## Tuesday, August 7, 2007

### Geometry Chapter 10.5 Segment Lengths in Circles

Chapter 10.5 Segment Lengths in Circles:

I) Vocabulary:
A) when two chords intersect in the interior of a circle, each chord is divided into two segments which are called segments of a chord.
B) Tangent Segment - is tangent to a circle at an endpoint.
C) Secant Segment - is a segment that has an endpoint on a circle.
D) External Segment - is part of a secant segment from an external point to the circle. It is does not have any part of the segment in the interior part of the circle.

II) Theorem:
A) 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.
Example: Given circle P with chord AB and chord CD that intersect at point E, AE = 9, ED = 12, CE = 15, EB = x, solve for x.
(9)(x) = (15)(12)
9x = 180
x = 20
B) 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.
Example: Given circle A with secant RPQ with PQ being a chord on the circle, and secant RST with ST being a chord on the circle, RP = 9, PQ = 11, RS = 10 and ST = x, solve for x.
(RP)(RQ) = (RS)(RT)
9 (11+9) = 10 (x + 10)
(9)(20) = 10x + 100
180 = 10x + 100
80 = 10x
8 = x
Example: Given circle A with secant RPQ with PQ being a chord on the circle, and secant RST with ST being a chord on the circle, RP = x, PQ = 65, RS = 15, and ST = 35. Solve for x.
(RP)(RQ) = (RS)(RT)
x (x + 65) = (15)(15 + 35)
x2 + 65x = 750
x2 + 65x - 750 = 0
(x + 75)(x - 10) = 0
x = - 75, x = 10
Since a length cannot be a negative number, the only answer would then be 10 units.
C) If a secant segment and a tangent segment share an endpoint ooutside 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.
Example: Given circle A with secant ECD with CD being a chord on the circle, and tangent EA, EC = x, CD = 12, and EA = 8, solve for x.
(x)(x + 12) = 82
x2 + 12x = 64
x2 + 12x - 64 = 0
(x + 16)(x - 4) = 0
x = -16, x = 4
Since a length cannot be a negative number, the only answer would then be 4 units.
Example: Given circle A with secant ECD with CD being a chord on the circle, and tangent EA, EC = 12, CD = 36, and EA = x, solve for x.
(12)(12 + 36) = x2
(12)(48) = x2
576 = x2
24 = x