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# INSCRIBED ANGLES PowerPoint PPT Presentation

INSCRIBED ANGLES. PROBLEM 1a. PROBLEM 1a. CONGRUENT AND INSCRIBED. INSCRIBED TO A SEMICIRCLE. PROBLEM 2. INSCRIBED AND CIRCUMSCRIBED. PROBLEM 3. PROBLEM 4. END SHOW. Standard 21:

INSCRIBED ANGLES

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INSCRIBED ANGLES

PROBLEM 1a

PROBLEM 1a

CONGRUENT AND INSCRIBED

INSCRIBED TO A SEMICIRCLE

PROBLEM 2

INSCRIBED AND CIRCUMSCRIBED

PROBLEM 3

PROBLEM 4

END SHOW

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.

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

A

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

B

C

ABC

KML

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

K

m

KL

1

2

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

M

L

m

is an inscribed angle

=

=

(40°)

BAC

3 3

m

BC

1

1

2

2

m

A

(3x+5)°

B

40°

(3X+5) =

C

3X + 5 = 20

-5 -5

3X = 15

X=5

K

J

(2x+7)°

=

54°

(54°)

L

JKL

2 2

m

JL

1

1

2

2

m

(2X+7) =

2X + 7 = 27

-7 -7

2X = 20

X=10

A

AB

B

P

D

C

ACB

ACB

and

intercept same arc

If

then

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.

AC

P

A

90°

m

C

B

ABC

ABC=

intercepts semicircle

If

then

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.

N

=

90°

4X°

K

L

M

(6X-10)°

10 10

+

(6X-10)°

4X°

10X-10 = 90

+10 +10

10X = 100

X=10

These are concentric circles and all circles are similar.

Estos son círculos concéntricos y todos los círculos son semejantes.

K

N

L

M

A

B

P

D

C

E

F

Q

G

H

A

B

P

D

C

E

F

X

Q

K

G

H

R

N

g

L

M

Quadrilateral ABCD is inscribed to circle P.

Cuadrilatero ABCD esta inscrito al círculo P.

Line g is TANGENT to circle X at point R.

Línea g es tangente al círculo X en punto R.

Quadrilateral EFGH is circumscribed tocircle 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.

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

A

F

4X+10

and

=

m

-3X+45

=

m

Find the following:

=

?

m

1.

ED

B

E

I

H

G

C

D

EBF

EBD

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

A

F

4X+10

and

=

m

-3X+45

=

m

Find the following:

=

?

m

1.

ED

B

E

I

H

G

C

D

EBF

EBD

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

A

F

4X+10

and

=

m

-3X+45

=

m

Find the following:

=

?

m

1.

ED

B

E

I

H

G

C

D

EBF

EBD

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

A

F

=

4X+10

and

=

m

-3X+45

=

m

Find the following:

60°

=

?

m

30°

1.

ED

B

E

I

H

30°

G

60°

90°

EFB

EDB

C

and FGDE is a rhombus so all sides

D

are congruent

because the

F

Reflexive Property

EFB

EBD

EFB

EBF

EBD

EBF

EBF

EBF

EBD

EBD

EBF

EDB

EFB

E

B

E

B

= 30°

=-3X+45

m

m

m

m

m

m

=-3( )+45

EAB

D

EB EB

EF ED

-7 -7

Since

is inscribed to SEMICIRCLE

then

are right,

and then

and

; and

therefore

then:

by HL.

And

by CPCTC.

30°

60°

So:

Then

=

-3X+45 = 4X+10

60°

30°

-45 -45

5

-3X = 4X – 35

= -15+45

-4X -4X

=30°

- 7X = - 35

So:

Take notes

X=5

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

A

F

4X+10

and

=

m

-3X+45

=

m

60°

Find the following:

B

E

I

H

G

C

D

(2)

(2)

30° =

EBD

EBD

EBF

=

m

?

2.

FE

m

m

m

m

m

ED

ED

FE

ED

ED

1

1

1

2

2

2

FE ED

m

EFED

=

?

m

30°

1.

ED

60°

If

=

60°

30°

60°

then:

30° =

= 60°

= 60°

Since

then

and

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

A

F

=

=

=

4X+10

and

=

m

-3X+45

=

m

Find the following:

B

E

I

60°

H

G

C

D

(2)

(2)

30° =

EBD

EBF

BGD

BED

EBD

BED

=

m

?

2.

FE

m

m

m

m

m

FE

ED

ED

ED

ED

1

1

1

4.

3.

?

?

60°

2

2

2

FE ED

m

m

m

m

EFED

60°

=

?

m

30°

1.

ED

60°

If

=

60°

30°

60°

then:

30° =

= 60°

= 60°

Since

then

and

From figure

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

A

F

=

=

=

=

=

=

4X+10

and

=

m

-3X+45

=

m

Find the following:

B

E

I

60°

H

G

C

D

(2)

(2)

30° =

DEG

BGD

EGD

BGD

BGD

BED

EBD

BED

EBF

EBD

BGD

=

m

?

2.

FE

m

m

m

m

m

ED

ED

ED

FE

ED

1

1

1

3.

4.

?

?

60°

2

2

2

FE ED

m

m

m

m

m

m

=

m

DGE

because EFGD is a rhombus

m

180°

+

m

m

EFED

180°

60° +

120°

60°

=

?

m

30°

1.

ED

60°

If

=

60°

30°

60°

then:

30° =

= 60°

= 60°

Since

then

and

From figure

and then

Take notes

-60° -60°

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

A

F

4X+10

and

=

m

-3X+45

=

m

Find the following:

B

E

I

H

G

m

5.

DB

=

120°

C

D

EBD

EBF

60°

120°

60°

120°

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

A

F

120°

4X+10

and

=

m

-3X+45

=

m

Find the following:

B

E

I

H

G

m

5.

DB

=

C

D

90°

7.

=

AIB

EBD

EBF

m

=

6.

m

DEB

60°

120°

60°

60°+60°+120°

120°

= 240°

Reasons

Statements

=

a.

a.

Alternate interior are

b.

b.

have the same measure

c.

c.

S

d.

d.

=

e.

e.

=

f.

f.

An inscribed is half its intercepted arc

ABD

CDB

CDB

ABD

ABD

CDB

DC

DC

AB

AB

=

g.

h.

g.

h.

=

m

m

BC

m

m

m

m

BC

BC

1

1

1

1

Arcs with the same measure are

2

2

2

2

m

m

An inscribed is half its intercepted arc

S

m

m

B

C

Given

A

D

Given:

Prove:

Transitive Property.

Division Property of Equality