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**In the diagram below, O circumscribes quadrilateral ABCD **

Inscribed Angles Lesson 12-3 Lesson Quiz . 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. m A = 100; m B = 75; m C = 80; m D = 105 45 Yes; each is formed by a tangent and a chord, and they intercept the same arc. m X = 80; m Y = 70; m Z = 90; m W = 120 5. Does a diagonal of quadrilateral ABCD intersect the center of the circle? Explain how you can tell. 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. . 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Check Skills You’ll Need (For help, go to Lessons 12-1 and 12-3.) In the diagram at the right, FE and FD are tangents to C. Find each arc measure, angle measure, or length. 1. mDE 2. mAED 3. mEBD 4. m EAD 5. m AEC 6. CE 7. DF 8. CF 9. m EFD . Check Skills You’ll Need 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Check Skills You’ll Need Solutions 1. mDE = m DCE = 57 2. mAED = m ACD = 180 3. mEBD = 360 – m DCE = 360 – 57 = 303 4. 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 = m EAD = 180 – 123 = m EAD = 57 ÷ 2 = 28.5 5. From Exercise 4, m AEC = m EAD = 28.5 6. Since all radii of the same circle are congruent, CE = CD = 4. 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Check Skills You’ll Need 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: a2 + b2 = c CE 2 + EF 2 = CF = CF = CF 2 20 = CF CF = = 9. 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 = m EFD = 360 m EFD = m EFD = 360 – 237 = 123 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Notes A secant is a line that intersects a circle at two points. 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Notes 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Notes 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Notes 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Notes 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Notes 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Additional Examples Finding Angle Measures Find the value of the variable. a. 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)). 1 2 x = 88 Simplify. 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Additional Examples (continued) b. 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)). 1 2 94 = x + 56 Distributive Property 1 2 38 = x Subtract. 1 2 Quick Check 76 = x Multiply each side by 2. 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Additional Examples Real-World Connection 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. 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Additional Examples Let mXY = x. (continued) Then mXAY = 360 – x. 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)). 1 2 72 = [(360 – x) – x] Substitute. 1 2 72 = (360 – 2x) Simplify. 1 2 72 = 180 – x Distributive Property x + 72 = Solve for x. x = 108 Quick Check A 108° arc will be in the advertising agency’s photo. 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Additional Examples Finding Segment Lengths Find the value of the variable. a. 5 • x = 3 • 7 Along a line, the product of the lengths of two segments from a point to a circle is constant (Theorem (1)). 5x = Solve for x. x = 4.2 b. 8(y + 8) = 152 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 = Quick Check 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Additional Examples Real-World Connection 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. Quick Check The perpendicular bisector of the chord AB contains the center of the circle. Because the radius is 125 ft, the diameter is 2 • 125 = 250 ft. The length of the other segment along the diameter is 250 ft – 50 ft, or 200 ft. x • x = 50 • 200 Along a line, the product of the lengths of the two segments from a point to a circle is constant (Theorem (1)). x2 = 10, Solve for x. x = 100 The shortest distance from point A to point B is 200 ft. 12-4

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**Angle Measures and Segment Lengths**

Lesson 12-4 Lesson Quiz 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. . 82 a = 60; b = 28 24 15.5 . 22 12-4

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M—06/01/09—HW #75: Pg 654-655: 1-20, skip 9, 16. Pg 821-822: 16—38 even, 39— 45 odd 2) C4) B6) C8) A10) 512) EF = 40, FG = 50, EH = 240 14) (3\2, 2); (5\2)18)

M—06/01/09—HW #75: Pg 654-655: 1-20, skip 9, 16. Pg 821-822: 16—38 even, 39— 45 odd 2) C4) B6) C8) A10) 512) EF = 40, FG = 50, EH = 240 14) (3\2, 2); (5\2)18)

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