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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°. 128 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. Discovering Geometry Solutions Manual ©2008 Key Curriculum Press 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 Discovering Geometry Solutions Manual ©2008 Key Curriculum Press 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