1 INSCRIBED ANGLES PROBLEM 1a CONGRUENT AND INSCRIBED INSCRIBED TO A SEMICIRCLE PROBLEM 2 INSCRIBED...

1

INSCRIBED ANGLES

PROBLEM 1a PROBLEM 1a

CONGRUENT AND INSCRIBED

INSCRIBED TO A SEMICIRCLE

PROBLEM 2

INSCRIBED AND CIRCUMSCRIBED

PROBLEM 3

Standard 21

PROBLEM 4

END SHOWPRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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2

Standard 21:

Students prove and solve problems regarding relationships among chords, secants, tangents, inscribed angles, and inscribed and circumscribed polygons of circles.

Los estudiantes prueban y resuelven problemas relacionados con cuerdas, secantes, tangentes, ángulos inscritos y polígonos inscritos y circunscritos a círculos.

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


3

CB

A

ABC is an inscribed angle

LM

K

KMLm = m KL1

2

Inscribed angles are angles formed by two chords whose vertex is on the circle.

Ángulos inscritos son ángulos formados por dos cuerdas cuyo vértice esta en el circulo

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

Si un ángulo en un círculo es inscrito entonces la medida de el ángulo es igual a la mitad de su arco intersecado.

Standard 21



4

C

B

A

(3x+5)°

40° (3X+5) =1

2(40°)

=BACm m BC1

2

3X + 5 = 20-5 -5

3X = 153 3

X=5

Standard 21



5

L

JK

(2x+7)°

54°(2X+7) =

1

2(54°)

=JKLm m JL1

2

2X + 7 = 27-7 -7

2X = 202 2

X=10

Standard 21



6

D

B

C

A

P

If ADB and ACB intercept same arc AB

then ADB ACB

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

Si dos ángulos inscritos de un círculo o de círculos congruentes intersecan el mismo arco o arcos congruentes entonces los ángulos son congruentes.

Standard 21



7

A

B

CP

If ABC intercepts semicircle AC

then ABC=m 90°

If an inscribed angle intercepts a semicircle, then the angle is a right angle.

Si un ángulo inscrito interseca a un semicírculo entonces el ángulo es recto.

Standard 21



8

4X°

(6X-10)°

L

N

M

K4X° + (6X-10)° = 90°

10X-10 = 90

+10 +10

10X = 10010 10

X=10

Standard 21



9

These are concentric circles and all circles are similar.

Estos son círculos concéntricos y todos los círculos son semejantes.

Standard 21



10

BA

D C

P

L

K

N

M

Q

H G

FE

Standard 21



11

BA

D C

P

Quadrilateral ABCD is inscribed to circle P.

Cuadrilatero ABCD esta inscrito al círculo P.

Quadrilateral EFGH is circumscribed to circle Q, having sides to be TANGENT at points K, L, M and N.

Cuadrilátero EFGH esta circunscrito al círculo Q, teniendo los lados TANGENTES en los puntos K, L, M y N.

L

K

N

M

Q

H G

FE

g

Line g is TANGENT to circle X at point R.

Línea g es tangente al círculo X en punto R.

R

X

Standard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


12

C

B

A

D

E

F

H IG1. EDm = ?


Inscribed to circle G are quadrilaterals EFBD and EABC, and quadrilaterals EFGD and GABC are rhombi.

EBFm = -3X+45 EBDm = 4X+10and

Find the following:


13

C

B

A

D

E

F

H IG1. EDm = ?




Find the following:

14

C

B

A

D

E

F

H IG1. EDm = ?




Find the following:

15

and FGDE is a rhombus so all sides C

B

A

D

E

F

H IG

EB EB

1. EDm = ?

Since is inscribed to SEMICIRCLEEFB

then EFBm =

EAB

90° and then are right,

E

F

B

EFB

are congruent

therefore ; and because the

Reflexive Property then: EFB EDB

by HL. EBF EBD by CPCTC.

Then = So:EBFm EBDm

-3X+45 = 4X+10-45 -45

-3X = 4X – 35 -4X -4X

- 7X = - 35-7 -7

X=5

EBFm =-3X+45

EBFm =-3( )+455= -15+45

So:

=30°

30°

30°

EBDm = 30°

30°

and

D

E B

EDB

And

60°

60°

Take notesStandard 21PRESENTATION CREATED BY SIMON PEREZ. All rights reserved



Find the following:

EF ED

60°

60° 30°

16

C

B

A

D

E

F

H IG1. EDm = ? 30°

30°60°60°

EBDm =

m ED1

2

If

then: 30° =

(2)(2) m ED1

230° =

m ED = 60°

2. FEm =? Since then and

m ED1

2

EF ED m FE = 60°FE ED

60°

60°




Find the following:


17

C

B

A

D

E

F

H IG1. EDm = ? 30°

30°60°60°

EBDm =

m ED1

2

If

then: 30° =

(2)(2) m ED1

230° =

m ED = 60°


m ED1

2

EF ED m FE = 60°

3. BEDm = ? BEDm =From figure

4. BGDm = ?

FE ED

60°

60°

60°

60°




Find the following:


18

C

B

A

D

E

F

H IG1. EDm = ? 30°

30°60°60°

EBDm =

m ED1

2

If

then: 30° =

(2)(2) m ED1

230° =

m ED = 60°


m ED1

2

EF ED m FE = 60°

3. BEDm = ? BEDm =From figure

4. BGDm = ?

FE ED

60°

60°

60°

60°


DGEmDEGm = because EFGD is a rhombus

and then

60° + BGDm = 180°-60° -60°

BGDm = 120°

Take notes

EGDm + BGDm = 180°



Find the following:


19

C

B

A

D

E

F

H IG

60°

60°

5. DBm = 120°


120°

120°



Find the following:


20

C

B

A

D

E

F

H IG

60°

60°

5. DBm =

6. DEB =m 60°+60°+120°

= 240°

7. AIBm =

120°


120°

120°

90°



Find the following:


21

Statements Reasons

a. a.

b. b.

c. c.

d. d.

e. e.

f. f.

g. g.

C

B

D

A

Given:

AB DC

Prove:

AD BC

Given

CDBmABDm =

Alternate interior are

S

S have the same measure

h. h.

m AD1

2m BC

1

2= Transitive Property.

ADm = BCm Division Property of Equality

AD BC Arcs with the same measure are

AB DC

ABD CDB

m AD1

2ABDm =

m BC1

2CDBm =

An inscribed is half its intercepted arcAn inscribed is half its intercepted arc



1 INSCRIBED ANGLES PROBLEM 1a CONGRUENT AND INSCRIBED INSCRIBED TO A SEMICIRCLE PROBLEM 2 INSCRIBED...

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