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10.5 Chord Length When two chords intersect in the interior of a circle, each chord is divided into segments. Theorem: If two chords intersect in the interior.

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Presentation on theme: "10.5 Chord Length When two chords intersect in the interior of a circle, each chord is divided into segments. Theorem: If two chords intersect in the interior."— Presentation transcript:

1 10.5 Chord Length When two chords intersect in the interior of a circle, each chord is divided into segments. Theorem: If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. A C B D E Therefore: AE • EB = CE • ED

2 PQ: external secant segment
Examples: find x. A C B D 18 9 12 x Q S P T R 9 6 3 x 3 • 6 = 9 • x 18 = 9x 2 = x 12 • 9 = 18 • x 108 = 18x 6 = x Remember: P Q R S PS: tangent segment PR: secant segment PQ: external secant segment

3 Theorem: If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment. E C D A B EA • EB = EC • ED

4 Theorem: If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the length of the secant segment and the length of its external segment equals the square of the length of the tangent segment. E C D A (EA)2 = EC • ED

5 Examples: find the value of x
R P Q S T 11 9 10 x F D G E H 10 11 12 x 11( ) = 12( 12 + x) 11(21) = x 231 = x 87 = 12x 7.25 = x

6 Doesn’t Factor: Use Quadratic Formula!!!!
B C D 5 x 4 Examples: B D C 30 x 24 A 52 = x (x + 4) 25 = x2 + 4x x2 + 4x – 25 = 0 Doesn’t Factor: Use Quadratic Formula!!!! 302 = x (x + 24) 900 = x2 + 24x x2 + 24x – 900 = 0 Use quadratic formula!!!


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