Download Use Inscribed Angles and Polygons

10.4
Use Inscribed Angles
and Polygons
Goal
Your Notes
p Use inscribed angles of circles.
VOCABULARY
Inscribed angle
Intercepted arc
Inscribed polygon
Circumscribed circle
THEOREM 10.7: MEASURE OF AN
INSCRIBED ANGLE
THEOREM
A
D
C
The measure of an inscribed angle
is one half the measure of its
1
m∠ADB 5 }
2
intercepted arc.
B
Use inscribed angles
Example 1
Find the indicated measure in (P.
C
a. m∠S
R
b. mRQ
608
Q
378
P
T
S
Solution
1
a. m∠S 5 }
2
1
5}
(
2
)5
C
52p
5
Because C
RQS is a semicircle,
RQ 5 1808 2
5 1808 2
mC
b. mQS 5 2m∠
278 Lesson 10.4 • Geometry Notetaking Guide
.
5
.
Copyright © Holt McDougal. All rights reserved.
10.4
Use Inscribed Angles
and Polygons
Goal
Your Notes
p Use inscribed angles of circles.
VOCABULARY
Inscribed angle An inscribed angle is an angle
whose vertex is on a circle and whose sides
contain chords of the circle.
Intercepted arc The arc that lies in the interior of
an inscribed angle and has endpoints on the angle
is called the intercepted arc of the angle.
Inscribed polygon A polygon is an inscribed
polygon if all of its vertices lie on a circle.
Circumscribed circle A circumscribed circle is a
circle that contains the vertices of an inscribed
polygon.
THEOREM 10.7: MEASURE OF AN
INSCRIBED ANGLE
THEOREM
A
D
C
The measure of an inscribed angle
B
is one half the measure of its
1
m∠ADB 5 }
mAB
2
intercepted arc.
C
Use inscribed angles
Example 1
Find the indicated measure in (P.
C
a. m∠S
R
b. mRQ
Solution
1
C
608
Q
378
P
T
S
1
a. m∠S 5 }
mRT 5 }
( 608 ) 5 308
2
2
C
Because C
RQS is a semicircle,
RQ 5 1808 2 mC
QS 5 1808 2
mC
b. mQS 5 2m∠ R 5 2 p 378 5 748 .
278 Lesson 10.4 • Geometry Notetaking Guide
748 5 1068 .
Copyright © Holt McDougal. All rights reserved.
Your Notes
Example 2
C
Find the measure of an intercepted arc
Find mHJ and m∠HGJ. What do you
notice about ∠HGJ and ∠HFJ ?
G
C
1
1
5}
(
2
J
.
)5
THEOREM 10.8
.
A
D
If two inscribed angles of a circle
intercept the same arc, then the
angles are congruent.
Example 3
398
F
Solution
From Theorem 10.7, you know that
5 2(
)5
mHJ 5 2m∠
Also, m∠HGJ 5 }
2
∠HFJ.
So ∠HGJ
H
B
C
∠ADB > ∠
Use Theorem 10.8
Q
Name two pairs of congruent angles
in the figure.
P
Solution
S
Notice that ∠QRP and ∠
intercept the
same arc, and so ∠QRP > ∠
by Theorem 10.8.
Also, ∠RQS and ∠
intercept the same arc, so
∠RQS > ∠
.
R
Checkpoint Find the indicated measure.
C
1. m∠GHJ
2. mCD
H
B
1008
Copyright © Holt McDougal. All rights reserved.
R
C
408
F
G
3. m∠RTS
Q
J
D
318
S
T
Lesson 10.4 • Geometry Notetaking Guide
279
Your Notes
Example 2
C
Find the measure of an intercepted arc
Find mHJ and m∠HGJ. What do you
notice about ∠HGJ and ∠HFJ ?
G
1
C
398
F
Solution
From Theorem 10.7, you know that
mHJ 5 2m∠ HFJ 5 2( 398 ) 5 788 .
C
H
J
1
Also, m∠HGJ 5 }
mHJ 5 }
( 788 ) 5 398 .
2
2
So ∠HGJ > ∠HFJ.
THEOREM 10.8
A
D
If two inscribed angles of a circle
intercept the same arc, then the
angles are congruent.
Example 3
B
C
∠ADB > ∠ ACB
Use Theorem 10.8
Q
Name two pairs of congruent angles
in the figure.
P
Solution
S
Notice that ∠QRP and ∠ QSP intercept the
same arc, and so ∠QRP > ∠ QSP by Theorem 10.8.
Also, ∠RQS and ∠ RPS intercept the same arc, so
∠RQS > ∠ RPS .
R
Checkpoint Find the indicated measure.
C
1. m∠GHJ
2. mCD
H
B
1008
Q
J
m∠GHJ 5 508
Copyright © Holt McDougal. All rights reserved.
R
C
408
F
G
3. m∠RTS
C
D
mCD 5 808
318
S
T
m∠RTS 5 318
Lesson 10.4 • Geometry Notetaking Guide
279
Your Notes
THEOREM 10.9
If a right triangle is inscribed in a
circle, then the hypotenuse is a
diameter of the circle. Conversely,
if one side of an inscribed triangle
is a diameter of the circle, then the
triangle is a right triangle and the
angle opposite the diameter is
the right angle.
Example 4
A
D
B
C
m∠ABC 5 908 if
and only if
is a diameter of
the circle.
Use a circumscribed circle
Security A security camera rotates 908
and needs to be able to view the width
of a wall. The camera is placed in a
spot where the only thing viewed when
rotating is the wall. You want to change
the camera’s position. Where else can
it be placed so that the wall is viewed
in the same way?
Solution
From Theorem 10.9, you know that if
a right triangle is inscribed in a circle,
then the hypotenuse of the triangle is
of the circle. So, draw the
a
circle that has the width of the wall as
. The wall fits perfectly with
a
your camera’s 908 rotation from any
in front of
point on the
the wall.
THEOREM 10.10
E
A quadrilateral can be inscribed in a
circle if and only if its opposite angles
are supplementary.
D, E, F, and G lie on (C if and only if
m∠D 1 m∠F 5 m∠E 1 m∠G 5
280 Lesson 10.4 • Geometry Notetaking Guide
F
C
D
G
.
Copyright © Holt McDougal. All rights reserved.
Your Notes
THEOREM 10.9
If a right triangle is inscribed in a
circle, then the hypotenuse is a
diameter of the circle. Conversely,
if one side of an inscribed triangle
is a diameter of the circle, then the
triangle is a right triangle and the
angle opposite the diameter is
the right angle.
Example 4
A
D
B
C
m∠ABC 5 908 if
}
and only if AC
is a diameter of
the circle.
Use a circumscribed circle
Security A security camera rotates 908
and needs to be able to view the width
of a wall. The camera is placed in a
spot where the only thing viewed when
rotating is the wall. You want to change
the camera’s position. Where else can
it be placed so that the wall is viewed
in the same way?
Solution
From Theorem 10.9, you know that if
a right triangle is inscribed in a circle,
then the hypotenuse of the triangle is
a diameter of the circle. So, draw the
circle that has the width of the wall as
a diameter . The wall fits perfectly with
your camera’s 908 rotation from any
point on the semicircle in front of
the wall.
THEOREM 10.10
A quadrilateral can be inscribed in a
circle if and only if its opposite angles
are supplementary.
D, E, F, and G lie on (C if and only if
m∠D 1 m∠F 5 m∠E 1 m∠G 5 1808 .
280 Lesson 10.4 • Geometry Notetaking Guide
E
F
C
D
G
Copyright © Holt McDougal. All rights reserved.
Your Notes
Example 5
Use Theorem 10.10
Find the value of each variable.
a.
b.
Q
3y8
888
P 1008
K
J 8x 8
y8 R
4x8
L
3y8
x8
M
S
Solution
a. PQRS is inscribed in a circle, so opposite angles
.
are
m∠P 1 m∠R 5
m∠Q 1 m∠S 5
1008 1 y8 5
888 1 x8 5
y5
x5
b. JKLM is inscribed in a circle, so opposite angles
are
.
m∠ J 1 m∠L 5
m∠K 1 m∠M 5
8x8 1 4x8 5
3y8 1 3y8 5
12x 5
6y 5
x5
y5
Checkpoint Complete the following exercises.
4. A right triangle is inscribed in a circle. The radius of
the circle is 5.6 centimeters. What is the length of
the hypotenuse of the right triangle?
Homework
5. Find the values of a and b.
B
A
5a8
4b8
C
968
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D
Lesson 10.4 • Geometry Notetaking Guide
281
Your Notes
Example 5
Use Theorem 10.10
Find the value of each variable.
a.
b.
Q
3y8
888
P 1008
K
J 8x 8
y8 R
4x8
L
3y8
x8
M
S
Solution
a. PQRS is inscribed in a circle, so opposite angles
are supplementary .
m∠P 1 m∠R 5 1808
m∠Q 1 m∠S 5 1808
1008 1 y8 5 1808
888 1 x8 5 1808
y 5 80
x 5 92
b. JKLM is inscribed in a circle, so opposite angles
are supplementary .
m∠ J 1 m∠L 5 1808
m∠K 1 m∠M 5 1808
8x8 1 4x8 5 1808
3y8 1 3y8 5 1808
12x 5 180
6y 5 180
x 5 15
y 5 30
Checkpoint Complete the following exercises.
4. A right triangle is inscribed in a circle. The radius of
the circle is 5.6 centimeters. What is the length of
the hypotenuse of the right triangle?
11.2 centimeters
Homework
5. Find the values of a and b.
a 5 18, b 5 21
B
A
5a8
4b8
C
968
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D
Lesson 10.4 • Geometry Notetaking Guide
281