EXAMPLE 4 Use Theorem 10.6

SOLUTION

Chords QR and ST are congruent, so by Theorem 10.6 they are equidistant from C. Therefore, CU = CV.

CU = CV

2x = 5x – 9

x = 3

So, CU = 2x = 2(3) = 6.

Use Theorem 10.6.

Substitute.

Solve for x.

In the diagram of C, QR = ST = 16. Find CU.

Since CU = CV. Therefore Chords QR and ST are equidistant from center and from theorem 10.6 QR is congruent to ST

SOLUTION

GUIDED PRACTICE for Example 4

QR = STQR = 32

Use Theorem 10.6.

Substitute.

6. QR

In the diagram in Example 4, suppose ST = 32, and CU = CV = 12. Find the given length.

Since CU is the line drawn from the center of the circle to the chord QR it will bisect the chord.

SOLUTION

GUIDED PRACTICE for Example 4

QU = 16

Substitute.

7. QU

2So QU = QR1

2So QU = (32)1

In the diagram in Example 4, suppose ST = 32, and CU = CV = 12. Find the given length.

Join the points Q and C. Now QUC is right angled triangle. Use the Pythagorean Theorem to find the QC which will represent the radius of the C

SOLUTION

GUIDED PRACTICE for Example 4

8. The radius of C

In the diagram in Example 4, suppose ST = 32, and CU = CV = 12. Find the given length.

SOLUTION

GUIDED PRACTICE for Example 4

In the diagram in Example 4, suppose ST = 32, and CU = CV = 12. Find the given length.8. The radius of C

So QC2 = 162 + 122

So QC2 = 256 + 144So QC2 = 400So QC = 20

So QC2 = QU2 + CU2 By Pythagoras Theorem

Substitute

Square

Add

Simplify

ANSWER The radius of C = 20