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Chapter 9.1 The Pythagorean theorem
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10.1-10.3 Quiz Exercise 1 The center of the circle is P, so the name of the circle is circle P and can be written as β ππ . Exercise 2 One of the radius of the circle is ππππ . Exercise 3 The diameter of the circle is πππΎπΎ . Exercise 4 The chord of the circle is π½π½π½π½ . Exercise 5 The secant of the circle is ππππ βοΏ½οΏ½οΏ½β . Exercise 6 The tangent of the circle is ππππ βοΏ½οΏ½οΏ½οΏ½β . Exercise 7 By the Pythagorean Theorem, ( π₯π₯ + 9) 2 = 15 2 + π₯π₯ 2 π₯π₯ 2 + 18 π₯π₯ + 81 = 225 + π₯π₯ 2 18 π₯π₯ + 81 = 225 18 π₯π₯ = 144 π₯π₯ = 8 So, the value of x is 8. Exercise 8 By External Tangent Congruence Theorem, 6 π₯π₯ β 3 = 3 π₯π₯ + 18 6 π₯π₯ β 3 π₯π₯ = 18 + 3 3 π₯π₯ = 21 π₯π₯ = 7 So, the value of x is 5. Exercise 9 π΄π΄π΄π΄ is a diameter, so πππ΄π΄π΄π΄π΄π΄ οΏ½ = 180 β . Then, πππ΄π΄π΄π΄ οΏ½ = 180 β β πππ΄π΄π΄π΄ οΏ½ πππ΄π΄π΄π΄ οΏ½ = 180 β β 36 β πππ΄π΄π΄π΄ οΏ½ = 144 β Then, π΄π΄π΄π΄ οΏ½ is a minor arc and πππ΄π΄π΄π΄ οΏ½ = 144 β . Exercise 10 π΄π΄π΄π΄ is a diameter, so πππ΄π΄π΄π΄π΄π΄ οΏ½ = 180 β . Then, πππ΅π΅π΅π΅ οΏ½ = 180 β β πππ΄π΄π΅π΅ οΏ½ β πππ΅π΅π΄π΄ οΏ½ πππ΅π΅π΅π΅ οΏ½ = 180 β β 67 β β 70 β πππ΅π΅π΅π΅ οΏ½ = 43 β
Then, π΅π΅π΅π΅ οΏ½ is a minor arc and πππ΅π΅π΅π΅ οΏ½ = 43 β . Exercise 11 πππ΄π΄π΅π΅ οΏ½ = πππ΄π΄π΅π΅ οΏ½ + πππ΅π΅π΅π΅ οΏ½ πππ΄π΄π΅π΅ οΏ½ = 67 β + 43 β πππ΄π΄π΅π΅ οΏ½ = 110 β Then, π΄π΄π΅π΅ οΏ½ is a minor arc and πππ΄π΄π΅π΅ οΏ½ = 110 β . Exercise 12 π΄π΄π΄π΄ is a diameter, so πππ΄π΄π΅π΅π΄π΄ οΏ½ = 180 β . Then, π΄π΄π΅π΅π΄π΄ οΏ½ is a semicircle and πππ΄π΄π΅π΅π΄π΄ οΏ½ = 180 β . Exercise 13 π΄π΄π΄π΄ is a diameter, so πππ΄π΄π΅π΅π΄π΄ οΏ½ = 180 β . Then, πππ΄π΄π΅π΅π΄π΄ οΏ½ = πππ΄π΄π΅π΅π΄π΄ οΏ½ + πππ΄π΄π΄π΄ οΏ½ πππ΄π΄π΅π΅π΄π΄ οΏ½ = 180 β + 36 β πππ΄π΄π΅π΅π΄π΄ οΏ½ = 216 β Then, π΄π΄π΅π΅π΄π΄ οΏ½ is a major arc and πππ΄π΄π΅π΅π΄π΄ οΏ½ = 216 β . Exercise 14 πππ΅π΅π΄π΄π΅π΅ οΏ½ = 360 β β πππ΅π΅π΅π΅ οΏ½ πππ΅π΅π΄π΄π΅π΅ οΏ½ = 360 β β 43 β πππ΅π΅π΄π΄π΅π΅ οΏ½ = 317 β Then, π΅π΅π΄π΄π΅π΅ οΏ½ is a major arc and πππ΅π΅π΄π΄π΅π΅ οΏ½ = 317 β . Exercise 15 The central angles of the minor arcs are vertical angles, so they are congruent. Because \hat{JM} and \hat{KL} lie on the same circle, by Congruent Central Angles Theorem, \hat{JM}\cong\hat{KL}. So, the red arcs are congruent. Exercise 16 From the figure, ππππππ οΏ½ = ππππππ οΏ½ . But, ππππ οΏ½ lies on the circle that the radius is 7 and ππππ οΏ½ lies on the circle that diameter is 15 or the radius is 7.5. Because the circles have different radius, the circles are not congruent. So, the red arcs are not congruent. Exercise 17 By Congruent Corresponding Chords Theorem, π΄π΄πΈπΈπΈπΈ οΏ½ β π΅π΅π΄π΄π΄π΄ οΏ½ . So, πππ΄π΄πΈπΈπΈπΈ οΏ½ = πππ΅π΅π΄π΄π΄π΄ οΏ½ πππ΄π΄πΈπΈπΈπΈ οΏ½ = 360 β β πππ΅π΅π΅π΅ οΏ½ β πππ΅π΅π΄π΄ οΏ½ πππ΄π΄πΈπΈπΈπΈ οΏ½ = 360 β β 150 β β 110 β πππ΄π΄πΈπΈπΈπΈ οΏ½ = 100 β So, the measure of the red arcs in β ππ is 100 β . Exercise 18 By Equidistant Chords Theorem, PG=PJ x+5=3x-1 x-3x=-1-5 -2x=-6 x=3 Then, by Perpendicular Chord Bisector Theorem, FJ=JD. Because FD=30, so
FJ=JD=15. Look at triangle PJD. By Pythagorean Theorem, πππ΄π΄ = οΏ½πππ½π½ 2 + π½π½π΄π΄ 2 πππ΄π΄ = οΏ½ (3 π₯π₯ β 1) 2 + 15 2 πππ΄π΄ = οΏ½ (3 β 3 β 1) 2 + 15 2 πππ΄π΄ = οΏ½ (9 β 1) 2 + 15 2 πππ΄π΄ = οΏ½ 8 2 + 15 2 πππ΄π΄ = β 64 + 225 πππ΄π΄ = β 289 πππ΄π΄ = 17 πππ΄π΄ is the radius of β ππ , so the radius of β ππ is 17. Exercise 19 a. Because a circular clock is divided into 12 congruent sections, the measure of each arc is 360 β 12 = 30 β . b. When the time is 7:00, the hour hand points number 7 and the minute hand points number 12. There are 5 sections between number 7 and number 12, so the measure of minor arc is 12 β 30 β = 360 β . c. The arc is congruent if there are 5 sections between the hour hand and the minute hand. If the minute hand points number 12, so the hour hand should point number 5 and the time is 5:00.
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