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In the diagram of C, QR = ST = 16 . Find CU. Chords QR and ST are congruent, so by Theorem 10.6 they are equidistant from C . Therefore, CU = CV. EXAMPLE 4. Use Theorem 10.6. SOLUTION. CU = CV. Use Theorem 10.6. 2 x = 5 x – 9. Substitute. Solve for x. x = 3.

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slide1
In the diagram of C, QR = ST = 16. Find CU.

Chords QRand STare congruent, so by Theorem 10.6 they are equidistant from C. Therefore, CU = CV.

EXAMPLE 4

Use Theorem 10.6

SOLUTION

CU = CV

Use Theorem 10.6.

2x = 5x – 9

Substitute.

Solve for x.

x =3

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

slide2
In the diagram in Example 4, suppose ST = 32, and CU= CV = 12. Find the given length.

for Example 4

GUIDED PRACTICE

6. QR

SOLUTION

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

QR = ST

Use Theorem 10.6.

QR= 32

Substitute.

slide3
In the diagram in Example 4, suppose ST = 32, and CU= CV = 12. Find the given length.

1

1

SoQU = QR

SoQU = (32)

2

2

for Example 4

GUIDED PRACTICE

7. QU

SOLUTION

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

Substitute.

QU= 16

slide4
In the diagram in Example 4, suppose ST = 32, and CU= CV = 12. Find the given length.

8.

The radius of C

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

for Example 4

GUIDED PRACTICE

SOLUTION

slide5
8.

The radius of C

ANSWER

The radius of C = 20

for Example 4

GUIDED PRACTICE

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

SOLUTION

So QC2 = QU2 + CU2

By Pythagoras Theorem

So QC2 = 162+ 122

Substitute

So QC2 = 256 + 144

Square

So QC2 = 400

Add

So QC = 20

Simplify

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