GEOMETRY CLASSWORK INSCRIBE ANGLES

(1)

For Exercises 10±13

,

refer to .

Name the center of the circle. 62/87,21

R

Is DUDGLXV"([SODLQ 62/87,21

A radius is a segment with endpoints at the center and on the circle. But KDVERWKWKHHQGSRLQWVRQ the circle, so it is a chord.

,

refer to .

Identify a chord that is not a diameter. 62/87,21

The chords GRQRWSDVVWKURXJKWKH

center. So, they are not diameters.

Is ? Explain.

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All radii of a circle are congruent. Since

DUHERWKUDGLLRI , .

Circle J has a radius of 10 units

,

KDVD radius of 8 units

,

and BC = 5.4 units. Find each

measure.

CK

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Since the radius of

K

is 8 units,

BK

= 8.

7KHUHIRUHCK = 2.6 units.

J

K

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First find

CK.

Since circle

K

has a radius of 8 units,

BK

Since circle J has a radius of 10 units, J

C

= 10.

Therefore, J

K

= 12.6 units.

PIZZA

Find the radius and circumference of the

pizza shown. Round to the nearest hundredth

,

if necessary.

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The diameter of the pizza is 16 inches.

So, the radius of the pizza is 8 inches.

The circumference

C

of a circle of diameter

d

is given by

Therefore, the circumference of the pizza is about 50.27 inches.

Find the diameter and radius of a circle by with the given circumference. Round to the nearest hundredth.

C

= 18 in.

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The circumference

C

of a circle with diameter

d

is given by

Here,

C

= 18 in. Use the formula to find the diameter. Then find the radius.

Therefore, the diameter is about 5.73 inches and the radius is about 2.86 inches.

C

= 375.3 cm

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The circumference

C

d

is given by

Here,

C

= 375.3 cm. Use the formula to find the diameter. Then find the radius.

Therefore, the diameter is about 119.46 centimeters and the radius is about 59.73 centimeters.

Find the exact circumference of each circle by using the given inscribed or circumscribed polygon.

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The hypotenuse of the right triangle is a diameter of the circle. Use the Pythagorean Theorem to find the diameter.

The diameter of the circle is 17 cm. The circumference

C

of a circle is given by

Therefore, the circumference of the circle is 17

FHQWLPHWHUV

62/87,21

Each diagonal of the inscribed rectangle will pass through the origin, so it is a diameter of the circle. Use the Pythagorean Theorem to find the diameter of the circle.

The diameter of the circle is inches. The circumference

C

Therefore, the circumference of the circle is

inches.

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A diameter perpendicular to a side of the square and the length of each side will measure the same. So,

d

= 25 millimeters.

The circumference

C

Therefore, the circumference of the circle is 25

millimeters.

,

refer to .

R

,

refer to .

Is ? Explain.

62/87,21

DUHERWKUDGLLRI , .

,

measure.

CK

62/87,21

Since the radius of

K

is 8 units,

BK

= 8.

J

K

62/87,21

First find

CK.

Since circle

K

BK

C

= 10.

Therefore, J

K

= 12.6 units.

PIZZA Find the radius and circumference of the pizza shown. Round to the nearest hundredth

,

if necessary.

62/87,21

The circumference

C

d

is given by

C

= 18 in.

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The circumference

C

d

is given by

Here,

C

= 375.3 cm

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The circumference

C

d

is given by

Here,

C

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C

Therefore, the circumference of the circle is 17

FHQWLPHWHUV

62/87,21

C

Therefore, the circumference of the circle is

inches.

62/87,21

d

= 25 millimeters.

The circumference

C

Therefore, the circumference of the circle is 25

millimeters.

(2)

,

refer to .

R

,

refer to .

Is ? Explain.

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DUHERWKUDGLLRI , .

,

measure.

CK

62/87,21

Since the radius of

K

is 8 units,

BK

= 8.

7KHUHIRUH

CK

= 2.6 units.

J

K

62/87,21

First find

CK.

Since circle

K

BK

C

= 10.

Therefore, J

K

= 12.6 units.

,

if necessary.

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The circumference

C

d

is given by

C

= 18 in.

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The circumference

C

d

is given by

Here,

C

= 375.3 cm

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The circumference

C

d

is given by

Here,

C

62/87,21

C

Therefore, the circumference of the circle is 17

FHQWLPHWHUV

62/87,21

C

Therefore, the circumference of the circle is

inches.

62/87,21

d

= 25 millimeters.

The circumference

C

Therefore, the circumference of the circle is 25

millimeters.

,

refer to .

R

,

refer to .

Is ? Explain.

62/87,21

DUHERWKUDGLLRI , .

,

measure.

CK

62/87,21

Since the radius of

K

is 8 units,

BK

= 8.

J

K

62/87,21

First find

CK.

Since circle

K

BK

C

= 10.

Therefore, J

K

= 12.6 units.

,

if necessary.

62/87,21

The circumference

C

d

is given by

C= 18 in. 62/87,21

The circumference

C

d

is given by

Here,

C

= 375.3 cm

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The circumference

C

d

is given by

Here,

C

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C

Therefore, the circumference of the circle is 17

FHQWLPHWHUV

62/87,21

C

Therefore, the circumference of the circle is

inches.

62/87,21

d

= 25 millimeters.

The circumference

C

Therefore, the circumference of the circle is 25

millimeters.

(3)

,

refer to .

R

,

refer to .

Is ? Explain.

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DUHERWKUDGLLRI , .

,

measure.

CK

62/87,21

Since the radius of

K

is 8 units,

BK

= 8.

J

K

62/87,21

First find

CK.

Since circle

K

BK

C

= 10.

Therefore, J

K

= 12.6 units.

PIZZA

Find the radius and circumference of the

,

if necessary.

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The circumference

C

d

is given by

C

= 18 in.

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The circumference

C

d

is given by

Here,

C

C= 375.3 cm 62/87,21

The circumference

C

d

is given by

Here,

C

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C

Therefore, the circumference of the circle is 17

FHQWLPHWHUV

62/87,21

C

Therefore, the circumference of the circle is

inches.

62/87,21

d

= 25 millimeters.

The circumference

C

Therefore, the circumference of the circle is 25

millimeters.

,

refer to .

R

,

refer to .

Is ? Explain.

62/87,21

DUHERWKUDGLLRI , .

,

measure.

CK

62/87,21

Since the radius of

K

is 8 units,

BK

= 8.

J

K

62/87,21

First find

CK.

Since circle

K

BK

C

= 10.

Therefore, J

K

= 12.6 units.

PIZZA

Find the radius and circumference of the

,

if necessary.

62/87,21

The circumference

C

d

is given by

C

= 18 in.

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The circumference

C

d

is given by

Here,

C

C= 375.3 cm 62/87,21

The circumference

C

d

is given by

Here,

C

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C

Therefore, the circumference of the circle is 17

FHQWLPHWHUV

62/87,21

C

Therefore, the circumference of the circle is

inches.

62/87,21

d

= 25 millimeters.

The circumference

C

Therefore, the circumference of the circle is 25

millimeters.

(4)

Find the value of x.

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The sum of the measures of the central angles of a circle with no interior points in common is 360.

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The sum of the measures of the central angles of a FLUFOHZLWKQRLQWHULRUSRLQWVLQFRPPRQLV

and are diameters of . Identify each arc as a m

a

j

o

r

ar

c,

m

i

n

o

r

ar

c,

or

se

m

ici

r

cle

. Then find its measure.

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Here, LVWKHVKRUWHVWDUFFRQQHFWLQJWKHSRLQWVC and

D

on 7KHUHIRUHLWLVDPLQRUDUF

m

(arc

C

F

G

)

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Here, LVDGLDPHWHU7KHUHIRUHDUFCFGLVD semicircle and

m

(arc

CFG

) = 180.

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Here, LVWKHORQJHVWDUFFRQQHFWLQJWKHSRLQWV

G

and

F

on 7KHUHIRUHLWLVDPDMRUDUF Arc

GCF

shares the same endpoints as minor arc

GF.

Therefore, the measure of arc

GCF

is 325.

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Here, LVWKHVKRUWHVWDUFFRQQHFWLQJWKHSRLQWV

A

and

G

The measure of a minor arc is equal to the measure of its related central angle.

AG

is 55.

SHOPPING The graph shows the results of a survey in which teens were asked where the best place was to shop for clothes.

a._{What would be the arc measures associated with} the mall and vintage stores categories?

b. Describe the kinds of arcs associated with the category ³Mall´and category ³None of these´. c._{Are there any congruent arcs in this graph?} Explain.

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a. The measure of the arc is equal to the measure of the central angle. The mall contributes 76% and the vintage stores contribute 4% of the total shopping. Find the 76% of 360 to find the central angle of the arc associated with the malls.

Find the 4% of 360 to find the central angle of the arc associated with the vintage stores.

b. The arc associated with the mall has a measure of 273.6. So, it is a major arc. The arc associated with none of these has a measure of 9% of 360 or 32.4. So, it is a minor arc.

c. Yes; the arcs associated with the online and none of these categories have the same arc measure since each category accounts for the same percentage of the circle, 9%.

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a

j

o

r

ar

c,

m

i

n

o

r

ar

c,

or

se

m

ici

r

cle

62/87,21

D

m

(arc

C

F

G

)

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Here, LVDGLDPHWHU7KHUHIRUHDUF

CFG

LVD semicircle and

m

(arc

CFG

) = 180.

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G

and

F

GCF

GF.

GCF

is 325.

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A

and

G

AG

is 55.

a. What would be the arc measures associated with the mall and vintage stores categories?

b. Describe the kinds of arcs associated with the category ³Mall´and category ³None of these´. c. Are there any congruent arcs in this graph? Explain.

62/87,21

a._{The measure of the arc is equal to the measure of} the central angle. The mall contributes 76% and the vintage stores contribute 4% of the total shopping. Find the 76% of 360 to find the central angle of the arc associated with the malls.

(5)

62/87,21

a

j

o

r

ar

c,

m

i

n

o

r

ar

c,

or

se

m

ici

r

cle

62/87,21

D

m

(arc

C

F

G

)

62/87,21

Here, LVDGLDPHWHU7KHUHIRUHDUFCFGLVD semicircle and

m

(arc

CFG

) = 180.

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G

and

F

GCF

GF.

GCF

is 325.

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A

and

G

AG

is 55.

a._{What would be the arc measures associated with} the mall and vintage stores categories?

b. Describe the kinds of arcs associated with the category ³Mall´and category ³None of these´. c._{Are there any congruent arcs in this graph?} Explain.

62/87,21

a. The measure of the arc is equal to the measure of the central angle. The mall contributes 76% and the vintage stores contribute 4% of the total shopping. Find the 76% of 360 to find the central angle of the arc associated with the malls.

(6)

ALGEBRA Findthe value of

x

.

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In the same circle or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent. Therefore,

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2

x

± 1 = 143 [= 72

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In the same circle or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent. Here,

7KHUHIRUH 5

x

± 1 = 4

x

+ 3

[ = 4

62/87,21

So, 7KHUHIRUH 3

x

= 7

x

± 44

4

x

= 44 [ = 11

In

,

the radius is 14 and

CD

=22. Find each measure. Roundto the nearesthundredth

,

if necessary.

CE

62/87,21

Radius is perpendicular to chord . So, by Theorem 10.3, bisects . Therefore,

CE

=

ED

.

By substitution,

CE

= (22) or 11 units.

In

,

the diameter is 18

, LM

= 12, and

. Find eachmeasure. Roundto the

nearesthundredth

,

ifnecessary.

62/87,21

Diameter is perpendicular to chord . So, by Theorem 10.3, bisects arc

LKM.

Therefore,

m

(arc

LK

)=

m

(arc

KM

).

By substitution,

m

(arc

LK

) = (84) or 42.

SNOWBOARDINGThe snowboarding rail shown is an arc of a circle in which LVSDUWRIWKH diameter. If is about 32% of a complete circle, what is ?

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The diameter containing

is perpendicular to

chord

. So, by Theorem 10.3,

bisects arc

ABC.

Therefore,

m

(arc

AB

) =

m

(arc

BC

). By

substitution,

m

(arc

AB

) = (115.2) or 57.6.

ALGEBRA In

,

,D

F

= 3

x

± 7

,

and

F

E

=

x

+ 9. What is

x

?

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In the same circle or in congruent circles, two chords are congruent if and only if they are equidistant from the center. Since ,

D

F

=

E

F

.

3

x

± 7 =

x

+ 9 2

x

= 16

x

= 8

x

.

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2

x

± 1 = 143 [ = 72

62/87,21

7KHUHIRUH 5

x

± 1 = 4

x

+ 3

[ = 4

62/87,21

So, 7KHUHIRUH 3

x

= 7

x

± 44

4

x

= 44 [ = 11

In

,

CD

,

if necessary.

CE

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CE

=

ED

.

By substitution,

CE

= (22) or 11 units.

In

,

the diameter is 18

, LM

= 12, and

nearesthundredth

,

ifnecessary.

62/87,21

LKM.

Therefore,

m

(arc

LK

)=

m

(arc

KM

).

By substitution,

m

(arc

LK

) = (84) or 42.

62/87,21

The diameter containing

is perpendicular to

chord

. So, by Theorem 10.3,

bisects arc

ABC.

Therefore,

m

(arc

AB

) =

m

(arc

BC

). By

substitution,

m

(arc

AB

) = (115.2) or 57.6.

ALGEBRA In

,

,D

F

= 3

x

± 7

,

and

F

E

=

x

+ 9. What is

x

?

62/87,21

D

F

=

E

F

.

3

x

± 7 =

x

+ 9 2

x

= 16

x

= 8

(7)

x

.

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2

x

± 1 = 143 [= 72

62/87,21

7KHUHIRUH 5

x

± 1 = 4

x

+ 3

[= 4

62/87,21

So, 7KHUHIRUH 3

x

= 7

x

± 44

4

x

= 44 [= 11

In

,

CD

,

if necessary.

CE

62/87,21

CE

=

ED

.

By substitution,

CE

= (22) or 11 units.

In

,

the diameter is 18

, LM

= 12, and

nearesthundredth

,

ifnecessary.

62/87,21

LKM.

Therefore,

m

(arc

LK

)=

m

(arc

KM

).

By substitution,

m

(arc

LK

) = (84) or 42.

62/87,21

The diameter containing

is perpendicular to

chord

. So, by Theorem 10.3,

bisects arc

ABC.

Therefore,

m

(arc

AB

) =

m

(arc

BC

). By

substitution,

m

(arc

AB

) = (115.2) or 57.6.

ALGEBRA In

,

,D

F

= 3

x

± 7

,

and

F

E

=

x

+ 9. What is

x

?

62/87,21

D

F

=

E

F

.

3

x

± 7 =

x

+ 9 2

x

= 16

x

= 8

x

.

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2

x

± 1 = 143 [= 72

62/87,21

7KHUHIRUH 5

x

± 1 = 4

x

+ 3

[= 4

62/87,21

So, 7KHUHIRUH 3

x

= 7

x

± 44

4

x

= 44 [= 11

In

,

CD

,

if necessary.

CE

62/87,21

CE

=

ED

.

By substitution,

CE

= (22) or 11 units.

In

,

the diameter is 18

, LM

= 12, and

nearesthundredth

,

ifnecessary.

62/87,21

LKM.

Therefore,

m

(arc

LK

)=

m

(arc

KM

).

By substitution,

m

(arc

LK

) = (84) or 42.

62/87,21

The diameter containing

is perpendicular to

chord

. So, by Theorem 10.3,

bisects arc

ABC.

Therefore,

m

(arc

AB

) =

m

(arc

BC

). By

substitution,

m

(arc

AB

) = (115.2) or 57.6.

ALGEBRA In

,

,D

F

= 3

x

± 7

,

and

F

E

=

x

+ 9. What is

x

?

62/87,21

D

F

=

E

F

.

3

x

± 7 =

x

+ 9 2

x

= 16

x

= 8

(8)

ALGEBRA Find each measure.

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If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its

intercepted arc. Therefore,

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If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc. Therefore,

62/87,21 Here,

The arc LVDVHPLFLUFOH6R 7KHQ

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If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. So,

Therefore,

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If two inscribed angles of a circle intercept the same arc or congruent arcs, then the angles are congruent. Since _Ә

C and

_Ә

D both intercept arc AB, m

_Ә

C =

m

_ӘD.

Therefore, m_Ә

C = 5(10) ± 3 or 47.

ALGEBRA Find each value.

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An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So,

The sum of the measures of the angles of a triangle

is 180. So,

Therefore,

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is 180. So,

Therefore,

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If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Therefore,

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Therefore,

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62/87,21 Here,

62/87,21

Therefore,

62/87,21

C and

_Ә

D both intercept arc AB, m

_Ә

C =

m

_ӘD.

Therefore, m_Ә

C = 5(10) ± 3 or 47.

62/87,21

is 180. So,

Therefore,

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is 180. So,

Therefore,

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Therefore,

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Therefore,

(9)

62/87,21

62/87,21 Here,

62/87,21

Therefore,

62/87,21

C and

_Ә

D both intercept arc AB, m

_Ә

C =

m

_ӘD.

Therefore, m_Ә

C = 5(10) ± 3 or 47.

62/87,21

is 180. So,

Therefore,

62/87,21

is 180. So,

Therefore,

62/87,21

Therefore,

62/87,21

Therefore,

62/87,21

62/87,21 Here,

62/87,21

Therefore,

62/87,21

C and

_Ә

D both intercept arc AB, m

_Ә

C =

m

_ӘD.

Therefore, m_Ә

C = 5(10) ± 3 or 47.

62/87,21

is 180. So,

Therefore,

62/87,21

is 180. So,

Therefore,

62/87,21

Therefore,

62/87,21

Therefore,

(10)

Copy each figure and draw the common

tangents. If no common tangent exists, state no

common t

a

ngent.

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Every tangent drawn to the small circle will intersect the larger circle in two points. Every tangent drawn to the large circle will not intersect the small circle at any point. Since a tangent must intersect the circle at exactly one point, no common tangent exists for these two circles.

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Two common tangents can be drawn to these two circles.

Determine whether each LVWDQJHQWWRWKH given circle. Justify your answer.

62/87,21 Yes;

62/87,21 yes;

Find x. Assume that segments that appear to be tangent are tangent. Round to the nearest tenth if necessary.

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By Theorem 10.10, . So, LVDULJKW triangle.

Use the Pythagorean Theorem.

Substitute.

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Substitute.

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If two segments from the same exterior point are tangent to a circle, then they are congruent.

Find the value of x. Then find the perimeter.

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Find the missing measures.

Since is circumscribed about , DQG DUHWDQJHQWWR , as are , , , and . Therefore, , , and

. So,

QU

=

QT

. 2

x

= 14

x

= 7

TS

=

VS

= 17

RU

=

RV

= 27 ± 17 = 10

To find the perimeter of a triangle, add the lengths of its sides.

Find x to the nearest hundredth. Assume that segments that appear to be tangent are tangent.

62/87,21

If two segments from the same exterior point are tangent to a circle, then they are congruent. Here,

TS

=

TR

and

TR

=

TQ

.

By the Transitive Property,

TS

=

TQ

.

common t

a

ngent.

62/87,21

62/87,21 Yes;

62/87,21 yes;

62/87,21

Substitute.

62/87,21

Substitute.

62/87,21

. So,

QU

=

QT

. 2

x

= 14

x

= 7

TS

=

VS

= 17

RU

=

RV

= 27 ± 17 = 10

62/87,21

TS

=

TR

and

TR

=

TQ

.

TS

=

TQ

.

(11)

common t

a

ngent.

62/87,21

62/87,21 Yes;

62/87,21 yes;

62/87,21

Substitute.

62/87,21

Substitute.

62/87,21

. So,

QU

=

QT

. 2

x

= 14

x

= 7

TS

=

VS

= 17

RU

=

RV

= 27 ± 17 = 10

62/87,21

TS

=

TR

and

TR

=

TQ

.

TS

=

TQ

.

common t

a

ngent.

62/87,21

62/87,21 Yes;

62/87,21 yes;

62/87,21

Substitute.

62/87,21

Substitute.

62/87,21

. So,

QU

=

QT

. 2

x

= 14

x

= 7

TS

=

VS

= 17

RU

=

RV

= 27 ± 17 = 10

62/87,21

TS

=

TR

and

TR

=

TQ

.

TS

=

TQ

.

(12)

Find eachmeasure. Assume that segments that appearto be tangentare tangent.

62/87,21

Ә

JMKDQG

_Ә

HMJ

form a linear pair.

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Arc

BD

and arc

BCD

are a minor and major arc that share the same endpoints.

62/87,21

By Theorem 10.13, .

Solve for .

We know that .

Substitute.

Find eachmeasure.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

m

(arc

JNM

)

62/87,21

So, the measure of arc

JNM

is 205.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

JEWELRYIn the circular necklace shown,

A

and

B

are tangent points. If

x

= 260

,

what is

y

?

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

x

.

62/87,21

By Theorem 10.14, .

Solve for

.

62/87,21

By Theorem 10.14, .

Solve for

.

62/87,21

Ә

JMKDQG

_Ә

HMJ

form a linear pair.

62/87,21

Arc

BD

and arc

BCD

62/87,21

By Theorem 10.13, .

Solve for .

We know that .

Substitute.

Find eachmeasure.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

m

(arc

JNM

)

62/87,21

JNM

is 205.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

A

and

B

x

= 260

,

what is

y

?

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

x

.

62/87,21

By Theorem 10.14, .

Solve for

.

62/87,21

By Theorem 10.14, .

Solve for

.

(13)

62/87,21

Ә

JMKDQG

_Ә

HMJ

form a linear pair.

62/87,21

Arc

BD

and arc

BCD

62/87,21

By Theorem 10.13, .

Solve for .

We know that .

Substitute.

Find eachmeasure.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

m

(arc

JNM

)

62/87,21

JNM

is 205.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

A

and

B

x

= 260

,

what is

y

?

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

x

.

62/87,21

By Theorem 10.14, .

Solve for

.

62/87,21

By Theorem 10.14, .

Solve for

.

62/87,21

Ә

JMKDQG

_Ә

HMJ

form a linear pair.

62/87,21

Arc

BD

and arc

BCD

62/87,21

By Theorem 10.13, .

Solve for .

We know that .

Substitute.

Find eachmeasure.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

m

(arc

JNM

)

62/87,21

JNM

is 205.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

A

and

B

x

= 260

,

what is

y

?

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

x

.

62/87,21

By Theorem 10.14, .

Solve for

.

62/87,21

By Theorem 10.14, .

Solve for

.

(14)

62/87,21

Ә

JMK

DQG

_Ә

HMJ

form a linear pair.

62/87,21

Arc

BD

and arc

BCD

62/87,21

By Theorem 10.13, .

Solve for .

We know that .

Substitute.

Find eachmeasure.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

m

(arc

JNM

)

62/87,21

JNM

is 205.

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

A

and

B

x

= 260

,

what is

y

?

62/87,21

By Theorem 10.14, .

Substitute.

Simplify.

x

.

62/87,21

By Theorem 10.14, .

Solve for

.

62/87,21

By Theorem 10.14, .

Solve for

.

(15)

Find

x

to the nearesttenth. Assume that segments thatappearto be tangentare tangent.

62/87,21

BRIDGES

What is the diameter of the circle

containing the arc of the Sydney Harbour Bridge? Round to the nearest tenth.

Refer to the photo on Page 740.

62/87,21

Since the height is perpendicular and bisects the chord, it must be part of the diameter. To find the diameter of the circle, first use Theorem 10.15 to find the value of x.

Add the two segments to find the diameter.

Therefore, the diameter of the circle is 528.01+ 60 or about 588.1 meters.

Find each variable to the nearesttenth. Assume that segments thatappearto be tangentare tangent.

62/87,21

Solve for x using the Quadratic Formula.

Since lengths cannot be negative, the value of x is 6.

62/87,21

First, find the length,

a

, of the tangent segment to both circles using the intersecting secant and tangent segments to the smaller circle.

Find x using the intersecting secant and tangent segments to the larger circle.

62/87,21

Use the intersecting tangent and secant segments to find q.

Since lengths cannot be negative, the value of q is 9. Use the intersecting secant segments to find r.

Solve for

r

using the Quadratic Formula.

Since lengths cannot be negative, the value of r is about 1.8.

Therefore, the values of the variables are q = 9 and r §

Find

x

62/87,21

BRIDGES

What is the diameter of the circle

Refer to the photo on Page 740.

62/87,21

a

62/87,21

Solve for

r

using the Quadratic Formula.

(16)

Find

x

62/87,21

BRIDGES

What is the diameter of the circle

Refer to the photo on Page 740.

62/87,21

a

62/87,21

Solve for

r

using the Quadratic Formula.

Find

x

62/87,21

BRIDGESWhat is the diameter of the circle containing the arc of the Sydney Harbour Bridge? Round to the nearest tenth.

Refer to the photo on Page 740.

62/87,21

a

62/87,21

Solve for

r

using the Quadratic Formula.

(17)

CCSS STRUCTURE Write the equation of each circle.

center at (6, 1), radius 7 62/87,21

Write the equation of each circle. center at (8, ±9), radius

62/87,21

center at (1, ±2), passes through (3, ±4) 62/87,21

Find the distance between the points to determine the radius.

Write the equation of the circle using

h

= 1, k =

±

2,

and

r

=

.

62/87,21

Write the equation of the circle using h = 3, k = 3,

and

r

=

.

GARDENING

A sprinkler waters a circular area

that has a diameter of 10 feet. The sprinkler is located 20 feet north of the house. If the house is located at the origin, what is the equation for the circle of area that is watered?

62/87,21

Find the equation of a circle with a center of (0, 20) and a radius of (10) or 5 feet.

Therefore, the equation for the circle of the area that

is watered is x

2

_{+ (}

_y

_{± 20)}

2

_{= 25.}

For each circle with the given equation, state the coordinates of the center and the measure of the radius. Then graph the equation.

62/87,21

The standard form of the equation of a circle with center at (

h

,

k

) and radius r is

.

Rewrite to find the center and

the radius.

So,

h

= 2,

k

= 1, and r = 2. The center is at (2, 1), and the radius is 2.

62/87,21

h

,

k

) and radius r is

.

Rewrite to find the center and the radius.

So,

h

= 8,

k

CCSS STRUCTURE Write the equation of each circle.

center at (6, 1), radius 7 62/87,21

Write the equation of each circle. center at (8,±9), radius

62/87,21

center at (1, ±2), passes through (3, ±4) 62/87,21

Write the equation of the circle using h = 1, k =

±

2,

and

r

=

.

62/87,21

Write the equation of the circle using h = 3, k = 3,

and

r

=

.

GARDENING

A sprinkler waters a circular area

that has a diameter of 10 feet. The sprinkler is located 20 feet north of the house. If the house is located at the origin, what is the equation for the circle of area that is watered?

62/87,21

Find the equation of a circle with a center of (0, 20) and a radius of (10) or 5 feet.

Therefore, the equation for the circle of the area that

is watered is x

2

+ (

y

± 20)

2

= 25.

For each circle with the given equation, state the coordinates of the center and the measure of the radius. Then graph the equation.

62/87,21

h

,

k

) and radius r is

.

Rewrite to find the center and

the radius.

So,

h

= 2,

k

= 1, and

r

= 2. The center is at (2,1), and the radius is 2.

62/87,21

h

,

k

) and radius r is

.

Rewrite to find the center and the radius.

So,

h

= 8,