ssignment ssignment for Lesson.1 Name Date Riding a Ferris Wheel Introduction to ircles 1. For each term, name all of the components of circle Y that are examples of the term. G R Y O T M a. hord GM, R, RO b. Radius YR, YO c. Diameter RO d. Tangent RT e. rc Sample nswer: RG, G, O f. Major arc Sample nswer: GRM, GRO, GR g. Minor arc Sample nswer: RG, G, O, OM h. Semicircle semicircle RO, semicircle RMO i. Secant R hapter ssignments 97
. On circle R, draw each item. a. tangent JB b. radius RB Sample nswer: B I c. diameter B d. chord I J R e. secant J 3. Use circle R from Question. Name all of the examples of each term. a. Major arc BI b. Minor arc I, BI Sample answers are based on the answer for Question. c. Semicircle semicircle BI 98 hapter ssignments
ssignment ssignment for Lesson. Name Date olding the Wheel entral ngles, Inscribed ngles, and Intercepted rcs Use circle S to complete Questions 1 through. F I S R 1. Suppose that m 9º. Find m F. 301º. Suppose that m SI 14º. Find m FI. 6º 3. Suppose that m º. Find m F. 7.º 4. Suppose that m FSI 71º. Find m I. 109º. What is m F? 180º 6. In circle shown below, m NG 74. Find m G and m NG. N G m G 74º 148º m NG 360º 148º 1º hapter ssignments 99
7. In circle shown below, m is 10º, m is 47º, and m T is 100º. Find m T, m T, m, and m T. T m T 1 (47º 10º) 76º m T 1 100º 0º m 1 (100º 108º) 104º m T 1 (108º 10º) 106.º 100 hapter ssignments
ssignment ssignment for Lesson.3 Name Date Manhole overs Measuring ngles Inside and Outside of ircles 1. In circle P shown below, m D 7 and m N 49. Find the value of x. D x 1 (7º 49º) x P N 1 14º 6º. In circle P shown below, m DN 144 and m N 68. Find m. m DN 180º 68º 11º D N P m DN 1 (m DN m ) 11º 1 (144º m ) 4º 144º m 80º m 3. In circle O shown below, m SN and m 3. Find m S. S Because SN is a semicircle, N So, m S 180º º 1º. O Because N is a semicircle, So, m N 180º 3º 14º. m S 1 (m S m N) m S m SN 180º. m m N 180º. 1 (1º 14º) 1 (70º) 13º So, m S is 13º. hapter ssignments 101
4. In circle O shown below, m S 11 and m MS 104. Find m DM. Because is a semicircle,. S O So, m mdm m DMS MS 180º DM 180º 104º 76º. D m DM 1 (m DM m S) M 1 (76º 11º) 1 (87º) 43.º So, m DM is 43.º.. In circle S shown below, m R 38 and m OT 11. Find m OUT. U R S O m OUT 1 (m OT m R) 1 (11º 38º) T 1 (83º) 41.º So, m OUT is 41.º. 10 hapter ssignments
Name Date 6. In circle S shown below, O is the diameter of the circle and m OT 13. Find m OUT. S O U T P Draw chord OT. Find the measure of exterior angle OTP. m OTP 1 (m OT) 1 (13º) Next, find m UOT. Because O is a diameter, So, m 66º T 180º 13º 48º. m OT m T 180º. m UOT 1 (m T) 1 (48º) 4º Because OTP is an exterior angle of OUT, m OTP m OUT m UOT. 66º m OUT 4º 4º m OUT So, m OUT is 4º. 7. In circle G shown below, O S, m O 41 and m 171. Find m U. U Because O S and m O 41º, m S 41º. m OS 360º 41º 41º 171º 107º O 41 G S 41 171 m U 1 (m m OS) 1 (171º 107º) 3º So, m U is 3º. hapter ssignments 103
8. In circle G shown below, m 99. Find m U. R G U Because m 99º, m R 360º 99º 61º. m U 1 (m R m ) 1 (61º 99º) 81º So, m U is 81º. 9. In circle T shown below, m R is 7º and m R is 141º. Find m BL. L B R T 1 The measure of R is equal to the difference of the measure of the arcs that are intercepted by the angle. So, m R 1 (m R m BL) 7º 1 (141º m BL) m BL 7º So, m BL is 7º. 104 hapter ssignments
ssignment ssignment for Lesson.4 Name Date olor Theory hords and ircles Use circle T below to complete Questions 1 through 7. I T G R N 1. Draw an inscribed right angle in circle T. Label each point where the angle intersects the circle. What is the name of the right angle? RIG. Draw the chord determined by the inscribed right angle. What is the name of the chord? RG 3. What else do you know about the chord determined by an inscribed right angle? The chord determined by an inscribed right angle is a diameter of the circle. 4. Draw a second inscribed right angle in circle T. Label each point where the angle intersects the circle. What is the name of the second right angle? N.. Draw the chord determined by the second inscribed right angle. What is the name of the chord? 6. What else do you know about the chord determined by the second inscribed right angle. The chord determined by the second inscribed right angle is also a diameter of the circle. 7. Do you think every inscribed right angle will determine the longest chord of the circle, which is the diameter of the circle? Yes. The arc of a semicircle is 180º, so the measure of an angle inscribed in a semicircle is half of 180º, or 90º. hapter ssignments 10
8. The figure below shows a section of a circle. Draw two chords and construct their perpendicular bisectors to locate the center of the circle. B G D F O The intersection of the perpendicular bisectors of two chords is the center of the circle. Draw chords B and D. onstruct line F, which is the perpendicular bisector of B. onstruct line G, which is the perpendicular bisector of D. Point O, which is the intersection of lines F and G, is the center of the circle. 9. In circle G shown below, MG = 1.84 centimeters, GL = 1.98 centimeters, m GL = 90, and m GMK 90. Determine which chord is longer, I or JK. J I L G M K Because point M is closer to the center of the circle than point L, chord JK is longer than chord I. 106 hapter ssignments
ssignment ssignment for Lesson. Name Date Solar clipses Tangents and ircles 1. In the space below, draw circle O with a tangent line drawn. Label the point of tangency as point. O B. Label another point on the tangent as point B. 3. Draw a second tangent line to the circle that passes through point B. Label this second point of tangency as point. 4. Draw the radii O and O.. What is m OB? xplain how you found your answer. The measure of OB is 90º, because a tangent to a circle is perpendicular to the radius that is drawn to the point of tangency. 6. What is m OB? xplain how you found your answer. The measure of OB is 90º, because a tangent to a circle is perpendicular to the radius that is drawn to the point of tangency. 7. Use a protractor to measure O. m O 114º 8. What is m B? xplain how you found this measure. 360º 90º 90º 114º = 66º The sum of the angles in any quadrilateral is 360º. Subtract the three known angles from 360º. hapter ssignments 107
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ssignment ssignment for Lesson.6 Name Date Gears rc Length 1. In circle shown below, describe the difference between the measure of minor arc B and the length of minor arc B. B The radius of circle is 3 centimeters. The circumference of circle is (3), or about 18.84 centimeters. The measure of minor arc B is determined by the measure of its central angle, B. So, m B is 140º and m B is 140º. The length of arc B is determined by the portion of the circumference that lies between points B and. 140º 140º rc B is of the circle. So, the length of B is 18.84 7.33 centimeters. 360º 360º. In circle shown below, the radius of the circle is 16 centimeters and m JSB is 40. Find the arc length of JB. S J B The circumference of circle is (16), or about 100.48 centimeters. The measure of JSB 1 is JB. So, m 1 m JSB JB m 40º 1 JB m 80º 80º m JB ; So, the arc length of JB is 100.48.33 centimeters. 360º hapter ssignments 109
3. In circle I shown below, the radius is 6 millimeters and m is 80. Find the arc length of S. S R I D 80 The circumference of circle I is (6), or about 37.68 millimeters. Because S is a diameter, S is a semicircle. Because is 80º,. The arc length of is m S 180º 80º 100º m S m S 360º 37.68 100º 360º 37.68 10.47 4. In circle shown below, the arc length of SJ is 4 centimeters and m JOS is 80. Find the length of a diameter of circle. S O J millimeters. Because m JOS is 80, m SJ (80º) 160º. Because the length of an arc is equal to the ratio of the measure of the arc and 360 times the circumference of the circle, 4 160º 360º r 8640 160º r 8640 30º r 7 r. So, the radius of circle is 7 centimeters and the diameter of circle is 4 centimeters. 110 hapter ssignments
ssignment ssignment for Lesson.7 Name Date Playing Darts reas of Parts of ircles In circle shown below, B is equilateral, = 10 inches, and m B is 60º. Use the figure to complete Questions 1 through 3. 10 in. B 1. Find the area of the sector B. The area of circle is ( ) or about 314 square inches. Because is 60, 60º m 10 m B B is 60. So, the area of sector B is 314 square inches, or about.3 360º square inches.. The height of B is about 8.66 inches. Find the area of B. The area of ΔB is 1 (8.66)10 43.3 square inches. 3. What is the shaded region called? Find the area of the shaded region. The shaded region is called a segment of the circle. The area of the segment of the circle is equal to the area of the sector minus the area of the triangle. So, the area of the segment is.3 43.3, or about 9.03 square inches. In circle shown below, the radius is 18 centimeters, B is equilateral, and m B is 60. Use the figure to complete Questions 4 and. B 18 cm 60 4. Find the area of the sector of the circle determined by radii B and. The area of the sector is 60º 360º (18 ) 169.6 square centimeters. hapter ssignments 111
. Find the area of the segment of the circle bounded by chord B. The area of the segment is the difference between the area of the sector and the area of the triangle. Because B is equilateral, B is 18 centimeters. The height of B is 9 3 centimeters. The area of B 1 is (18)(9 3) 140.30 square centimeters. So, the area of the segment is 169.6 140.30 = 9.6 square centimeters. In circle shown below, the radius is centimeters and m RT is 90. Use the figure to complete Questions 6 and 7. R cm T 6. Find the area of the sector of the circle determined by radii R and T. The area of the sector is 90º 360º ( ) 379.94 square centimeters. 7. Find the area of the segment of the circle bounded by chord RT. The area of the segment is the difference between the area of the sector and the area of the triangle. 1 The area of RT is ()() 4 square centimeters. So, the area of the segment is 379.94 4 137.94 square centimeters. 11 hapter ssignments
ssignment ssignment for Lesson.8 Name Date rop ircles ircle Measurements and Relationships Use the diagram and the given information to answer Questions 1 through. xplain your reasoning. The radius of circle P in the diagram below is 8 inches and segment GF is a diameter of the circle. Segments K and GK are tangents to circle P. The measure of angle GK is 43 degrees. J F K P L G 1. Find GF. The radius of circle P is given as 8 inches. Segment GF is a diameter, so the length of segment GF is 8 16 inches.. Find m GK. Tangent segments to the same circle are equal in length. So, triangle KG is an isosceles triangle. The measure of angle GK is given as 43 degrees. So, the measure of GK is equal to 180 ( 43 ), or 94. 3. Find m FG. The measure of angle FGK is 90 degrees because GK is tangent to circle P, and you are given that the measure of angle GK is 43 degrees. So, the measure of angle GF is 90 43, or 47. The base of triangle GF is the diameter of circle P, so triangle GF is a right triangle, and the measure of angle GF is 90 degrees. So, the measure of angle FG is 180 (90 47 ), or 43. hapter ssignments 113
4. Find G. Triangle GF is a right triangle. The length of the hypotenuse is 16 inches, and you found in Question 3 that the measure of angle FG is 43. Use a trigonometric ratio to find the length G. sin FG sin 43º G 16 G hypotenuse G 16 sin 43º G 10.9 inches. Find m G. The measure of an intercepted arc is twice the measure of its inscribed angle. From Question 3, the measure of angle FG is 43. So, the measure of arc G is 43, or 94. Use the diagram and the given information to answer Questions 6 through 11. xplain your reasoning. In the diagram below, line SU is tangent to circle at point U. Segment VT is a diameter of the circle. The measure of arc VW is 33 and the measure of arc WX is 8. R X V W S Y Z U T 6. Find m VXT. The base of triangle VXT is the diameter of circle, so triangle VXT is a right triangle. The measure of angle VXT is 90. 114 hapter ssignments
Name Date 7. Find m VTX. The measure of angle VTX is equal to half the measure of its intercepted arc, which is arc VX. The measure of arc VW is 33 and the measure of arc WX is 8, so the measure of arc VX is 33 8, or 61. So, the measure of angle VTX is 8. Find m TX. 9. Find m VYZ. 10. Find m ZYX. 1 61, or 30.. The measure of arc TX is equal to twice the measure of its intercepted angle, which is angle TVX. In Questions 6 and 7 you found that the measure of angle VXT is 90 and the measure of angle VTX is 30., so the measure of angle TVX is 180 (90 30. ), or 9.. So, the measure of arc TX is 9., or 119. First, find the measure of angle VZY. The measure of angle VZY is equal to half the sum of the measures of the two intercepted arcs, VW and TU. The measure of arc VW is 33. ngle TU is a right angle, so the measure of arc TU is 90. The measure 1 of angle VZY is (33 90 ), or 61.. In Question 8 you found that the measure of angle TVX is 9.. So, the measure of angle VYZ is 180 (61. 9. ), or 9. ngles ZYX and VYZ form a linear pair. In Question 9 you found that the measure of angle VYZ is 9, so the measure of angle ZYX is 180 9, or 11. 11. Find m ZUS. Because line SU is tangent to circle at point U, angle US is a right angle. So, the measure of angle ZUS is equal to 90 minus the measure of angle UZ. The measure of angle ZU is 90 and the measure of angle ZU is 61. (angles ZU and VZY are vertical angles and thus congruent). So, the measure of angle UZ is 180 (90 61. ), or 8.. This means that the measure of angle ZUS is 90 8., or 61.. hapter ssignments 11
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