of the angle equals one half the measure of its
intercepted arc. Therefore,
ANSWER: 126
10-4 Inscribed Angles
Find each measure.
1. 3. SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
SOLUTION: Here,
is a semi-circle. So,
intercepted arc. Therefore,
If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
ANSWER: 30
intercepted arc. So,
Therefore,
2. ANSWER: 66
4. SCIENCE The diagram shows how light bends in a
raindrop to make the colors of the rainbow. If
, what is
?
SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. Therefore,
ANSWER: 126
SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
3. intercepted arc. Therefore,
ANSWER: 72
ALGEBRA Find each measure.
SOLUTION: Here,
5. is a semi-circle. So,
If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. So,
Therefore,
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ANSWER: 66
SOLUTION: Page 1
If two inscribed angles of a circle intercept the same
arc or congruent arcs, then the angles are congruent.
So,
intercepted arc. Therefore,
Therefore,
10-4ANSWER: Inscribed Angles
72
ANSWER: 36
ALGEBRA Find each measure.
5. 7. PROOF Write a two-column proof.
Given:
bisects
.
Prove:
SOLUTION: If two inscribed angles of a circle intercept the same
arc or congruent arcs, then the angles are congruent.
So,
SOLUTION: Given:
Prove:
bisects
.
Therefore,
ANSWER: 54
Proof:
Statements (Reasons)
1.
bisects
. (Given)
2.
(Def. of segment bisector)
6. 3.
intercepts .
(Def. of intercepted arc)
4.
(Inscribed
.)
5.
(Vertical 6.
(AAS)
SOLUTION: If two inscribed angles of a circle intercept the same
arc or congruent arcs, then the angles are congruent.
So,
intercepts .
of same arc are
are
.)
ANSWER: Given:
Prove:
bisects
.
Therefore,
ANSWER: 36
Proof:
Statements (Reasons)
1.
bisects
. (Given)
2.
(Def. of segment bisector)
7. PROOF Write a two-column proof.
Given:
bisects
.
Prove:
3.
intercepts .
(Def. of intercepted arc)
4.
(Inscribed
.)
5.
(Vertical 6.
(AAS)
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SOLUTION: Given:
Prove:
bisects
.
intercepts of same arc are
are
.)
CCSS STRUCTURE Find each value.
8. .
Page 2
(Def. of intercepted arc)
4.
(Inscribed
.)
10-45.Inscribed Angles (Vertical 6.
(AAS)
is 180. So,
of same arc are
are
ANSWER: 25
.)
CCSS STRUCTURE Find each value.
10. and
8. SOLUTION: If a quadrilateral is inscribed in a circle, then its
opposite angles are supplementary.
SOLUTION: 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
Solve the two equations to find the values of x and y.
is 180. So,
Use the values of the variables to find
.
and
Therefore,
ANSWER: 62
ANSWER: 122; 80
9. x
ALGEBRA Find each measure. 11. SOLUTION: An inscribed angle of a triangle intercepts a diameter
or semicircle if and only if the angle is a right angle.
So,
Since The sum of the measures of the angles of a triangle
SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. Therefore,
ANSWER: 162
is 180. So,
ANSWER: 25
10. 12. and
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SOLUTION: SOLUTION: Page 3
If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. Therefore,
intercepted arc. Therefore,
ANSWER: 10-4162
Inscribed Angles
12. ANSWER: 70
14. SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. Therefore,
intercepted arc. Therefore,
ANSWER: 48
ANSWER: 46
15. 13. SOLUTION: Here,
SOLUTION: The sum of the measures of the central angles of a
circle with no interior points in common is 360. So,
The arc
is a semicircle. So, If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. So,
Therefore,
If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
ANSWER: 140
intercepted arc. Therefore,
ANSWER: 70
16. 14. SOLUTION: Here,
SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
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intercepted
arc. Therefore,
ANSWER: The arc
is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its Page 4
intercepted arc. Therefore,
intercepted arc. So,
arc or congruent arcs, then the angles are congruent.
So,
Therefore,
10-4ANSWER: Inscribed Angles
140
16. ANSWER: 32
18. SOLUTION: If two inscribed angles of a circle intercept the same
arc or congruent arcs, then the angles are congruent.
So,
SOLUTION: Here,
The arc
is a semicircle. So, Then, If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
Therefore,
ANSWER: 34
intercepted arc. Therefore,
ANSWER: 66
19. 17. SOLUTION: If two inscribed angles of a circle intercept the same
arc or congruent arcs, then the angles are congruent.
Since ∠A and ∠B both intercept arc CD , m∠A =
m∠B.
SOLUTION: If two inscribed angles of a circle intercept the same
arc or congruent arcs, then the angles are congruent.
So,
ANSWER: 32
Therefore, m∠A = 5(4) or 20.
ANSWER: 20
18. 20. SOLUTION: If two inscribed angles of a circle intercept the same
arc or congruent arcs, then the angles are congruent.
So,
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Therefore,
SOLUTION: If two inscribed angles of a circle intercept the same
Page 5
arc or congruent arcs, then the angles are congruent.
Since ∠C and ∠D both intercept arc AB, m ∠C =
m∠D. Therefore, m∠A = 5(4) or 20.
Therefore, m∠C = 5(10) – 3 or 47.
10-4ANSWER: Inscribed Angles
20
ANSWER: 47
PROOF Write the specified type of proof.
21. paragraph proof
20. Given:
Prove:
SOLUTION: If two inscribed angles of a circle intercept the same
arc or congruent arcs, then the angles are congruent.
Since ∠C and ∠D both intercept arc AB, m ∠C =
m∠D. SOLUTION: Proof: Given
means that
. Since Therefore, m∠C = 5(10) – 3 or 47.
ANSWER: 47
and , the equation becomes . Multiplying each side of the
PROOF Write the specified type of proof.
21. paragraph proof
equation by 2 results in
.
Given:
ANSWER: Prove:
Proof: Given
means that
. Since and , the equation becomes . Multiplying each side of the
SOLUTION: Proof: Given
equation by 2 results in
means that
. Since and , the equation becomes .
22. two-column proof
Given:
Prove:
. Multiplying each side of the
equation by 2 results in
.
ANSWER: Proof: Given
means that
. Since and eSolutions Manual - Powered by Cognero
, the equation becomes SOLUTION: Statements (Reasons):
1.
(Given)
2.
(Inscribed intercepting same arc Page 6
are .)
3.
(Vertical
are .)
. Multiplying each side of the
10-4equation
Inscribed
by 2Angles
results in
ANSWER: 30
.
22. two-column proof
Given:
Prove:
24. SOLUTION: 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,
SOLUTION: Statements (Reasons):
1.
(Given)
2.
(Inscribed intercepting same arc are .)
3.
(Vertical
are .)
4.
(AA Similarity)
Therefore,
ANSWER: 60
25. x
SOLUTION: 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
ANSWER: Statements (Reasons):
1.
(Given)
2.
(Inscribed intercepting same arc are .)
3.
(Vertical
are .)
4.
(AA Similarity)
is 180. So,
ANSWER: 12.75
ALGEBRA Find each value.
26. SOLUTION: 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
23. x
SOLUTION: 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,
is 180. So,
ANSWER: 51.75
Therefore,
CCSS STRUCTURE Find each measure.
ANSWER: 30
24. SOLUTION: An inscribed angle of a triangle intercepts a diameter
or semicircle if and only if the angle is a right angle.
So,
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Page 7
27. SOLUTION: Therefore,
Therefore,
10-4ANSWER: Inscribed Angles
51.75
ANSWER: 80
CCSS STRUCTURE Find each measure.
CCSS STRUCTURE Find each measure.
27. SOLUTION: If a quadrilateral is inscribed in a circle, then its
opposite angles are supplementary.
29. SOLUTION: If a quadrilateral is inscribed in a circle, then its
opposite angles are supplementary.
Therefore,
ANSWER: 135
Therefore,
28. SOLUTION: If a quadrilateral is inscribed in a circle, then its
opposite angles are supplementary.
Therefore,
ANSWER: 80
ANSWER: 106
30. SOLUTION: If a quadrilateral is inscribed in a circle, then its
opposite angles are supplementary.
Therefore,
CCSS STRUCTURE Find each measure.
ANSWER: 93
31. PROOF Write a paragraph proof for Theorem 10.9.
SOLUTION: Given: Quadrilateral ABCD is inscribed in
Prove:
and are supplementary.
and
are supplementary.
29. .
SOLUTION: If a quadrilateral is inscribed in a circle, then its
opposite angles are supplementary.
Therefore,
ANSWER: 106 Manual - Powered by Cognero
eSolutions
30. Proof: By arc addition and the definitions of arc
measure and the sum of central angles,
. Since by Theorem 10.6 Page 8
and angles of a quadrilateral is 360,
. But
, so
them supplementary also.
10-4 Inscribed Angles
Proof: By arc addition and the definitions of arc
measure and the sum of central angles,
SIGNS: A stop sign in the shape of a regular
octagon is inscribed in a circle. Find each
. Since by Theorem 10.6
, making
and , but
, so
or
180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior
angles of a quadrilateral is 360,
. But
, so
, making
them supplementary also.
ANSWER: Given: Quadrilateral ABCD is inscribed in
Prove:
and are supplementary.
and
are supplementary.
measure.
32. SOLUTION: Since all the sides of the stop sign are congruent
chords of the circle, the corresponding arcs are
congruent. There are eight adjacent arcs that make
up the circle, so their sum is 360. Thus, the measure
.
of each arc joining consecutive vertices is
or 45.
of 360
Therefore, measure of arc NPQ is 135.
ANSWER: 135
Proof: By arc addition and the definitions of arc
measure and the sum of central angles,
33. SOLUTION: Since all the sides of the stop sign are congruent
chords of the circle, all the corresponding arcs are
congruent. There are eight adjacent arcs that make up the circle, so their sum is 360. Thus, the measure
. Since by Theorem 10.6
and , but
of each arc joining consecutive vertices is
, so
of 360
or 45. Since ∠RLQ is inscribed in the circle, its
measure equals one half the measure of its
intercepted arc QR.
or 180. This makes ∠C and ∠A supplementary. Because the sum of the measures of the interior
angles of a quadrilateral is 360,
. But
, so
, making
them supplementary also.
Therefore, the measure of ∠RLQ is 22.5.
SIGNS: A stop sign in the shape of a regular
octagon is inscribed in a circle. Find each
ANSWER: 22.5
34. eSolutions Manual - Powered by Cognero
measure.
32. SOLUTION: Page 9
If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
Therefore, the measure of ∠RLQ is 22.5.
10-4ANSWER: Inscribed Angles
22.5
b.
34. SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. So,
c.
The sum of the measures of the central angles of a
circle with no interior points in common is 360. Here,
360 is divided into 8 equal arcs and each arc
measures
Then,
d.
Therefore,
ANSWER: 112.5
35. SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. So,
The sum of the measures of the central angles of a
circle with no interior points in common is 360. Here,
360 is divided into 8 equal arcs and each arc
SOLUTION: a. The sum of the measures of the central angles of
a circle with no interior points in common is 360.
Here, 360 is divided into 5 equal arcs and each arc
measures
If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. So, the measure of each inscribed
angle is 36.
b. Here, 360 is divided into 6 equal arcs and each arc
measures
measures
Then,
Therefore,
ANSWER: 135
36. ART Four different string art star patterns are
shown. If all of the inscribed angles of each star
shown are congruent, find the measure of each
inscribed angle.
a.
by intercepting alternate vertices of the star. So, each
intercepted arc measures 120. If an angle is inscribed
in a circle, then the measure of the angle equals one
half the measure of its intercepted arc. So, the
measure of each inscribed angle is 60.
c. Here, 360 is divided into 7 equal arcs and each arc
measures
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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. So, the measure
of each inscribed angle is about 25.7.
d. The sum of the measures of the central angles of
a circle with no interior points in common is 360.
Here, 360 is divided into 8 equal arcs and each arc
measures
b.
Each inscribed angle is formed
Each inscribed angle is formed by intercepting alternate vertices of the star. So, each
intercepted arc measures 90. If an angle is inscribed
in a circle, then the measure of the angle equals one
half the measure of its intercepted arc. So, the Page 10
measure of each inscribed angle is 45.
a circle with no interior points in common is 360.
Here, 360 is divided into 8 equal arcs and each arc
measures
Each inscribed angle is formed 10-4 Inscribed Angles
by intercepting alternate vertices of the star. So, each
intercepted arc measures 90. If an angle is inscribed
in a circle, then the measure of the angle equals one
half the measure of its intercepted arc. So, the
measure of each inscribed angle is 45.
c.
or about 25.7
d. 45
PROOF Write a two-column proof for each case
of Theorem 10.6.
37. Case 2
Given: P lies inside
.
is a diameter.
Prove:
ANSWER: a. 36
b. 60
c.
or about 25.7
d. 45
PROOF Write a two-column proof for each case
of Theorem 10.6.
37. Case 2
Given: P lies inside
.
is a diameter.
Prove:
SOLUTION: Proof:
Statements (Reasons)
1. m∠ABC = m∠ABD + m∠DBC (
Postulate)
2. m∠ABD = m(arc AD) Addition
m∠DBC = m(arc DC) (The measure of an SOLUTION: Proof:
Statements (Reasons)
1. m∠ABC = m∠ABD + m∠DBC (
Postulate)
2. m∠ABD = m(arc AD) Addition
m∠DBC = m(arc DC) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)).
3. m∠ABC = m(arc AD) + m(arc DC)
(Substitution)
4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)
5. m(arc AD) + m(arc DC) = m(arc AC) (Arc
Addition Postulate)
6. m∠ABC = m(arc AC) (Substitution)
ANSWER: Proof:
Statements (Reasons)
1. m∠ABC = m∠ABD + m∠DBC (
Postulate)
2. m∠ABD = m(arc AD) Addition
inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)).
3. m∠ABC = m(arc AD) + m(arc DC)
(Substitution)
4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)
5. m(arc AD) + m(arc DC) = m(arc AC) (Arc
Addition Postulate)
6. m∠ABC = m(arc AC) (Substitution)
ANSWER: Proof:
Statements (Reasons)
1. m∠ABC = m∠ABD + m∠DBC (
Postulate)
2. m∠ABD = m(arc AD) m∠DBC = m(arc DC) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)).
3. m∠ABC = m(arc AD) + m(arc DC)
(Substitution)
4. m∠ABC = [m(arc AD) + m(arc DC)] (Factor)
5. m(arc AD) + m(arc DC) = m(arc AC) (Arc
Addition Postulate)
6. m∠ABC = m(arc AC) (Substitution)
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m∠DBC = m(arc DC) (The measure of an inscribed
whose side is a diameter is half the Addition
Page 11
38. Case 3
Given: P lies outside
.
is a diameter.
∠
5. m(arc AD) + m(arc DC) = m(arc AC) (Arc
Postulate)
10-4Addition
Inscribed
Angles
6. m∠ABC = m(arc AC) (Substitution)
38. Case 3
Given: P lies outside
.
is a diameter.
Prove:
measure of the intercepted arc (Case 1)).
3. m∠ABC = m(arc DC) – m(arc DA)
(Substitution)
4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)
5. m(arc DA) + m(arc AC) = m(arc DC) (Arc
Addition Postulate)
6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction
Property of Equality)
7. m∠ABC = m(arc AC) (Substitution)
PROOF Write the specified proof for each
theorem.
39. Theorem 10.7, two-column proof
SOLUTION: SOLUTION: Proof:
Statements (Reasons)
1. m∠ABC = m∠DBC – m∠DBA ( Addition
Postulate, Subtraction Property of Equality)
2. m∠DBC = m(arc DC) and Given:
Prove:
Proof:
are inscribed; m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)).
3. m∠ABC = m(arc DC) – m(arc DA)
(Substitution)
4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)
5. m(arc DA) + m(arc AC) = m(arc DC) (Arc
Addition Postulate)
6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction
Property of Equality)
7. m∠ABC = m(arc AC) (Substitution)
ANSWER: Proof:
Statements (Reasons)
1. m∠ABC = m∠DBC – m∠DBA ( Addition
Postulate, Subtraction Property of Equality)
2. m∠DBC = m(arc DC) Statements (Reasons)
1.
and
are inscribed;
(Given)
2.
(Measure
;
of an inscribed
3.
= half measure of intercepted arc.)
(Def. of
arcs)
(Mult. Prop. of Equality)
4.
5.
6.
(Substitution)
(Def. of )
ANSWER: Given:
Prove:
Proof:
and are inscribed; m∠DBA = m(arc DA) (The measure of an inscribed whose side is a diameter is half the measure of the intercepted arc (Case 1)).
3. m∠ABC = m(arc DC) – m(arc DA)
(Substitution)
4. m∠ABC = [m(arc DC) – m(arc DA)] (Factor)
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5. m(arc DA) + m(arc AC) = m(arc DC) (Arc
Addition Postulate)
6. m(arc AC) = m(arc DC) – m(arc DA) (Subtraction
Statements (Reasons)
1.
and
are inscribed;
(Given)
Page 12
is a right angle, m
= 90. Then
m
= 2(90) or 180. So by definition,
semicircle.
10-4 Inscribed Angles
is a
ANSWER: Statements (Reasons)
1.
and
are inscribed;
(Given)
2.
(Measure
;
of an inscribed
= half measure of intercepted arc.)
3.
(Def. of
arcs)
(Mult. Prop. of Equality)
4.
5.
6.
Part I: Given:
is a semicircle.
Prove:
is a right angle.
(Substitution)
(Def. of )
Proof: Since 40. Theorem 10.8, paragraph proof
is a semicircle, then or 90. So, by definition, then
SOLUTION: is an inscribed angle,
. Since is a right angle.
Part II: Given:
Prove:
is a right angle.
is a semicircle.
Proof: Since is an inscribed angle, then
and by the Multiplication
Part I: Given:
is a semicircle.
Prove:
is a right angle.
Property of Equality, m
Proof: Since = 2(90) or 180. So by definition,
m
semicircle.
is a right angle, m
is a semicircle, then is an inscribed angle,
. Since or 90. So, by definition, then
is a right angle.
Part II: Given:
Prove:
is a right angle.
is a semicircle.
Proof: Since is an inscribed angle, then
and by the Multiplication
Property of Equality, m
is a right angle, m
. Because
= 90. Then
= 2m
m
= 2(90) or 180. So by definition,
semicircle.
ANSWER: eSolutions Manual - Powered by Cognero
is a
= 2m
. Because
= 90. Then
is a
41. MULTIPLE REPRESENTATIONS In this
problem, you will investigate the relationship between
the arcs of a circle that are cut by two parallel
chords.
a. GEOMETRIC Use a compass to draw a circle
with parallel chords
and
. Connect points A
and D by drawing segment
.
b. NUMERICAL Use a protractor to find
and
. Then determine
.
and What is true about these arcs? Explain.
c. VERBAL Draw another circle and repeat parts a
and b. Make a conjecture about arcs of a circle that
are cut by two parallel chords.
d. ANALYTICAL Use your conjecture to find
and in the figure at the right. Verify by using inscribed angles to find the measures of the
arcs.
Page 13
are cut by two parallel chords.
d. ANALYTICAL Use your conjecture to find
and Angles
in the figure at the right. Verify by 10-4 Inscribed
using inscribed angles to find the measures of the
arcs.
b. Sample answer:
,
;
,
; The arcs are congruent
because they have equal measures.
c. Sample answer: In a circle, two parallel chords cut
congruent arcs. See students’ work.
d. 70; 70
SOLUTION: a.
CCSS ARGUMENTS Determine whether the
quadrilateral can always, sometimes, or never
be inscribed in a circle. Explain your reasoning.
42. square
b. Sample answer:
,
;
,
; The arcs are congruent
because they have equal measures.
c. Sample answer: SOLUTION: Squares have right angles at each vertex, therefore
each pair of opposite angles will be supplementary
and inscribed in a circle. Therefore, the statement is
always true.
ANSWER: Always; squares have right angles at each vertex,
therefore each pair of opposite angles will be
supplementary and inscribed in a circle.
43. rectangle
m∠A = 15, m∠D = 15, = 30.
= 30, and The arcs are congruent.
In a circle, two parallel chords cut congruent arcs.
d. In a circle, two parallel chords cut congruent arcs.
So, =
.
SOLUTION: Rectangles have right angles at each vertex,
therefore each pair of opposite angles will be
supplementary and inscribed in a circle. Therefore,
the statement is always true.
ANSWER: Always; rectangles have right angles at each vertex,
therefore each pair of opposite angles will be
supplementary and inscribed in a circle.
44. parallelogram
= 4(16) + 6 or 70
= 6(16) – 26 or 70
Therefore, the measure of arcs QS and PR are each
70.
ANSWER: a.
SOLUTION: The opposite angles of a parallelogram are always
congruent. They will only be supplementary when
they are right angles. So, a parallelogram can be
inscribed in a circle as long as it is a rectangle.
Therefore, the statement is sometimes true.
ANSWER: Sometimes; a parallelogram can be inscribed in a
circle as long as it is a rectangle.
45. rhombus
b. Sample answer:
,
; The arcs
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;
are congruent
because they have equal measures.
c. Sample answer: In a circle, two parallel chords cut
SOLUTION: The opposite angles of a rhombus are always
congruent. They will only be supplementary when
they are right angles. So, a rhombus can be inscribed
Page 14
in a circle as long as it is a square. Since the opposite
angles of rhombi that are not squares are not
supplementary, they can not be inscribed in a circle.
Therefore, the statement is sometimes true.
ANSWER: a parallelogram can be inscribed in a
10-4Sometimes;
Inscribed Angles
circle as long as it is a rectangle.
45. rhombus
SOLUTION: The opposite angles of a rhombus are always
congruent. They will only be supplementary when
they are right angles. So, a rhombus can be inscribed
in a circle as long as it is a square. Since the opposite
angles of rhombi that are not squares are not
supplementary, they can not be inscribed in a circle.
Therefore, the statement is sometimes true.
ANSWER: Sometimes; a rhombus can be inscribed in a circle as
long as it is a square. Since the opposite angles of
rhombi that are not squares are not supplementary,
they can not be inscribed in a circle.
Therefore, the statement is sometimes true.
ANSWER: Sometimes; as long as the angles that compose the
pair of congruent opposite angles are right angles.
47. CHALLENGE A square is inscribed in a circle.
What is the ratio of the area of the circle to the area
of the square?
SOLUTION: A square with side s is inscribed in a circle with
radius r. Using the Pythagorean Theorem, 46. kite
SOLUTION: Exactly one pair of opposite angles of a kite are
congruent. To be supplementary, they must each be
a right angle. If one pair of opposite angles for a
quadrilateral is supplementary, the other pair must
also be supplementary. So, as long as the angles that
compose the pair of congruent opposite angles are
right angles, a kite can be inscribed in a circle.
Therefore, the statement is sometimes true.
ANSWER: Sometimes; as long as the angles that compose the
pair of congruent opposite angles are right angles.
47. CHALLENGE A square is inscribed in a circle.
What is the ratio of the area of the circle to the area
of the square?
SOLUTION: A square with side s is inscribed in a circle with
radius r. 2
The area of a square of side s is A = s and the area
of a circle of radius r is A =
.
Therefore, the ratio of the area of the circle to the
area of the inscribed square is
ANSWER: 48. WRITING IN MATH A
right triangle is inscribed in a circle. If the radius of the
circle is given, explain how to find the lengths of the
right triangle’s legs.
SOLUTION: Sample answer: A
triangle will have two inscribed angles of 45 and one of 90. The
hypotenuse is across from the 90, or right angle.
According to theorem 10.8, an inscribed angle of a
triangle intercepts a diameter if the angle is a right
angle. Therefore, the hypotenuse is a diameter and
has a length of 2r. Use trigonometry to find the
length of the equal legs.
Using the Pythagorean Theorem, 2
The area of a square of side s is A = s and the area
of a circle of radius r is A =
.
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area of the inscribed square is
SOLUTION: See students’ work.
ANSWER: ANSWER: See students’ work.
10-4 Inscribed Angles
48. WRITING IN MATH A
right triangle is inscribed in a circle. If the radius of the
circle is given, explain how to find the lengths of the
right triangle’s legs.
SOLUTION: Sample answer: A
triangle will have two inscribed angles of 45 and one of 90. The
hypotenuse is across from the 90, or right angle.
According to theorem 10.8, an inscribed angle of a
triangle intercepts a diameter if the angle is a right
angle. Therefore, the hypotenuse is a diameter and
has a length of 2r. Use trigonometry to find the
length of the equal legs.
Therefore, the length of each leg of the 45-45-90
triangle can be found by multiplying the radius of the
circle in which it is inscribed by .
50. WRITING IN MATH Compare and contrast
inscribed angles and central angles of a circle. If they
intercept the same arc, how are they related?
SOLUTION: An inscribed angle has its vertex on the circle. A
central angle has its vertex at the center of the circle.
If a central angle intercepts arc AB, then the
measure of the central angle is equal to m(arc AB). If
an inscribed angle also intercepts arc AB, then the
measure of the inscribed angle is equal to m(arc
AB). So, if an inscribed angle and a central angle
intercept the same arc, then the measure of the
inscribed angle is one-half the measure of the central
angle.
ANSWER: An inscribed angle has its vertex on the circle. A
central angle has its vertex at the center of the circle.
If an inscribed angle and a central angle intercept the
same arc, then the measure of the inscribed angle is
one-half the measure of the central angle.
51. In the circle below,
What is
?
and .
ANSWER: Sample answer: According to theorem 10.8, an
inscribed angle of a triangle intercepts a diameter if
the angle is a right angle. Therefore, the hypotenuse
is a diameter and has a length of 2r. Using
trigonometry, each leg = sin
.
or 49. OPEN ENDED Find and sketch a real-world logo
with an inscribed polygon.
SOLUTION: See students’ work.
ANSWER: See students’ work.
50. WRITING IN MATH Compare and contrast
inscribed angles and central angles of a circle. If they
intercept the same arc, how are they related?
SOLUTION: An inscribed angle has its vertex on the circle. A
central angle has its vertex at the center of the circle.
If a central angle intercepts arc AB, then the
measure of the central angle is equal to m(arc AB). If
an inscribed angle also intercepts arc AB, then the
measure of the inscribed angle is equal to m(arc
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AB). So, if an inscribed angle and a central angle
intercept the same arc, then the measure of the
inscribed angle is one-half the measure of the central
A 42
B 61
C 80
D 84
SOLUTION: If an angle is inscribed in a circle, then the measure
of the angle equals one half the measure of its
intercepted arc. Therefore,
But,
Therefore,
The correct choice is A.
ANSWER: A
52. ALGEBRA Simplify
2
4(3x – 2)(2x + 4) + 3x + 5x – 6.
2
F 9x + 3x – 14
G 9x2 + 13x – 14
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Therefore, the circumference of the circle is
Therefore,
The correct choice is A.
ANSWER: d = 17 in., r = 8.5 in .,
10-4ANSWER: Inscribed Angles
A
52. ALGEBRA Simplify
2
4(3x – 2)(2x + 4) + 3x + 5x – 6.
2
F 9x + 3x – 14
G 9x2 + 13x – 14
2
H 27x + 37x – 38
J 27x2 + 27x – 26
or about 53.4 in.
54. SAT/ACT The sum of three consecutive integers is
–48. What is the least of the three integers?
A –15
B –16
C –17
D –18
E –19
SOLUTION: Use the Distributive Property to simplify the first
term.
SOLUTION: Let the three consecutive integers be x , x +1,and x +
2. Then,
Therefore, the correct choice is H.
So, the three integers are –17, –16, and –15 and the
least of these is –17.
Therefore, the correct choice is C.
ANSWER: H
ANSWER: C
In
, FL = 24, HJ = 48, and
each measure.
53. In the circle below,
is a diameter, AC = 8
inches, and BC = 15 inches. Find the diameter, the
radius, and the circumference of the circle.
SOLUTION: Use the Pythagorean Theorem to find the
hypotenuse of the triangle which is also the diameter
of the circle.
. Find
55. FG
SOLUTION: If a diameter (or radius) of a circle is perpendicular
to a chord, then it bisects the chord and its arc. So,
bisects Therefore, The radius is half the diameter. So, the radius of the
circle is about 8.5 in.
The circumference C of a circle of a circle of
diameter d is given by
Therefore, the circumference of the circle is
ANSWER: 48
56. ANSWER: d = 17 in., r = 8.5 in .,
or about 53.4 in.
54. SAT/ACT The sum of three consecutive integers is
–48. What is the least of the three integers?
A –15
B –16
C –17
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D –18
E –19
SOLUTION: If a diameter (or radius) of a circle is perpendicular
to a chord, then it bisects the chord and its arc. So,
bisects Therefore, ANSWER: 65
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57. NJ
SOLUTION: ANSWER: 10-448
Inscribed Angles
ANSWER: 130
Find x.
56. SOLUTION: If a diameter (or radius) of a circle is perpendicular
to a chord, then it bisects the chord and its arc. So,
bisects Therefore, 59. ANSWER: 65
SOLUTION: The sum of the measures of the central angles of a
circle with no interior points in common is 360. So,
57. NJ
SOLUTION: If a diameter (or radius) of a circle is perpendicular
to a chord, then it bisects the chord and its arc. So,
bisects Therefore, ANSWER: 107
ANSWER: 24
58. SOLUTION: If a diameter (or radius) of a circle is perpendicular
to a chord, then it bisects the chord and its arc. So,
bisects 60. SOLUTION: The sum of the measures of the central angles of a
circle with no interior points in common is 360. So,
Therefore, ANSWER: 130
ANSWER: 162
Find x.
59. SOLUTION: The sum of the measures of the central angles of a
circle with no interior points in common is 360. So,
61. SOLUTION: The sum of the measures of the central angles of a
circle with no interior points in common is 360. So,
ANSWER: 107
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ANSWER: 144
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62. PHOTOGRAPHY In one of the first cameras
10-4ANSWER: Inscribed Angles
162
61. SOLUTION: The sum of the measures of the central angles of a
circle with no interior points in common is 360. So,
ANSWER: 144
62. PHOTOGRAPHY In one of the first cameras
invented, light entered an opening in the front. An
image was reflected in the back of the camera,
upside down, forming similar triangles. Suppose the
image of the person on the back of the camera is 12
inches, the distance from the opening to the person is
7 feet, and the camera itself is 15 inches long. How
tall is the person being photographed?
ANSWER: 144
62. PHOTOGRAPHY In one of the first cameras
invented, light entered an opening in the front. An
image was reflected in the back of the camera,
upside down, forming similar triangles. Suppose the
image of the person on the back of the camera is 12
inches, the distance from the opening to the person is
7 feet, and the camera itself is 15 inches long. How
tall is the person being photographed?
SOLUTION: The height of the person being photographed is the
length of the base of the larger triangle. The two
triangles are similar. So, their corresponding sides will
be proportional. One foot is equivalent to 12 in. Then,
the height of the larger triangle is 7(12) = 84 in. Let x
be the length of the base of the larger triangle.
Therefore, the person being photographed is 5.6 ft
tall.
ANSWER: 5.6 ft
SOLUTION: The height of the person being photographed is the
length of the base of the larger triangle. The two
triangles are similar. So, their corresponding sides will
be proportional. One foot is equivalent to 12 in. Then,
the height of the larger triangle is 7(12) = 84 in. Let x
be the length of the base of the larger triangle.
ALGEBRA Suppose B is the midpoint of
.
Use the given information to find the missing
measure.
63. AB = 4x – 5, BC = 11 + 2x, AC = ?
SOLUTION: Since B is the midpoint of
Therefore, the person being photographed is 5.6 ft
tall.
ANSWER: 5.6 ft
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ALGEBRA Suppose B is the midpoint of
.
Use the given information to find the missing
ANSWER: 54
64. AB = 6y – 14, BC = 10 – 2y, AC = ?
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Therefore, the person being photographed is 5.6 ft
tall.
ANSWER: 10-4ANSWER: Inscribed Angles
5.6 ft
ALGEBRA Suppose B is the midpoint of
.
Use the given information to find the missing
measure.
63. AB = 4x – 5, BC = 11 + 2x, AC = ?
66. AB = 10s + 2, AC = 40, s = ?
SOLUTION: Since B is the midpoint of
SOLUTION: Since B is the midpoint of
ANSWER: 1.8
ANSWER: 54
64. AB = 6y – 14, BC = 10 – 2y, AC = ?
SOLUTION: Since B is the midpoint of
ANSWER: 8
65. BC = 6 – 4m, AC = 8, m = ?
SOLUTION: Since B is the midpoint of
ANSWER: 66. AB = 10s + 2, AC = 40, s = ?
SOLUTION: eSolutions
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Since
B is the
midpoint
of
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