5-4 Inscribed Angles
Find each measure.
1.
SOLUTION:
ANSWER:
72°
2. m∠N
SOLUTION:
ANSWER:
31°
eSolutions Manual - Powered by Cognero
Page 1
5-4 Inscribed Angles
3.
SOLUTION:
ANSWER:
226°
4.
SOLUTION:
ANSWER:
46°
eSolutions Manual - Powered by Cognero
Page 2
5-4 Inscribed Angles
5. m∠E
SOLUTION:
ANSWER:
81°
6. m∠R
SOLUTION:
ANSWER:
50°
eSolutions Manual - Powered by Cognero
Page 3
5-4 Inscribed Angles
Find each measure.
7. m∠N
SOLUTION:
So, m∠N = 3(3) + 8 or 17°.
ANSWER:
17°
eSolutions Manual - Powered by Cognero
Page 4
5-4 Inscribed Angles
8. m∠L
SOLUTION:
So, m∠L = 5(9) + 9 or 54°.
ANSWER:
54°
eSolutions Manual - Powered by Cognero
Page 5
5-4 Inscribed Angles
9. m∠C
SOLUTION:
So, m∠C = 3(21) – 20 or 43°.
ANSWER:
43°
eSolutions Manual - Powered by Cognero
Page 6
5-4 Inscribed Angles
10. m∠A
SOLUTION:
So, m∠A = 5(18) – 1 or 89°.
ANSWER:
89°
eSolutions Manual - Powered by Cognero
Page 7
5-4 Inscribed Angles
PROOF Write the specified type of proof.
11. paragraph proof
Given:
=
Prove:
SOLUTION:
Proof: It is given that
=
=
and
. This means that
=
equation by 2 results in
=
. By substitution,
. By the Inscribed Angle Theorem,
. Multiplying each side of the
.
ANSWER:
Proof: It is given that
=
and
equation by 2 results in
eSolutions Manual - Powered by Cognero
=
. This means that
=
. By substitution,
=
. By the Inscribed Angle Theorem,
. Multiplying each side of the
.
Page 8
5-4 Inscribed Angles
12. two-column proof
Given: ⊙C
Prove:
SOLUTION:
Proof:
Statements
1. ⊙C
2. ∠H ≅ ∠L
3. ∠KML ≅ ∠JMH
4.
ANSWER:
Proof:
Statements
1. ⊙C
2. ∠H ≅ ∠L
3. ∠KML ≅ ∠JMH
4.
eSolutions Manual - Powered by Cognero
Reasons
1. Given
2. Inscribed angles intercepting the same
arc are ≅.
3. Vertical angles are congruent.
4. AA Similarity
Reasons
1. Given
2. Inscribed angles intercepting the same
arc are ≅.
3. Vertical angles are congruent.
4. AA Similarity
Page 9
5-4 Inscribed Angles
Find each value.
13. x
SOLUTION:
is a right triangle because ∠Y inscribes a semicircle.
ANSWER:
39
eSolutions Manual - Powered by Cognero
Page 10
5-4 Inscribed Angles
14. m∠W
SOLUTION:
is a right triangle because ∠Y inscribes a semicircle.
So, m∠W = 39 + 12 or 51°.
ANSWER:
51°
eSolutions Manual - Powered by Cognero
Page 11
5-4 Inscribed Angles
15. x
SOLUTION:
is a right triangle because ∠S inscribes a semicircle.
ANSWER:
16
eSolutions Manual - Powered by Cognero
Page 12
5-4 Inscribed Angles
16. m∠T
SOLUTION:
is a right triangle because ∠S inscribes a semicircle.
So, m∠T = 4(16) – 6 or 58°.
ANSWER:
58°
eSolutions Manual - Powered by Cognero
Page 13
5-4 Inscribed Angles
17. m∠J
SOLUTION:
is a right triangle because ∠H inscribes a semicircle.
So, m∠J = 5(12) – 2 or 58°.
ANSWER:
58°
eSolutions Manual - Powered by Cognero
Page 14
5-4 Inscribed Angles
18. m∠K
SOLUTION:
is a right triangle because ∠H inscribes a semicircle.
So, m∠K = 2(12) + 8 or 32°.
ANSWER:
32°
eSolutions Manual - Powered by Cognero
Page 15
5-4 Inscribed Angles
19. m∠A
SOLUTION:
is a right triangle because ∠B inscribes a semicircle.
So, m∠A = 3(21) + 6 or 69°.
ANSWER:
69°
eSolutions Manual - Powered by Cognero
Page 16
5-4 Inscribed Angles
20. m∠C
SOLUTION:
is a right triangle because ∠B inscribes a semicircle.
So, m∠C = 21°.
ANSWER:
21°
eSolutions Manual - Powered by Cognero
Page 17
5-4 Inscribed Angles
Find each measure.
21. m∠R
SOLUTION:
Because PQRS is inscribed in a circle, its opposite angles are supplementary.
So, m∠R = 3(35) or 105°.
ANSWER:
105°
eSolutions Manual - Powered by Cognero
Page 18
5-4 Inscribed Angles
22. m∠S
SOLUTION:
Because PQRS is inscribed in a circle, its opposite angles are supplementary.
So, m∠S = 2(35) or 70°.
ANSWER:
70°
eSolutions Manual - Powered by Cognero
Page 19
5-4 Inscribed Angles
23. m∠W
SOLUTION:
Because WXYZ is inscribed in a circle, its opposite angles are supplementary.
So, m∠W = 54 + 18 or 72°.
ANSWER:
72°
eSolutions Manual - Powered by Cognero
Page 20
5-4 Inscribed Angles
24. m∠X
SOLUTION:
Because WXYZ is inscribed in a circle, its opposite angles are supplementary.
So, m∠X = 3(31) – 7 or 86°.
ANSWER:
86°
eSolutions Manual - Powered by Cognero
Page 21
5-4 Inscribed Angles
25. USE ESTIMATION Darius bought a circular picture frame with a geometric design. The frame has a
quadrilateral inscribed in a circle.
a. Estimate the value of x.
b. Find the exact value of x and m∠J.
c. Is your answer reasonable? Justify your argument.
SOLUTION:
a. 22.5
b. Because JKLM is inscribed in a circle, its opposite angles are supplementary.
So, m∠J = 4(23) – 2 or 90°.
c. Yes; sample answer: Because my estimate for the value of x is close to the exact value and ∠J appears to be a
right angle, my answer is reasonable.
ANSWER:
a. 22.5
b. 23; 90°
c. Yes; sample answer: Because my estimate for the value of x is close to the exact value and ∠J appears to be a
right angle, my answer is reasonable.
eSolutions Manual - Powered by Cognero
Page 22
5-4 Inscribed Angles
Find each measure.
26.
SOLUTION:
ANSWER:
102°
eSolutions Manual - Powered by Cognero
Page 23
5-4 Inscribed Angles
27.
SOLUTION:
ANSWER:
82°
eSolutions Manual - Powered by Cognero
Page 24
5-4 Inscribed Angles
28. m∠QPR
SOLUTION:
ANSWER:
66°
eSolutions Manual - Powered by Cognero
Page 25
5-4 Inscribed Angles
29.
SOLUTION:
Because TUVW is inscribed in a circle, its opposite angles are supplementary.
ANSWER:
158°
eSolutions Manual - Powered by Cognero
Page 26
5-4 Inscribed Angles
30. m∠AFE
SOLUTION:
ANSWER:
96°
eSolutions Manual - Powered by Cognero
Page 27
5-4 Inscribed Angles
31. m∠KJH
SOLUTION:
ANSWER:
65°
PROOF Write the specified type of proof to prove each theorem.
32. two-column proof Theorem 10.8
Given:
; ∠JKL and ∠RQP are inscribed.
Prove: ∠JKL ≅ ∠RQP
eSolutions Manual - Powered by Cognero
Page 28
5-4 Inscribed Angles
SOLUTION:
Proof:
Statements
1.
Reasons
;∠JKL and ∠RQP are
1. Given
inscribed.
2. m∠JKL =
;m∠RQP =
2. The measure of an inscribed angle
equals one half the measure of its
intercepted arc.
3.
3. Definition of congruent arcs
4.
4. Multiplication Property of Equality
5. m∠JKL = m∠RQP
6. ∠JKL ≅ ∠RQP
5. Substitution
6. Definition of congruent angles
ANSWER:
Proof:
Statements
1.
Reasons
; ∠JKL and ∠RQP are
1. Given
inscribed.
2. m∠JKL =
; m∠RQP =
3.
eSolutions Manual - Powered by Cognero
2. The measure of an inscribed angle
equals one half the measure of its
intercepted arc.
3. Definition of congruent arcs
Page 29
5-4 Inscribed Angles
4.
4. Multiplication Property of Equality
5. m∠JKL = m∠RQP
6. ∠JKL ≅ ∠RQP
5. Substitution
6. Definition of congruent angles
33. paragraph proof Theorem 10.10
Given: Quadrilateral DEFG is inscribed in ⊙C.
Prove: ∠D and ∠F are supplementary.
∠E and ∠G are supplementary.
SOLUTION:
Proof: By arc addition and the definitions of arc measure and the sum of central angles,
.
By Theorem 10.7, m∠D =
and m∠F =
So, m∠D + m∠F =
or
By substitution, m∠D + m∠F =
or 180°.
.
.
By the definition of supplementary angles, ∠D and ∠F are supplementary.
Because the sum of the measures of the interior angles of a quadrilateral is
360°, m∠D + m∠F + m∠E + m∠G = 360°.
By substitution, 180 + m∠E + m∠G = 360°.
By the Subtraction Property of Equality, m∠E + m∠G = 180°.
Therefore, ∠E and ∠G are supplementary by the definition of supplementary angles.
ANSWER:
Proof: By arc addition and the definitions of arc measure and the sum of central angles,
eSolutions Manual - Powered by Cognero
.
Page 30
5-4 Inscribed Angles
By Theorem 10.7, m∠D =
and m∠F =
So, m∠D + m∠F =
or
By substitution, m∠D + m∠F =
or 180°.
.
.
By the definition of supplementary angles, ∠D and ∠F are supplementary.
Because the sum of the measures of the interior angles of a quadrilateral is 360°, m∠D + m∠F + m∠E +
m∠G = 360°.
By substitution, 180 + m∠E + m∠G = 360°.
By the Subtraction Property of Equality, m∠E + m∠G = 180°.
Therefore, ∠E and ∠G are supplementary by the definition of supplementary angles.
eSolutions Manual - Powered by Cognero
Page 31
5-4 Inscribed Angles
34. PROOF Write a paragraph proof to prove each part of Theorem 10.9.
a. Given:
is a semicircle.
Prove: ∠XYZ is a right angle.
b. Given: ∠XYZ is a right angle.
Prove:
is a semicircle.
SOLUTION:
a. Because
is a semicircle,
. Then, m∠XYZ =
because ∠XYZ is an inscribed angle. So, ∠XYZ is a right angle by definition or right angle.
b. Because ∠XYZ is an inscribed angle, m∠XYZ =
. By the Multiplication Property of Equality,
Because ∠XYZ is a right angle, m∠XYZ = 90°. So,
= 2(90°) = 180°. So,
is a semicircle by definition of semicircle.
ANSWER:
a. Because
is a semicircle,
. Then, m∠XYZ =
because ∠XYZ is an inscribed angle. So, ∠XYZ is a right angle by definition of right angle.
b. Because ∠XYZ is an inscribed angle, m∠XYZ =
. By the Multiplication Property of Equality,
Because ∠XYZ is a right angle, m∠XYZ = 90°. So,
= 2(90°) = 180°. So,
is a semicircle by definition of semicircle.
eSolutions Manual - Powered by Cognero
Page 32
5-4 Inscribed Angles
35. PROOF Write a paragraph proof to prove that is PQRS is an inscribed quadrilateral and ∠Q ≅ ∠S, then
is
a diameter of the circle.
SOLUTION:
Proof: Because PQRS is inscribed in a circle, ∠Q is supplementary to ∠S because they are opposite angles.
Because ∠Q ≅ ∠S, ∠Q and ∠S must be right angles.
So, ∠Q intercepts
, m∠Q = 90° and
= 180° by the Inscribed Angle Theorem.
This means that
is a semicircle, so
must be a diameter.
ANSWER:
Proof: Because PQRS is inscribed in a circle, ∠Q is supplementary to ∠S because they are opposite angles.
Because ∠Q ≅ ∠S, ∠Q and ∠S must be right angles.
So, ∠Q intercepts
, m∠Q = 90° and
= 180° by the Inscribed Angle Theorem.
This means that
is a semicircle, so
eSolutions Manual - Powered by Cognero
must be a diameter.
Page 33
5-4 Inscribed Angles
36. REASONING A landscaping crew is installing a circular garden with three paths that form a triangle. The figure
shows the plan for the garden and the paths. The leader of the crew wants to determine the radius they should use
when they install the circular garden.
a. Find the measures of the angles in △ABD.
b. Use trigonometry to find the radius of the circular garden. Round to the nearest tenth meter, if necessary.
c. Given that the path will go along the outside of the garden, as well as along the three interior paths, find the total
length of the path that will need to be installed.
SOLUTION:
a. Since
is a semicircle, ∠ABD is a right angle; by definition of right angle, m ∠ABD = 90°.
By the Inscribe Angle Theorem, since
Since
is a semicircle, then
substitution,
b.
, m∠BAD =
.
. Therefore, by arc addition postulate,
So, by the Inscribed Angle Theorem, m∠BDA =
By
.
. Since the diameter is about 21.2 m, then the radius is about
10.6 m.
c. Sample answer: The circumference of the circle is (2π)(10.6) = 66.6 m.
AD = 2(10.6) = 21.2 m. AB = 16 m.
; the total is 66.6 + 21.2 + 16 + 13.9 =
117.7 m.
ANSWER:
a. m∠ABD = 90°, m∠BAD = 41°, and m∠BDA = 49°
b. 10.6 m
c. Sample answer: The circumference of the circle is (2π)(10.6) = 66.6 m.
AD = 2(10.6) = 21.2 m. AB = 16 m.
; the total is 66.6 + 21.2 + 16 + 13.9 =
117.7 m.
eSolutions Manual - Powered by Cognero
Page 34
5-4 Inscribed Angles
37. ARENA A concert arena is lit by five lights equally spaced around the perimeter. What is m∠1?
SOLUTION:
The lights are 360° ÷ 5, or 72° apart. The measure of the arc that is intercepted by ∠1 has a measure of 2(72°),
or 144°.
So, by the Inscribed Angle Theorem, m∠1 =
.
ANSWER:
72°
38. JEWELRY Alyssa makes earrings by bending wire into various shapes. She often bends the wire to form a circle
with an inscribed quadrilateral as shown. She would like to know how she can find
if she
knows m∠ADC. Write a formula for finding
given that m∠ADC = x.
SOLUTION:
By the Inscribed Angle Theorem,
. Therefore,
.
ANSWER:
eSolutions Manual - Powered by Cognero
Page 35
5-4 Inscribed Angles
39. TAPESTRY Helga is creating a design for a tapestry she is making for her art class. The design shown uses a kite
inscribed in a circle. Find m∠SRU and the value of x.
SOLUTION:
ANSWER:
m∠SRU = 80°; x = 12
eSolutions Manual - Powered by Cognero
Page 36
5-4 Inscribed Angles
ANALYZE Determine whether the quadrilateral can always, sometimes, or never be inscribed in a circle.
Justify your argument.
40. square
SOLUTION:
Squares have right angles at each vertex, therefore each pair of opposite angles will be supplementary and
inscribed in a circle.
So, a square can always be inscribed in a circle.
ANSWER:
Always; sample answer: Squares have right angles at each vertex, therefore each pair of opposite angles will be
supplementary and
inscribed in a circle.
41. rectangle
SOLUTION:
Rectangles have right angles at each vertex, therefore each pair of opposite angles will be supplementary
and inscribed in a circle.
So, a rectangle can always be inscribed in a circle.
ANSWER:
Always; sample answer: Rectangles have right angles at each vertex, therefore each pair of opposite angles will be
supplementary and inscribed in a circle.
42. parallelogram
SOLUTION:
A parallelogram can be inscribed in a circle as long as it is a rectangle.
So, a parallelogram can sometimes be inscribed in a circle.
ANSWER:
Sometimes; sample answer: A parallelogram can be inscribed in a circle as long as it is a rectangle.
eSolutions Manual - Powered by Cognero
Page 37
5-4 Inscribed Angles
43. rhombus
SOLUTION:
A rhombus can be inscribed in a circle as long as it is a square. Because the opposite angles of rhombi that are not
squares are not supplementary, they cannot be inscribed in a circle.
So, a rhombus can sometimes be inscribed in a circle.
ANSWER:
Sometimes; sample answer: A rhombus can be inscribed in a circle as long as it is a square. Because the opposite
angles of rhombi that are not squares are not supplementary, they cannot be inscribed in a circle.
44. kite
SOLUTION:
As long as the angles that compose the pair of congruent opposite angles are right angles.
So, a kite can sometimes be inscribed in a circle.
ANSWER:
Sometimes; sample answer: as long as the angles that compose the pair of congruent opposite angles are right
angles.
45. PERSEVERE A square is inscribed in a circle. What is the ratio of the area of the circle to the area of the
square?
SOLUTION:
Sample answer: The area of the circle is πr2. The diagonal of the square splits the square into two 45-45-90
triangles. The triangles have a diameters of the circle as their hypotenuse, which measures 2r. Using the 45-45-90
relationship of right triangles that have a hypotenuse of 2r, each leg of the triangles (sides of the square) have length
of
. The area of the square equal
, or 2r2.
The ratio of the area of the circle to the area of the square is
.
ANSWER:
eSolutions Manual - Powered by Cognero
Page 38
5-4 Inscribed Angles
46. WRITE A 45°-45°-90° right triangle is inscribed in a circle. If the radius of the circle is given, explain how to find
the lengths of the legs of the triangle.
SOLUTION:
According to Theorem 10.9, 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 45° • 2r or
.
ANSWER:
Sample answer: According to Theorem 10.9, 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 45° •
2r or
.
eSolutions Manual - Powered by Cognero
Page 39
5-4 Inscribed Angles
47. CREATE Draw an inscribed polygon. Then find the measure of each intercepted arc.
SOLUTION:
A sample drawing is shown.
ANSWER:
Sample answer:
eSolutions Manual - Powered by Cognero
Page 40
5-4 Inscribed Angles
48. WRITE 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 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.
ANALYZE Determine whether each statement is always, sometimes, or never true. Justify your argument.
49. If
is a major arc of a circle, then ∠PQR is obtuse.
SOLUTION:
. By the Inscribed Angle Theorem, m∠PQR < 90°.
is a minor arc, so
So, the statement is never true.
ANSWER:
Never; sample answer:
50. If
is a minor arc, so
. By the Inscribed Angle Theorem, m∠PQR < 90°.
is a diameter of circle O, and X is any point on circle O other than A or B, then
is a right triangle.
SOLUTION:
∠AXB is an inscribed angle that intercepts a diameter, so it is a right angle.
So, the statement is always true.
ANSWER:
Always; sample answer: ∠AXB is an inscribed angle that intercepts a diameter, so it is a right angle.
eSolutions Manual - Powered by Cognero
Page 41
5-4 Inscribed Angles
51. When an equilateral triangle is inscribed in a circle it partitions the circle into three minor arcs that each measure
120°.
SOLUTION:
Each angle of the triangle measures 60°, so each intercepted arc measures 120°.
So, the statement is always true.
ANSWER:
Always; sample answer: Each angle of the triangle measures 60°, so each intercepted arc measures 120°.
eSolutions Manual - Powered by Cognero
Page 42