Find the lengths of segments formed by lines that intersect circles

Holt McDougal Geometry

11-6Segment Relationships in Circles

Find the lengths of segments formed by lines that intersect circles.

Use the lengths of segments in circles to solve problems.

Objectives

Holt McDougal Geometry

11-6Segment Relationships in Circles

Holt McDougal Geometry

11-6Segment Relationships in Circles

Holt McDougal Geometry

11-6Segment Relationships in Circles

Holt McDougal Geometry

11-6Segment Relationships in Circles

Holt McDougal Geometry

11-6Segment Relationships in Circles

Example 1: Applying the Chord-Chord Product Theorem

Find the value of x.

EJ JF = GJ JH

10(7) = 14(x)

70 = 14x

5 = x

EF = 10 + 7 = 17

GH = 14 + 5 = 19

J

Holt McDougal Geometry

11-6Segment Relationships in Circles

Check It Out! Example 1

Find the value of x and the length of each chord.

DE EC = AE EB

8(x) = 6(5)

8x = 30

x = 3.75

AB = 6 + 5 = 11

CD = 3.75 + 8 = 11.75

Holt McDougal Geometry

11-6Segment Relationships in Circles

Check It Out! Example 2

Suppose the length of chord AB that the archeologists measured was 12 in. Find QR.

AQ QB = PQ QR

6(6) = 3(QR)

12 = QR

12 + 3 = 15 = PR

6 in.

Holt McDougal Geometry

11-6Segment Relationships in Circles

Find the value of x and the length of each secant segment.

Example 3: Applying the Secant-Secant Product Theorem

16(7) = (8 + x)8

112 = 64 + 8x

48 = 8x

6 = x

ED = 7 + 9 = 16

EG = 8 + 6 = 14

Holt McDougal Geometry

11-6Segment Relationships in Circles

Check It Out! Example 3

Find the value of z and the length of each secant segment.

39(9) = (13 + z)13

351 = 169 + 13z

182 = 13z

14 = z

LG = 30 + 9 = 39

JG = 14 + 13 = 27

Holt McDougal Geometry

11-6Segment Relationships in Circles

Example 4: Applying the Secant-Tangent Product Theorem

Find the value of x.

The value of x must be 10 since it represents a length.

ML JL = KL2

20(5) = x2

100 = x2

±10 = x

Holt McDougal Geometry

11-6Segment Relationships in Circles

Check It Out! Example 4

Find the value of y.

DE DF = DG2

7(7 + y) = 102

49 + 7y = 100

7y = 51