ch 6 worksheet l1 key - palisades high school · ch 6 worksheet l1 key.doc name _____ s. stirling...

Ch 6 Worksheet L1 Key.doc Name ___________________________

S. Stirling of 18

Lesson 6.1 Tangent Properties

Investigation 1 Tangent Conjecture “If you draw a tangent to a circle, then…” Draw a radius to the point of tangency. What do you notice? perpendicular Would this be true for all tangent lines? Yes

Investigation 2 Tangent Segments Conjecture Draw tangent segments to circle E from point N. What do you notice about these segments? They’re congruent. Measure them. Write the Tangent Segments Conjecture in your notes. Draw Kite ANGE. Do you know any of the angles of this kite? What relationships can you make between the angles of this kite? Make sure you can justify your answers with properties!

O

T

N

E

A

G

Converse of the Tangent Theorem

Draw a line perpendicular to OT at point T, call

it TA��

.

What type of line isTA��

? tangent Would this work for any radius? Yes Write the Converse of the Tangent Conjecture in your notes.

T

O

Write the Tangent Conjecture in your notes.

90m A∠ = and 90m G∠ = because tangents are perpendicular to the radii at the point of tangency.

Sum of the angles of a quadrilateral are 360°.

So 360 90 90 m AEG m N= + + ∠ + ∠ and 180 m AEG m N= ∠ + ∠ Always? Yes

A

4.5 cm

4.5 cm


S. Stirling of 18

EXERCISES Lesson 6.1 Page 313-314 #1 – 5, 8 – 10. Show how you are finding your answers! State the properties you are applying and show calculations!!.

w = 180 – 130 = 50

50

Or Tangent ⊥ radius

Quad ∠ sum – 360

w = 360 – 90 – 90 – 130 = 50

Tangents from a point outside a ⊙ =.

Isos. △ base angles =

and △ sum = 180

x = (180 – 70)/2 = 55

60

Linear pair supplementary.

Tangent ⊥ radius

△ sum = 180

y = 180 – 60 – 90 = 30

Tangent ⊥ radius

Quad. sum = 360 z = 360 – 180 – 75 = 105

Tangents from a point outside a ⊙ =.

OR = OA = AP = PC = 13 TC = TD = DS = SR and TD = ½ of 12 = 6 Perim = 4 * 13 + 4 * 6 = 76

13

13 13

13

6

6 6

6


S. Stirling of 18

10. Draw an obtuse triangle ABC inscribed in the circle given below.

Is the longest side of triangle ABC longer or shorter than the diameter?

Now do page 313-316 #18 and 21 from your book. Write your answers on a separate sheet of paper.

r t

diameter

r

t

X

Y

Z

Various lines.

Tangents must be ⊥ radii!

A

B

C

Various triangles. Shorter


S. Stirling of 18

Lesson 6.2 Investigation 3 Chord Properties If two chords in a circle are congruent, then… Investigate the following:

• the central angles associated with those chords • the intercepted arcs associated with those chords • how far the chords are from the center of the circle.

Write the 3 Chord Conjectures in your notes. Investigation 4 Perpendiculars & Chords

If AB CD≅ , then If “equal chords” then Write your observations:

116m BOA m DOC∠ = ∠ = equal central angles � � 116mAB mCD= =

equal intercepted arcS The chords are the same distance from the center. Fyi: 3.4 cmAB CD= =

A

O

BD

C

What if the chords are not congruent?

EF GH≅ none of the measures are equal Fyi:

75m FPE∠ = 101m GPH∠ =

…

P

E

F

H

G

Draw a perpendicular segment from the center to each of the chords (and extend it through the circle). What do you notice about the chord and its intercepted arcs? perpendicular bisects the chord and the arc Write the Perpendicular to a Chord Conjecture in your notes.

C

AB D

Draw the entire circle. Hint: Draw any chord and its perpendicular bisector. Draw any other chord and its perpendicular bisector. What have you found? Can you complete the circle? you find the center Write the Perpendicular Bisector of Chord Conjecture in your notes.

1 cm 116

116

1 cm

1 cm

1.6 cm


S. Stirling of 18

EXERCISES Lesson 6.2 Pages 320 – 321 # 1 – 12, 18 Write the properties you are using as you are finding the missing measures. (You don’t need to name them, you just need to state them.)

8

= chords are equidistant from the center.

8

165

128

70

Central angle = intercepted arc. x = 165

= chords cut = arcs. Circle’s arcs = 360. z = 360 – 276 = 84

84

70

70

= chords cut = arcs and = Central angles = intercepted arc. w = 70

Various: Linear pair supplementary.

65m AOI∠ = Central angle = intercepted arc. w = 115 = chords cut = arcs and = central angles. x = 115 and y = 65 115

115

65 65

65

3

3

Various:

Radius ⊥ chord

bisects the chord. 3AP PB= = and 4DQ QC= = .

Perim = 3 + 4 + 3 + 4 + 6 = 20 cm

4 3 4

4 6

Various: Central angle = intercepted arc. � 68mAC =

Linear pair supp. & Radii = so △COB isos.

& base ∠ =

△ sum = 180

(180 – 112)/2 = 34 34m B∠ =

112

68

34 34


S. Stirling of 18

110

Various: � 130mAC = so � 130 48 82mAB = − =

Central angle = intercepted arc. x = 48, y = 82, w = 110 Circle’s arcs add to 360. 360 – 48 – 82 – 110 = 120 z = 120 48

82

82

120120

Various: Circle’s arcs add to 360, = chords cut = arcs and Arcs = central angles. � 360 72 288mFAT = − =

288 ÷ 3 = 96 = x y = 96 Radii = so △FOE isos. &

base ∠ = , △ sum = 180

(180 – 96)/2 = 42 = z

96

96

96

96 96

42

42

Various: Radius = 18 so the diameter = 36. The diameter would have to be the longest chord of the circle, so the chord can’t be greater than 36.

Various: Central angle = intercepted arc. ||, so corresponding angles =. x = 66 Since radii of a circle =,

AOB∆ isos. & base angles =. △ sum = 180, so

180 – 66 – 66 = 48 = y 180 114 66m AOC∠ = − =

and z = 66.

66

66

66 66 48

66

66

If a chord has a ⊥

bisector, then it should pass throughthe center of the circle.

Page 321 #18. See book. Find center.

⊥ bisector of a chord passes

through the center.

C


S. Stirling of 18

Lesson 6.3 Arcs and Angles

Investigation 5 The Big Question: What is the measure of an inscribed angle?

What is the relationship between an inscribed angle and its intercepted arc? Write the Inscribed Angle Conjecture in your notes. Investigation 6

O

A

B

What is the measure of �mAB ? 76°

Draw an inscribed angle, AXB∠ .

What is m AXB∠ ? 38°

What is the measure of �mCD ? 144°

Draw an inscribed angle, CYD∠ .

What is m CYD∠ ? 72°

Draw 3 different inscribed angles for �CD . Call them CED∠ , CFD∠ and CGD∠ . What can you say about inscribed angles that intercept the same arc? they’re equal

Given � � 40mCD mEF= = ° . Draw inscribed angle CXD∠ . Draw inscribed angle EYF∠ . What can you say bout the measures of these angles? Intercept the same sized arcs so they’re equal. Write the Inscribed Angles Intercepting Arcs Conjecture in your notes.

C

PD

P

C

DE

F

X

76

76

38

38

P

C

D

Y

144

144

72 72

inscribed angle = ½ intercepted arc

F

60

30

30

30 E

G

X 20

40

20 Y

40


S. Stirling of 18

Investigation 7: Angles Inscribed in a Semicircle Draw any inscribed angle in either semicircle. Call it ACB∠ . Draw another inscribed angle in either semicircle. Call it AFB∠ . Draw another inscribed angle in either semicircle. Call it AEB∠ .

What about the measures of ACB∠ , AFB∠ and AEB∠ ? all 90°

Could you prove this to be true? How? Write the Angles Inscribed in a Semicircle Conjecture in your notes.

Investigation 8: Inscribed Quadrilaterals Use your notes and draw a cyclic quadrilateral in P○ . Remember each angle must be an inscribed angle and each side must be a chord. Label your quadrilateral ABCD. Measure all of the angles of your quadrilateral. Are there any relationships between the angles? Write the Cyclic Quadrilateral Conjecture in your notes. Try to draw a cyclic parallelogram in circle O. What type of parallelogram can be inscribed in a circle? Write the Cyclic Parallelogram Conjecture in your notes.

B

A

Q

P

O

Inscribed angle = ½ of the intercepted arc

A semi-circle = 180°

1 180 902 = °i .

.

The opposite angles are supplementary in a cyclic quadrilateral.

Only rectangles (and squares) can be inscribed in a circle. .

C

F

E

90

90

90


S. Stirling of 18

Investigation 9 Given AB CF ED� � . Examine the measures of the arcs. What could you conclude about the intercepted arcs? Write the Parallel Lines (Secants) Intercepted Acrs Conjecture in your notes. EXERCISES Lesson 6.3 Pages 327 – 328 # 1 – 16, 22 Write the properties you are using as you are finding the missing measures. (You don’t need to name them, you just need to state them.)

D

BA

F

P

C

E

� �mAF mBC= � �mEF mDC=

.

Inscribed angle = ½ intercepted arc.

65

Various: Inscribed angle = ½ intercepted arc. Semi circle

measures 180 °

180 – 120 = 60 60 ÷ 2 = 30

120 60

30

Inscribed angle = ½ intercepted arc. 95 * 2 = 190 c = 190 – 120 = 70

Various:

Radius ⊥ tangent.

△ sum = 180

180 – 90 – 40 = 50 Central angle = intercepted arc. h = 50

50 70

22 22

54

22 22

54 54 54

140

Various: Inscribed angle = ½ intercepted arc &

Semi circle = 180 °

20 * 2 = 40

d = 180 – 40 = 140 180 – 96 = 84

e = 84 ÷ 2 = 42 84

42

40

Various: Inscribed angle = ½ intercepted arc 75 * 2 = 150

Circle’s arcs = 360°

g = 360 – 150 – 110 = 100 x = (110 + 100)/2 = 105

Quad. sum = 360°

f = 360 – 75 – 105 – 90 = 90

150

100

90

x 105


S. Stirling of 18

Various: �NDO is a semicircle. 180 – 136 = 44 Kite, so = chords make = arcs so y = 44

44

Various:

Central ∠ = arc &

vertical ∠s =.

Radius ⊥ tangent.

Quad. sum = 360 °

w = 360 – 180 – 130 = 50

130 130

50

Various: Opposite sides of a rectangle =. And = chords cut = arcs. Circle = 360º 2 64 360

2 296

148

x

x

x

+ ===

32

x

Various: Inscribed angle = ½ intercepted arc. 38 * 2 = 76

Circle = 360° and =

chords cut = arcs k = (360 – 76)/2 = 142

76

142 142

Various: Inscribed angle = ½ intercepted arc. 90 = ½ (98 + x) 180 = 98 + x, x = 82 m = 360 – 220 = 140 n = ½ (140 + 82) = 111

82 x

140

Various:

Circle = 360° & =

chords cut = arcs s = 360/6 = 60 Inscribed angle = ½ intercepted arc. r = ½ (60 * 4) = 120

60

60 60

60

Various: Inscribed angle = ½ intercepted arc. Circle = 360º.

( )2 2 2 2 2 360

2 360

180

a b c d e

a b c d e

a b c d e

+ + + + =+ + + + =+ + + + =

Sum = 180º 2a

2b

2c

Various: Isos ∆, base ∡s =.

180 9841

2q

−= =

Central ∡ =

intercepted arc. Parallel secants cut = arcs. Circle = 360º. 2 218 360

2 142

71

p

p

p

+ ===

41

p

120

98

41

2d

2e


S. Stirling of 18

22.

Various: Inscribed angle = ½ intercepted arc Exterior ∡ = sum

of 2 remote interior ∡s.

OR Since vertex inside: Angle = ½ sum arcs

( )1 70 80275

y

y

= +

=

75

40 35

Various: Inscribed angle = ½ intercepted arc

37 2 74x = • = But 35 2 70x = • = The arc can’t have two different measures!

x

For p, look for the big triangle involving p, e and c.

180 108 36 36p = − − =

108

108

108

108 72

72 108

36 108

72

108

90 = h

( )5 2 180108

5a

−= =

36

108

m = 90 – 72= 18

36

180 – 108 – 36 = 36

36 54


S. Stirling of 18

EXERCISES Lesson 6.5 Pages 337 – 340 # 1 – 13, 15, 19. On all problems, show algebraic procedures: write the formula, substitute in known information, then solve. On #1 – 6, leave your answers in terms of π. On #7 – 9, use the π approximation on the calculator and round final answers to 3 decimal places. For #10 – 15, see your book for the problem statement.

1. If C = 5π cm, find d.

5

5

C d

d

d

ππ π

===

2. If r = 5 cm, find C.

( )2

2 5

10

C r

C

C

πππ

===

3. If C = 24 cm, find r.

2

24 2

24 2

2 212

C r

r

r

r

πππ

π π

π

==

=

=

4. If d = 5.5 cm, find C.

5.5

C d

C

ππ

==

5. If a circle has a diameter of 12 cm, what is its circumference?

12

C d

C

ππ

==

6. If a circle has a circumference of 46π, what is its diameter?

46

46

C d

d

d

ππ π

===

7. If d = 5 cm, find C.

5

15.708

C d

C

C

ππ

==≈

8. If r = 4 cm, find C.

( )

2

2 4

8

25.133

C r

C

C

C

πππ

===≈

9. If C = 44 m, find r.

2

44 2

44 2

2 222

7.003

C r

r

r

r

πππ

π π

π

==

=

= ≈

10. A bicycle tire with a 27 inch diameter, find C.

27

84.823 in

C d

C

C

ππ

==≈

11. Ferris wheel with r = 24 cm, find distance traveled by a seat in one revolution.

( )

2

2 24

48

150.796

C r

C

C

C

πππ

===≈


S. Stirling of 18

#19

H K

M

N

P

R

S

b = 90 c = 42 d = 70 e = 48 f = 132 g = 52

84

52

(180 – 76)/2 = 52

52

84/2 = 42 48

48

132

90

70

360 – 90 – 68 – 132 = 70

180 – 42 – 48 = 90

180 – 42 – 90 = 48

12. Circle inscribed in a square with perimeter 24 cm, find C.

4

24 4

6

p s

s

s

===

15. Find number of 1 inch tiles to put around the edge of the pool. The circular ends:

18

56.549

C d

C

C

ππ

==≈

13. Circle with C = 16π inches is circumscribed about a square, find length of the diagonal.

16

16

C d

d

d

ππ π

===

6

PP

6

18.850

C d

C

C

ππ

==≈

6 16

18

30

Sides of the rectangle are =.

( )56.549 2 30 116.549 ftperim = + =

116.549 * 12 = 1398.6 one-inch tiles So need 1399 one-inch tiles


S. Stirling of 18

Investigation 10: Arc Length So far the measure of an arc = the measure of its central angle (in degrees).

In the diagram, � � 120m AB mCD= = . If you are thinking in terms of “turn” or degrees, it makes sense that if you are

standing at point O you will turn 120° to get

from A to B and you would turn the same amount of degrees to turn from C to D. But if you are on the circle itself, and if you are traveling from point A to point B did you travel the same distance from point C to point D? NO! The distance from C to D is longer than the distance from A to B. How can you explain this? The distance would be part of the circumference, but what part? What part (fraction) of the circle are we talking about?

120 1Fraction =

360 3=

If 4 cmOA = and 12 cmOC = , how far is it from A to B? How far is it from C to D? Think part of the circumference!

� ( )21 16length of = 4 16.755

3 3AB π π= ≈i cm

� ( )21length of = 12 48 150.796

3CD π π= ≈i cm

So if you are looking at the length of the arc and not the amount of turn (or degree of the arc), then it makes complete sense.

B

CO A

D

120

4 cm 8 cm

OC = 12 cm


S. Stirling of 18

EXERCISES Lesson 6.7 Pages 351 # 1 – 8, 9, 13 – 14 On all problems, show algebraic procedures: write the formula, substitute in known information, then solve. Leave your answers in terms of π!!

( )802 3

3602

694

34.189

length π

π

π

=

= •

=

≈

( )1202 12

3601

2438

25.133

length π

π

π

=

= •

=≈

( )602 18

3601

3666

18.850

length π

π

π

=

= •

=≈

80

( )802 9

3602

1894

12.566

length π

π

π

=

= •

=≈

160 160

12360

9 9 412

4 4 9

27

d

d

r

π π

πππ π

=

=

=

( )2102 12

3607

241214

43.982

length π

π

π

=

= •

=≈

210

1206 2

3603 3 2

62 2 3

9

r

r

r

π π

πππ π

=

=

=

60

7240 2

3605 40 5 2

2 1 2 5

100

r

r

r

π π

π ππ π

=

=

=

72

72


S. Stirling of 18

Read the problems from the book pages 352 – 353.

9. Completes 4 laps in 6 minutes. Calculate average speed in meters per minute. Round to two decimal places! Make a drawing.

2 100

40 200

325.66 meters

P d

P

P

ππ

= += +≈

i

100 meters

40 meters

325.66 m m13.57 min24 min≈

13. 1 revolution in 20 seconds, what is the angular velocity?

360 deg18 sec20 sec

° =

Since all of the horses rotate 360º in one revolution, they all have the same angular velocity.

14. 2 horses complete 1 revolution in 20 seconds. The horses are 8 m and 6 m from the center. What are the tangential velocities of the two horses? Round to two decimal places!

Horse #1:

( )1 rev 2 8 16 meters

16 m m2.51 sec20 sec

π ππ

= =

≈

Horse #2:

( )1 rev 2 6 12 meters

12 m m1.88 sec20 sec

π ππ

= =

≈

The horse on the outside is moving faster because he has to travel further to make one revolution in the same amount of time (20 seconds).


S. Stirling of 18

EXERCISES Chapter 6 Review Pages 359 – 360 # 4 – 19, 21, 22 Write the properties you are using as you are finding the missing measures. (You don’t need to name them. You just need to state them.) Mark diagrams with the information as you go!

The degree measure describes the amount of turn, based on the central angle. The arc length is part of the circumference. Measured in a unit of length, like inches.

Various:

Tangent ⊥ Radius

Central angle = intercepted arc. △ sum = 180

180 – 90 – 35 = 55 b = 55

90 b

Various: Inscribed angle = ½ intercepted arc. 110 * 2 = 220 a = 220 – 155 = 65

Various:

Circle = 360°

= chords cut = arcs. c = (360 – 104)/2 = 128

( )2

2 20

40

125.664

C r

C

C

C

πππ

===≈

132

132

42.017

d

d

d

π

π

=

=

≈

160

Various: Equal chords cut = arcs. (360 – 220)/2 = 70

( )702 36

3607

723614

43.982

length π

π

π

=

= •

=≈

x

Various: Vertex inside so

( )1 60 64262

x

x

= +

=

Linear pair supp 180 62 118e = − =

Various: Inscribed angle = ½ intercepted arc. 90 * 2 = 180

Circle = 360°

d = 360 – 180 – 89 = 91

180

92

Various: Linear pair supp 180 88 92− = Vertex inside so

( )192 1182184 118

66

f

f

f

= +

= +=

92

100

Various: Equal chords cut = arcs.

( )1002 27

3605

541815

47.124

length π

π

π

=

= •

=≈

70


S. Stirling of 18

Various:

△ sum = 180.

x = 180 – 35 – 57 = 108 Inscribed angle = ½ intercepted arc. 108 * 2 = 216 but the angle intercepts a semi-circle which = 180. x should = 90.

x

Various: Semi-⊙ = 180. 180 – 108 = 72

Inscribed angle = ½ intercepted arc. 72 ÷ 2 = 36

Alternate interior ∠s =,

so lines ||.

72

36

Various: Circle’s arcs = 360. 360 – 152 – 56 = 152 = chords cut = arcs. So JI IM= and

JIM∆ is isos.

Various: Inscribed angle = ½ intercepted arc. � 140mKIM = & � 140 70 70mKI = − =

= chords cut = arcs. so KIM∆ is isos.

70

Various: Parallel secants cut = arcs, so x = 56. ⊙s arcs add to 360º, but

84 + 56 + 56 + 158 = 354, not 360.

x =56

72 152

(Place answers below.) Need at least 2 perpendicular bisectors. Point V is equidistant from the triangle’s vertices.

A

B

C

V

Need at least 2 angle bisectors and a radius drawn perpendicular to a side (for the radius). Point S is equidistant from the triangle’s sides.

C

B

A

S

ch 6 worksheet l1 key - palisades high school · ch 6 worksheet l1 key.doc name _____ s. stirling...

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