Chapter 9.1 The Pythagorean theorem

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 ⊙ 𝑃𝑃 . Exercise 2 One of the radius of the circle is 𝑃𝑃𝑃𝑃 . Exercise 3 The diameter of the circle is 𝑃𝑃𝐾𝐾 . Exercise 4 The chord of the circle is 𝐽𝐽𝐽𝐽 . Exercise 5 The secant of the circle is 𝑄𝑄𝑄𝑄 ⃖���⃗ . Exercise 6 The tangent of the circle is 𝑄𝑄𝑄𝑄 ⃖����⃗ . Exercise 7 By the Pythagorean Theorem, ( 𝑥𝑥 + 9) 2 = 15 2 + 𝑥𝑥 2 𝑥𝑥 2 + 18 𝑥𝑥 + 81 = 225 + 𝑥𝑥 2 18 𝑥𝑥 + 81 = 225 18 𝑥𝑥 = 144 𝑥𝑥 = 8 So, the value of x is 8. Exercise 8 By External Tangent Congruence Theorem, 6 𝑥𝑥 − 3 = 3 𝑥𝑥 + 18 6 𝑥𝑥 − 3 𝑥𝑥 = 18 + 3 3 𝑥𝑥 = 21 𝑥𝑥 = 7 So, the value of x is 5. Exercise 9 𝐴𝐴𝐴𝐴 is a diameter, so 𝑚𝑚𝐴𝐴𝐴𝐴𝐴𝐴 � = 180 ∘ . Then, 𝑚𝑚𝐴𝐴𝐴𝐴 � = 180 ∘ − 𝑚𝑚𝐴𝐴𝐴𝐴 � 𝑚𝑚𝐴𝐴𝐴𝐴 � = 180 ∘ − 36 ∘ 𝑚𝑚𝐴𝐴𝐴𝐴 � = 144 ∘ Then, 𝐴𝐴𝐴𝐴 � is a minor arc and 𝑚𝑚𝐴𝐴𝐴𝐴 � = 144 ∘ . Exercise 10 𝐴𝐴𝐴𝐴 is a diameter, so 𝑚𝑚𝐴𝐴𝐴𝐴𝐴𝐴 � = 180 ∘ . Then, 𝑚𝑚𝐵𝐵𝐵𝐵 � = 180 ∘ − 𝑚𝑚𝐴𝐴𝐵𝐵 � − 𝑚𝑚𝐵𝐵𝐴𝐴 � 𝑚𝑚𝐵𝐵𝐵𝐵 � = 180 ∘ − 67 ∘ − 70 ∘ 𝑚𝑚𝐵𝐵𝐵𝐵 � = 43 ∘

Then, 𝐵𝐵𝐵𝐵 � is a minor arc and 𝑚𝑚𝐵𝐵𝐵𝐵 � = 43 ∘ . Exercise 11 𝑚𝑚𝐴𝐴𝐵𝐵 � = 𝑚𝑚𝐴𝐴𝐵𝐵 � + 𝑚𝑚𝐵𝐵𝐵𝐵 � 𝑚𝑚𝐴𝐴𝐵𝐵 � = 67 ∘ + 43 ∘ 𝑚𝑚𝐴𝐴𝐵𝐵 � = 110 ∘ Then, 𝐴𝐴𝐵𝐵 � is a minor arc and 𝑚𝑚𝐴𝐴𝐵𝐵 � = 110 ∘ . Exercise 12 𝐴𝐴𝐴𝐴 is a diameter, so 𝑚𝑚𝐴𝐴𝐵𝐵𝐴𝐴 � = 180 ∘ . Then, 𝐴𝐴𝐵𝐵𝐴𝐴 � is a semicircle and 𝑚𝑚𝐴𝐴𝐵𝐵𝐴𝐴 � = 180 ∘ . Exercise 13 𝐴𝐴𝐴𝐴 is a diameter, so 𝑚𝑚𝐴𝐴𝐵𝐵𝐴𝐴 � = 180 ∘ . Then, 𝑚𝑚𝐴𝐴𝐵𝐵𝐴𝐴 � = 𝑚𝑚𝐴𝐴𝐵𝐵𝐴𝐴 � + 𝑚𝑚𝐴𝐴𝐴𝐴 � 𝑚𝑚𝐴𝐴𝐵𝐵𝐴𝐴 � = 180 ∘ + 36 ∘ 𝑚𝑚𝐴𝐴𝐵𝐵𝐴𝐴 � = 216 ∘ Then, 𝐴𝐴𝐵𝐵𝐴𝐴 � is a major arc and 𝑚𝑚𝐴𝐴𝐵𝐵𝐴𝐴 � = 216 ∘ . Exercise 14 𝑚𝑚𝐵𝐵𝐴𝐴𝐵𝐵 � = 360 ∘ − 𝑚𝑚𝐵𝐵𝐵𝐵 � 𝑚𝑚𝐵𝐵𝐴𝐴𝐵𝐵 � = 360 ∘ − 43 ∘ 𝑚𝑚𝐵𝐵𝐴𝐴𝐵𝐵 � = 317 ∘ Then, 𝐵𝐵𝐴𝐴𝐵𝐵 � is a major arc and 𝑚𝑚𝐵𝐵𝐴𝐴𝐵𝐵 � = 317 ∘ . 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, 𝑚𝑚𝑃𝑃𝑄𝑄 � = 𝑚𝑚𝑄𝑄𝑄𝑄 � . But, 𝑃𝑃𝑄𝑄 � lies on the circle that the radius is 7 and 𝑄𝑄𝑄𝑄 � 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, 𝐴𝐴𝐸𝐸𝐸𝐸 � ≅ 𝐵𝐵𝐴𝐴𝐴𝐴 � . So, 𝑚𝑚𝐴𝐴𝐸𝐸𝐸𝐸 � = 𝑚𝑚𝐵𝐵𝐴𝐴𝐴𝐴 � 𝑚𝑚𝐴𝐴𝐸𝐸𝐸𝐸 � = 360 ∘ − 𝑚𝑚𝐵𝐵𝐵𝐵 � − 𝑚𝑚𝐵𝐵𝐴𝐴 � 𝑚𝑚𝐴𝐴𝐸𝐸𝐸𝐸 � = 360 ∘ − 150 ∘ − 110 ∘ 𝑚𝑚𝐴𝐴𝐸𝐸𝐸𝐸 � = 100 ∘ So, the measure of the red arcs in ⊙ 𝑄𝑄 is 100 ∘ . 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, 𝑃𝑃𝐴𝐴 = �𝑃𝑃𝐽𝐽 2 + 𝐽𝐽𝐴𝐴 2 𝑃𝑃𝐴𝐴 = � (3 𝑥𝑥 − 1) 2 + 15 2 𝑃𝑃𝐴𝐴 = � (3 ⋅ 3 − 1) 2 + 15 2 𝑃𝑃𝐴𝐴 = � (9 − 1) 2 + 15 2 𝑃𝑃𝐴𝐴 = � 8 2 + 15 2 𝑃𝑃𝐴𝐴 = √ 64 + 225 𝑃𝑃𝐴𝐴 = √ 289 𝑃𝑃𝐴𝐴 = 17 𝑃𝑃𝐴𝐴 is the radius of ⊙ 𝑃𝑃 , so the radius of ⊙ 𝑃𝑃 is 17. Exercise 19 a. Because a circular clock is divided into 12 congruent sections, the measure of each arc is 360 ∘ 12 = 30 ∘ . 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 ⋅ 30 ∘ = 360 ∘ . 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.