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CHAPTER 6 REVIEW
EXERCISES
1. Possible answer: The Inscribed Angle Conjecture is
very important because several other conjectures
build on it, and it can be used in many different
situations.
2. Possible answers:
With compass and straightedge: Draw two nonparallel chords, and construct their perpendicular
bisectors. The intersection of their perpendicular
bisectors is the center of the circle.
With patty paper: Fold the paper along a diameter
so that two semicircles coincide. Repeat with a
different diameter. The center is the intersection of
the two folds.
With the right-angled corner of a carpenter’s
square: Place the corner in the circle so that it is an
inscribed right angle. Trace the sides of the corner.
Use the square to construct the hypotenuse of the
right triangle (which is the diameter of the circle).
Repeat. The center is the intersection of the two
diameters.
3. The velocity vector is always perpendicular to the
radius at the point of tangency to the object’s
circular path.
4. Sample answer: An arc measure is between 0° and
360°. An arc length is proportional to arc measure
and depends on the radius of the circle.
5. 55°. Draw the radius to the point of tangency. By
the Tangent Conjecture, the radius is perpendicular
to the tangent, so a right triangle has been formed.
Therefore the measure of the central angle that
intercepts the arc of measure b is 90° 2 35° 5 55°,
so b 5 55° by the definition of the measure of
an arc.
6. 65°. The 110° angle is an inscribed angle that intercepts an arc of measure a 1 155°, so by the
Inscribed Angle Conjecture, 110° 5 }12}(a 1 155°);
220° 5 a 1 155°, and a 5 65°.
7. 128°. Congruent chords intercept congruent arcs
(Chord Arcs Conjecture), so the unmarked arc has
measure c. Then 2c 1 104° 5 360°, so 2c 5 256°,
and c 5 128°.
8. 118°. First find the measure of either of the two
vertical angles that form a linear pair with the angle
of measure e. The measure of an angle formed by
two intersecting chords is half the sum of the measures of the intercepted arcs (Intersecting Chords
Conjecture), so the measure of either one of these
angles is }12}(60° 1 64°) 5 62°. Then e 5 180° 2 62°
(Linear Pair Conjecture), so e 5 118°.
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CHAPTER 6
9. 91°. The angle marked as a right angle intercepts a
semicircle and is therefore inscribed in the opposite
semicircle, so d 1 89° 5 180°, and d 5 91°. (You
could also use the Cyclic Quadrilateral Conjecture
to see that the angle opposite the marked right
angle is the supplement of a right angle, and therefore, it is also a right angle. Then the intercepted arc
of this right angle must measure 180°, so d 1 89° 5
180°.)
10. 66°. Look at either of the angles that form a linear
pair with the 88° angle. The supplement of an 88°
angle is a 92° angle. Because the measure of an
angle formed by two intersecting chords is one-half
the sum of the measures of their intercepted arcs
(Intersecting Chords Conjecture), 92° 5 }21}(f 1 118°),
so 184° 5 f 1 118°, and f 5 66°.
11. 125.7 cm. C 5 2pr 5 2p(20) < 125.7 cm.
132
}
12. 42.0 cm. C 5 pd, so 132 5 pd, and d 5 }
p <
42.0 cm.
X 5 100° (Chord Arcs Conjecture),
13. 15p cm. mAB
so by the Arc Length Conjecture, the length of
5
100°
X is }
}
}}
AB
360° C 5 18 (2p ? 27) 5 15p cm.
14. 14p ft. By the Intersecting Secants Conjecture,
1
X 2 60°), so 100° 5 mDL
X 2 60°,
50° 5 }2}(mDL
X
and mDL 5 160°. By the Chord Arcs Conjecture,
X 5 mOL
X, so 2mCD
X 1 160° 1 60° 5 360°.
mCD
X
X 5 70°. Now apply
Then 2mCD 5 140°, and mCD
X is
the Arc Length Conjecture. The length of CD
70°
7
}}C 5 }}(2p ? 36 ft) 5 14p ft.
360°
36
15. Look at the inscribed angles with measures 57° and
35°. By the Inscribed Angle Conjecture, the sum of
the measures of their intercepted arcs is 2(57°) 1
2(35°) 5 184°. However, the sum of these two arcs
is a semicircle, and the measure of a semicircle is
180°, so this is impossible.
Another way to look at this is to add the angle
measures in the triangle. The third angle of the
triangle must be a right angle because it is inscribed
in a semicircle (Angles Inscribed in a Semicircle
Conjecture). Therefore, the sum of the three angles
of the triangle would be 57° 1 35° 1 90° 5 182°,
which is impossible by the Triangle Sum Conjecture.
16. By the Parallel Lines Intercepted Arcs Conjecture,
the unmarked arc measures 56°. Then the sum of
the arcs would be 84° 1 56° 1 56° 1 158° 5 354°,
but this is impossible because the sum of the measures of the arcs of a circle must be 360°.
X 5 }1}(180° 2 108°) 5 36° 5
17. m/EKL 5 }1}mEL
2
2
# i YL
# by the Converse of the
m/KLY. Therefore, KE
Parallel Lines Conjecture.
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X 5 360° 2 56° 2 152° 5 152° 5 mMI
X. Therefore,
18. mJI
m/J 5 m/M (Inscribed Angles Intercepting Arcs
Conjecture), and nJIM is isosceles by the Converse
of the Isosceles Triangle Conjecture.
X 5 2m/KEM 5 140°. Then mKI
X 5 140° 2
19. mKIM
X
X5
70° 5 70° 5 mMI . Therefore, m/IKM 5 }12}mMI
1 X
}}mKI 5 m/IMK, so /IKM > /IMK. So, nKIM is
the intersection of the arcs as O. Construct RHOM.
The intersection of the diagonals is the center of the
inscribed circle. Construct a perpendicular to a side
to find the radius. It is not possible to construct the
circumscribed circle unless the rhombus is a square.
H
O
2
isosceles by the Converse of the Isosceles Triangle
Conjecture.
20. Ertha can trace the incomplete circle on paper. She
can lay the corner of the pad on the circle to trace
an inscribed right angle. Then Ertha should mark
the endpoints of the intercepted arc and use the pad
to construct the hypotenuse of the right triangle,
which is the diameter of the circle.
21. Sample answer: Construct
perpendicular bisectors of two
sides of the triangle. The point
at which they intersect (the
circumcenter) is the center of
the circle. The distance from
the circumcenter to each vertex
is the radius.
22. Sample answer: Construct the incenter (from the
angle bisectors) of the triangle. From the incenter,
which is the center of the circle, construct a perpendicular to a side. The distance from the incenter to
the foot of the perpendicular is the radius.
23. Sample answer: Construct a right angle, and label
# and RT
# with any lengths.
the vertex R. Mark off RE
From point E, swing an arc with radius RT. From
point T, swing an arc with radius RE. Label the
intersection of the arcs as C. Construct the diago# and RC
#. Their intersection is the center
nals ET
of the circumscribed circle. The circle’s radius is
the distance from the center to a vertex. It is not
possible to construct the inscribed circle unless the
rectangle is a square.
E
C
R
T
24. Sample answer: Construct acute angle R. Mark off
equal lengths RM and RH. From points M and H,
swing arcs of lengths equal to RM and RH. Label
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R
M
25. 4x 1 3y 5 32 or y 5 2}43}x 1 }332}. A tangent is
perpendicular to a radius drawn to the point of
tangency (Tangent Conjecture). Here, the slope of
421
3
}
}}
the radius to the point of tangency is }
5 2 1 5 4 , so
the slope of the tangent is the opposite reciprocal of
4
3
}}, which is 2}3}. The tangent is the line with slope
4 4
2}3} passing through (5, 4). Find an equation for
this line.
y24
4 24
}} 5 2}} 5 }}
3
3
x25
3(y 2 4) 5 24(x 2 5)
3y 2 12 5 24x 1 20
4x 1 3y 5 32
or
3y 5 24x 1 32
4
32
y 5 2}3}x 1 }3}
26. (23, 2). This is equivalent to finding the circumcenter of the triangle with the three given points
as vertices. For reference, label the three points as
A(27, 5), B(0, 6), and C(1, 21). Draw the triangle
on a coordinate grid.
y
A (–7, 5)
B (0, 6)
x
C (1, –1)
It appears from the drawing that nABC is a right
triangle with /B as the right angle. Verify this by
#5
finding slopes of the two sides of /B: Slope AB
1
2
1
2
6
2
7
625
# 5 }} 5 }} 5 27.
}} 5 }}, and slope BC
7
120
1
0 2 (27)
# ' BC
#, so /B is a right angle, and AC
#
Therefore, AB
is the hypotenuse of the right triangle. Thus, the
circumcenter of the triangle (or the center of the
CHAPTER 6
129