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EXAMPLE 4

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.

austin-morris
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EXAMPLE 4

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  1. 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.

  2. 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.

  3. 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

  4. 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

  5. 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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