1 = 150 150150 ft = a 150 ft = a

5.

45º

a

6. 6.

150 ft 6.


w sin 55º = w 14 14 ww 14 =º = w 8192 8192 55= 1. ..sin 14 14 14 w (14 ) ..8192 = w w 14 8192 = w .8192 = 55º 11..4 w= = 11 414 688 ft ft w 688 (14) .8192hh = w 2. sin sin 4 4ºº = = 2. 3 ft w = 11.4688 3 h 0698 = h 0698 2. ..sin 4º = 3 3 =h h 3) ..0698 0698h= (.0698 3 = 3 h = . 2094 mi h = .2094 mi (2094 = h) = 3) .0698 =1 1,,105 105..632 632 ft 5,280 ft (5,280 ..2094 h = .2094 mi p p 3. 3. sin sin 34 34ºº = = 43.5 .2094 (5,280 ft 4.5) = 1,105.632 h p p 5592 5592 =º =4.5p 3. ..sin 34= 4.45.5 ( 44..55) ..5592 p= 5592 =p p .5592 = 4 . 5 2 . 5164 p = p = 2.5164 mi mi or or 13,286.592 13,286.592 ft ft ( 4.5) .5592 gg= p 4. co 4. c os s 55 55ºº = = 14 p = 2.516414 mi or 13,286.592 ft gg g 5736 =º = 4. ..c5736 os 55= 14 1414 (4.5 ) 14 . 5736 g= g 14 . 5736 .5736 = = g 14 ft g= =8 8..0304 0304 ft p g (14) .5736 gg= g 5. cos 5. 34º 7ºº = = 500 cos 7 g = 8.0304 ft 500 g gg 9925 == 500 9925 5. ..cos 7º= 500 500 (500 g= 500) ..9925 9925 =g g .9925 = luti ons 224 so 500 . ft g = 496 25 g = 496.25 ft (500) .992530 =g cos 60 6. cos 60ºº = = 30 6. Lft g = 496.25 L 30 30 30 = 60º = 5= 5 6. ..cos L L L ..5 L = 30 5L =30 30 .5 = L= = 60 60L ft ft L 1.

1. sin 55º = Lesson 30

3mi h = .2094 (5,280 ) = 1,105.632 ft .2094 h 0698 (=5,280 ) = 1,105.632 ft ..2094 p sin 34º =3 4 . 5 p (3) .0698 sin 34º = = h p4.5 .h5592 = mi = .2094 4p.5 5592 (=5,280 ) = 1,105.632 ft ..(2094 4.5= p 4.5) .5592 pmip or 13,286.592 ft (psin )34 4=.52 ..5164 5592 º= = .5 or 13,286.592 ft p = 2.51644gmi cos 55º = p .5592 = 14 g cos 55º =4.5 g14 .(5736 = 14 4.5) .5592 =p 14 g .p5736 = .5164 (14= )2.5736 14=mi g or 13,286.592 ft g (gc14 ) 5736 55º .0304 o=s 8.55 º = =ftg g = 8.030414 gft g cos 7º = g .5736 = 500 g cos 7º = 14 500 g .(14 = =g 9925 ) .5736 500 g .g9925 = = 8).0304 ft= g (500 500 .9925 g= g (500 g = 496 7) .º9925 =.25 ft cos 500 ft g = 496.25 cos 60º = g30 .9925 = 30 L cos 60º =500 500 L 30 .(5 = ) .9925 = 7º 500 g L 30 .g5== 496.25 ft g L .5L = 30 30 .L 5 L = 30 = 60 60ft º= cos L = 60 ft L L .5 = 30 L .5L = 30 60º L = 60 ft 30

geometry