FeatureLesson

GeometryGeometry

LessonMain

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.

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Lesson 12-3

Inscribed AnglesInscribed Angles

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5. Does a diagonal of quadrilateral ABCD intersect the center of the circle? Explain how you can tell.

Lesson Quiz

12-4

FeatureLesson

GeometryGeometry

LessonMain

(For help, go to Lessons 12-1 and 12-3.)

Lesson 12-4

In the diagram at the right, FE and FD aretangents 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

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Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Check Skills You’ll Need

Check Skills You’ll Need

12-4

FeatureLesson

GeometryGeometry

LessonMain

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 123 + 2m EAD = 180 2m EAD = 180 – 123 = 57 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.

Solutions

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Check Skills You’ll Need

12-4

FeatureLesson

GeometryGeometry

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Lesson 12-4

Angle Measures and Segment LengthsAngle 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: a2 + b2 = c2 CE 2 + EF 2 = CF 2 42 + 22 = CF 2 16 + 4 = CF 2

20 = CF 2 CF = 20 = 2 5 4.5

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 = 360 m EFD + 90 + 57 + 90 = 360m EFD + 237 = 360 m EFD = 360 – 237 = 123

Check Skills You’ll Need

12-4

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Notes

12-4

A secant is a line that intersects a circle at two points.

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Notes

12-4

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Notes

12-4

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Notes

12-4

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Notes

12-4

FeatureLesson

GeometryGeometry

LessonMain

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Notes

12-4

FeatureLesson

GeometryGeometry

LessonMain

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 12-11 (2)).

12

x = 88 Simplify.

a.

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Find the value of the variable.

Additional Examples

12-4

Finding Angle Measures

FeatureLesson

GeometryGeometry

LessonMain

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 12-11 (1)).

12

76 = x Multiply each side by 2.

b.

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

(continued)

94 = x + 56 Distributive Property12

38 = x Subtract.12

Quick Check

Additional Examples

12-4

FeatureLesson

GeometryGeometry

LessonMain

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 LengthsAngle Measures and Segment Lengths

Additional Examples

12-4

Real-World Connection

FeatureLesson

GeometryGeometry

LessonMain

72 = 180 – x Distributive Propertyx + 72 = 180 Solve for x.

x = 108A 108° arc will be in the advertising agency’s photo.

72 = [(360 – x) – x] Substitute.12

72 = (360 – 2x) Simplify.12

Then mXAY = 360 – x.Let mXY = x.

(continued)

Lesson 12-4

Angle Measures and Segment LengthsAngle 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 12-11 (2)).

12

Quick Check

Additional Examples

12-4

FeatureLesson

GeometryGeometry

LessonMain

5 • x = 3 • 7 Along a line, the product of the lengths of two segments from a point to a circle is constant (Theorem 12-12 (1)).

5x = 21 Solve for x.

x = 4.2

Find the value of the variable.

a.

8(y + 8) = 152 Along a line, the product of the lengths of two segments from a point to a circle is

constant (Theorem 12-12 (3)).

8y + 64 = 225 Solve for y.

8y = 161

y = 20.125

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

b.

Quick Check

Additional Examples

12-4

Finding Segment Lengths

FeatureLesson

GeometryGeometry

LessonMain

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.

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 = 50 • 200 Along a line, the product of the lengths of the two segments from a point to a circle is constant (Theorem 12-12 (1)).

x2 = 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 LengthsAngle 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

FeatureLesson

GeometryGeometry

LessonMain

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

22

15.5

a = 60; b = 28

24.

Lesson 12-4

Angle Measures and Segment LengthsAngle Measures and Segment Lengths

Lesson Quiz

12-4