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Warm Up Section 3.2 Draw and label each of the following in a circle with center P.PowerPoint Presentation

Warm Up Section 3.2 Draw and label each of the following in a circle with center P.

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Draw and label each of the following in a circle with center P.

(1). Radius: (2). Diameter:

(3). Chord that is NOT a diameter:

(4). Secant:

(5). Tangent:

Draw and label each of the following in a circle with center P.

(1). Radius:

(2). Diameter:

(3). Chord that is NOT

a diameter:

(4). Secant:

(5). Tangent:

T

A

E

P

R

D

C

Section 3.2

Standard: MCC9-12.G.C.2

Essential Question: Can I understand and use properties of chords to solve problems?

Parts of circles are called Arcs.

If the part of the circle is less than half the circle it is called a minor arc. If the part of the circle is more than half the circle it is called a major arc. And if it is exactlyhalf the circle it is called a semicircle.

The circle below has center P. Draw a diameter on your circle. Label the endpoints of the diameter H and K. Put another point on your circle and label it T.

There are two semicircles pictured in your drawing. One can be symbolized HK and the other HTK .

K

T

P

H

K circle. Label the endpoints of the diameter

T

P

Name two minor arcs(you only need two letters to name a minor arc because you always travel the shortest distance unless told otherwise.) _______ , _______

Name two major arcs(you must use three letters to name a major arc because you always travel the shortest distance unless told otherwise.) _______ , _______

H

HT

TK

KHT

HKT

The circle below has center M. Mark two points on your circle and label them R and S. Now draw the

following rays: and

Color the inside of RMS.

Name the arc that is inside the colored part of the angle. This is called the intercepted arc. Arc: ________

The angle is called acentral anglebecause its vertex is the center of the circle.

S

R

M

RS

S circle and label them

R

M

Arcs are measured in degrees. The measure of the intercepted arc is equal to the measure of the central angle that forms that arc.

So, if mRMS = 60o, then m RS = 60o.

In this section you will learn to use relationships circle and label them

of arcs and chords in a circle.

In the same circle, or congruent circles, two minor

arcs are congruent if and only if their corresponding

chords are congruent.

C

B

AB DC if and only if

_____ ______.

CD

AB

D

A

1. In the diagram, A circle and label them D, , and

m EF = 125o. Find m BC.

E

B

A

D

F

C

chords

Because BC and EF are congruent ________ in

congruent _______, the corresponding minor arcs

BC and EF are __________ .

So, m ______ = m ______ = ______o.

circles

congruent

125

BC

EF

If one chord is a perpendicular bisector of another circle and label them

chord, then the first chord is a diameter.

If QS is a perpendicular bisector of TR, then ____

is a diameter of the circle.

T

S

P

Q

R

QS

If a diameter of a circle is perpendicular to a chord, circle and label them

then the diameter bisects the chord and its arc.

If EG is a diameter and EG DF, then HD HF

and ____ ____ .

F

E

H

G

D

GF

GD

2. If m TV = 121 circle and label them o, find m RS.

T

6

S

V

6

R

If the chords are congruent, then the arcs are congruent. So, m RS =

121o

3. Find the measure of CB, BE, and CE. circle and label them

C

Since BD is a diameter,

it bisects, the chord and the arcs.

4xo

A

B

D

4x = 80 – x

5x = 80

x = 16

(80 – x)o

E

4(16) = 64 so mCB = m BE = 64o

mCE = 2(64o) = 128o

In the same circle, or in congruent circles, two chords circle and label them

are congruent if and only if they are equidistant

from the center.

if and only if

_____ ____

C

GE

FE

G

E

D

A

F

B

4. In the diagram of F, AB = CD = 12. Find EF. circle and label them

Chords and are

congruent, so they are

equidistant from F.

Therefore EF = 6

G

A

B

7x – 8

F

3x

D

E

C

7x – 8 = 3x

4x = 8

x = 2

So, EF = 3x = 3(2) = 6

- In the diagram of F, suppose AB = 27 and circle and label them
- EF = GF = 7. Find CD.

Since and are

both 7 units from the

center, they are

congurent.

G

A

B

F

D

E

C

So, AB = CD = 27.

- In S, SP = 5, MP = 8, ST = SU, circle and label them and
- NRQ is a right angle. Show that PTS NRQ.

N

Step 1: Look at PTS !!

Since MP = 8, and NQ bisects

MP, we know MT= PT = 4.

Since , PTN is

a right angle.

4

T

4

M

P

U

3

5

S

R

Q

Use the Pythagorean

Theorem to find TS.

42 + TS2 = 52

TS2 = 25 – 16

TS2 = 9 So, TS = 3.

- In F, SP = 5, MP = 8, ST = SU, circle and label them and
- NRQ is a right angle. Show that PTS NRQ.

N

Step 2: Now, look at NRQ!

Since the radius of the circle

is 5, QN = 10.

Since ST = SU, MP and RN

are equidistantfrom the center.

Hence, MP = RN = 8.

8

T

M

P

U

10

S

R

6

Q

Use the Pythagorean

Theorem to find RQ.

RQ2 + 82 = 102

RQ2 = 100 – 64

RQ2 = 36 So, RQ = 6.

- In F, SP = 5, MP = 8, ST = SU, circle and label them and
- NRQ is a right angle. Show that PTS NRQ.

N

Step 3: Identify ratios of

corresponding sides. In PTS,

PT = 4, TS = 3, and PS = 5.

In NRQ, NR = 8, RQ = 6,

and QN = 10.

Find the corresponding ratios:

T

M

P

U

S

R

Q

Because the corresponding sides lengths are proportional, circle and label them

(all have a ratio of ½), PTS NRQ by SSS Thrm.

7. In S, QN = 26, NR = 24, ST = SU, circle and label them and

NRQ is a right angle. Show that PTS NRQ.

N

N

24

T

M

P

U

26

S

R

R

10

Q

Q

242 + RQ2 = 262

RQ2 = 676 – 576

RQ2 = 100

So, RQ = 10.

7. In S, QN = 26, NR = 24, ST = SU, circle and label them and

NRQ is a right angle. Show that PTS NRQ.

N

12

T

P

5

T

M

P

13

U

S

S

Cords equidistant from center

MP = RN = 24. So, PT = ½(24).

SP is half of the diameter QN,

so SP = ½(26) = 13.

R

Q

122 + ST2 = 132

ST2 = 169 –144

ST2 = 25 So, ST = 5.

Step 3: Identify ratios of circle and label them

corresponding sides. In PTS,

PT = 12, TS = 5, and PS = 13,

In NRQ, NR = 24, RQ = 10,

and QN = 26.

Find the corresponding ratios:

N

T

M

P

U

S

R

Q

Because the corresponding sides lengths are proportional

(all have a ratio of ½), PTS NRQ by SSS Thrm.

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