Download Reteach 12.4

Name ________________________________________ Date __________________ Class__________________
LESSON
12-4
Reteach
Inscribed Angles
Inscribed Angle Theorem
The measure of an inscribed
angle is half the measure of its
intercepted arc.
∠ABC is an
inscribed angle.
p is an
AC
intercepted arc.
1
p
m∠ABC = m AC
2
Inscribed Angles
An inscribed angle
subtends a semicircle
if and only if the angle
is a right angle.
If inscribed angles of
a circle intercept the
same arc, then the
angles are congruent.
∠ABC and ∠ADC intercept
p, so ∠ABC ≅ ∠ADC.
AC
Find each measure.
q
1. m∠LMP and mMN
p
2. m∠GFJ and mFH
_________________________________________
Find each value.
3. x
________________________________________
4. m∠FJH
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
12-30
Holt McDougal Geometry
Name ________________________________________ Date __________________ Class__________________
LESSON
12-4
Reteach
Inscribed Angles continued
Inscribed Angle Theorem
∠A and ∠C are supplementary.
If a quadrilateral is inscribed
in a circle, then its opposite
angles are supplementary.
∠B and ∠D are supplementary.
ABCD is inscribed in :E.
Find m∠G.
Step 1 Find the value of z.
m∠E + m∠G = 180°
EFGH is inscribed in a circle.
4z + 3z + 5 = 180
Substitute the given values.
7z = 175
z = 25
Simplify.
Divide both sides by 7.
Step 2 Find the measure of ∠G.
m∠G = 3z + 5
= 3(25) + 5 = 80°
Substitute 25 for z.
Find the angle measures of each quadrilateral.
5. RSTV
6. ABCD
_________________________________________
7. JKLM
________________________________________
8. MNPQ
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
12-31
Holt McDougal Geometry
11. 90°; 90°; 90°; 90°
congruent. Draw any inscribed angle. Use
the compass to copy the arc that this angle
intercepts. Mark off the same arc from the
vertex of the inscribed angle. Connect the
points.
12. 68°; 95°; 112°; 85°
13. 59°; 73°; 121°; 107°
Practice C
1. Possible answer: It is given that
AC ≅ AD. In a circle, congruent chords
intercept congruent arcs, so
q ≅ AED
q. DC
p is congruent to itself by
ABC
the Reflexive Property of Congruence. By
the Arc Addition Postulate and the
Addition Property of Congruence,
q ≅ ADC
q . ∠ABC intercepts ADC
q , so
ACD
1 q
. ∠AED intercepts
m∠ABC = m ADC
2
q , so m∠AED = 1 m ACD
q . By
ACD
2
substitution, m∠ABC = m∠AED.
Therefore ∠ABC ≅ ∠AED.
2. Possible answer: It is given that
p ≅ RS
p. By the definition of congruent
PQ
p = m RS
p . ∠PSQ intercepts
arcs, m PQ
p , and ∠RQS intercepts RS
p . So
PQ
m∠PSQ must equal m∠RQS. Therefore
∠PSQ ≅ ∠RQS. ∠PSQ and ∠RQS are
congruent alternate interior angles of QR
6. cannot be inscribed in a circle
Reteach
q = 96°
1. m∠LMP = 18°; m MN
p = 72°
2. m∠GFJ = 55°; m FH
3. 16.4
4. 45°
5. 70°; 88°; 110°; 92°
6. 120°; 75°; 60°; 105°
7. 132°; 90°; 48°; 90°
8. 101°; 86°; 79°; 94°
Challenge
1. chord; inscribed
2. a. 45°
b. 67.5°
and PS. So QR & PS.
c. |180 − 11.25n|
3. cannot be inscribed in a circle
d. 0 < |n − 16| < 16
4. Can be inscribed in a circle; possible
answer: The two congruent angles of the
kite are opposite, so they must be right
angles. Draw a diameter. Draw segments
from opposite ends of the diameter to any
point on the circle. Use the compass to
copy one of the segments across the
diameter. Draw the fourth side.
⎡
⎛
2 n ⎞⎤
⎢180 ⎜ 1 −
⎟⎥ ,
p ⎠⎦
⎝
⎣
p p
where 0 < n − n − <
2 2
e. 180 −
360 n
, or
p
f. Answers will vary. Students may
choose any values of n and p for which
n 5
=
. Sample answer: n = 15, p = 36
p 12
Problem Solving
1. 160°
2. 112°; 52°; 68°; 128°
5. Can be inscribed in a circle; possible
answer: The pairs of base angles of a
trapezoid inscribed in a circle must be
3. C
4. G
5. B
6. G
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A52
Holt McDougal Geometry