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FeatureLesson Geometry Lesson Main In the diagram below, O circumscribes quadrilateral ABCD and is inscribed in quadrilateral XYZW. 1.Find the measure of each inscribed angle. 2.Find m DCZ. 3.Are XAB and XBA congruent? Explain. 4.Find the angle measures in quadrilateral XYZW. Yes; each is formed by a tangent and a chord, and they intercept the same arc. 45 m A = 100; m B = 75; m C = 80; m D = 105 m X = 80; m Y = 70; m Z = 90; m W = 120 No; the diagonal would be a diameter of O and the inscribed angle would be a right angle, which was not found in Exercise 1 above.. Lesson 12-3 Inscribed Angles. 5.Does a diagonal of quadrilateral ABCD intersect the center of the circle? Explain how you can tell. Lesson Quiz 12-4

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FeatureLesson Geometry Lesson Main (For help, go to Lessons 12-1 and 12-3.) Lesson 12-4 In the diagram at the right, FE and FD are tangents to C. Find each arc measure, angle measure, or length. 1. mDE2. mAED3. mEBD 4. m EAD5. m AEC6. CE 7. DF8. CF9. m EFD. Angle Measures and Segment Lengths Check Skills Youll Need 12-4

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FeatureLesson Geometry Lesson Main 1.mDE = m DCE = 57 2.mAED = m ACD = mEBD = 360 – m DCE = 360 – 57 = Since radii of the same circle are congruent, ACE is isosceles and m ACE = 180 – 57 = 123. The sum of the angles of a triangle is 180 and the base angles of an isosceles triangle are congruent, so m EAD = 180 2m EAD = 180 – 123 = 57 m EAD = 57 ÷ 2 = From Exercise 4, m AEC = m EAD = Since all radii of the same circle are congruent, CE = CD = 4. Solutions Lesson 12-4 Angle Measures and Segment Lengths Check Skills Youll Need 12-4

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FeatureLesson Geometry Lesson Main Lesson 12-4 Angle Measures and Segment Lengths Solutions (continued) 7.By Theorem 11-3, two tangents tangent to a circle from a point outside the circle are congruent, so DF = EF = 2. 8.Draw CF. Since FE is tangent to the circle, FE CE. By def. of, CEF is a right angle. By def. of right angle, m CEF = 90, so CEF is a right triangle. From Exercise 6, CE = 4. Also, EF = 2. Use the Pythagorean Theorem: a 2 + b 2 = c 2 CE 2 + EF 2 = CF = CF = CF 2 20 = CF 2 CF = 20 = CEFD is a quadrilateral, so the sum of its angles is 360. From Exercise 8, m ACE = 90. Similarly, m CDF = 90. Also, m DCE = 57. So, m EFD + m FEC + m DCE + m CDF = 360 m EFD = 360 m EFD = 360 m EFD = 360 – 237 = 123 Check Skills Youll Need 12-4

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FeatureLesson Geometry Lesson Main Lesson 12-4 Angle Measures and Segment Lengths Notes 12-4 A secant is a line that intersects a circle at two points.

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FeatureLesson Geometry Lesson Main Lesson 12-4 Angle Measures and Segment Lengths Notes 12-4

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FeatureLesson Geometry Lesson Main Lesson 12-4 Angle Measures and Segment Lengths Notes 12-4

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FeatureLesson Geometry Lesson Main Lesson 12-4 Angle Measures and Segment Lengths Notes 12-4

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FeatureLesson Geometry Lesson Main Lesson 12-4 Angle Measures and Segment Lengths Notes 12-4

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FeatureLesson Geometry Lesson Main Lesson 12-4 Angle Measures and Segment Lengths Notes 12-4

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FeatureLesson Geometry Lesson Main x = (268 – 92)The measure of an angle formed by two lines that intersect outside a circle is half the difference of the measures of the intercepted arcs (Theorem (2)) x = 88Simplify. a. Lesson 12-4 Angle Measures and Segment Lengths Find the value of the variable. Additional Examples 12-4 Finding Angle Measures

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FeatureLesson Geometry Lesson Main 94 = (x + 112)The measure of an angle formed by two lines that intersect inside a circle is half the sum of the measures of the intercepted arcs (Theorem (1)) = xMultiply each side by 2. b. Lesson 12-4 Angle Measures and Segment Lengths (continued) 94 = x + 56Distributive Property = xSubtract Quick Check Additional Examples 12-4

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FeatureLesson Geometry Lesson Main An advertising agency wants a frontal photo of a flying saucer ride at an amusement park. The photographer stands at the vertex of the angle formed by tangents to the flying saucer. What is the measure of the arc that will be in the photograph? In the diagram, the photographer stands at point T. TX and TY intercept minor arc XY and major arc XAY. Lesson 12-4 Angle Measures and Segment Lengths Additional Examples 12-4 Real-World Connection

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FeatureLesson Geometry Lesson Main 72 = 180 – x Distributive Property x + 72 = 180 Solve for x. x = 108 A 108° arc will be in the advertising agencys photo. 72 = [(360 – x) – x] Substitute = (360 – 2x) Simplify Then mXAY = 360 – x. Let mXY = x. (continued) Lesson 12-4 Angle Measures and Segment Lengths m T = (mXAY – mXY)The measure of an angle formed by two lines that intersect outside a circle is half the difference of the measures of the intercepted arcs (Theorem (2)) Quick Check Additional Examples 12-4

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FeatureLesson Geometry Lesson Main 5 x = 3 7Along a line, the product of the lengths of two segments from a point to a circle is constant (Theorem (1)). 5x = 21 Solve for x. x = 4.2 Find the value of the variable. a. 8(y + 8) = 15 2 Along a line, the product of the lengths of two segments from a point to a circle is constant (Theorem (3)). 8y + 64 = 225 Solve for y. 8y = 161 y = Lesson 12-4 Angle Measures and Segment Lengths b. Quick Check Additional Examples 12-4 Finding Segment Lengths

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FeatureLesson Geometry Lesson Main Because the radius is 125 ft, the diameter is = 250 ft. The length of the other segment along the diameter is 250 ft – 50 ft, or 200 ft. A tram travels from point A to point B along the arc of a circle with a radius of 125 ft. Find the shortest distance from point A to point B. x x = Along a line, the product of the lengths of the two segments from a point to a circle is constant (Theorem (1)). x 2 = 10,000 Solve for x. x = 100 The shortest distance from point A to point B is 200 ft. Lesson 12-4 Angle Measures and Segment Lengths The perpendicular bisector of the chord AB contains the center of the circle. Quick Check Additional Examples 12-4 Real-World Connection

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FeatureLesson Geometry Lesson Main Use M for Exercises 1 and 2. 1.Find a. 2.Find x.. Use O for Exercises 3–5. 3.Find a and b. 4.Find x to the nearest tenth. 5.Find the diameter of O a = 60; b = Lesson 12-4 Angle Measures and Segment Lengths Lesson Quiz 12-4

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