9.6. Segments of Chords, Secants, and Tangents http://www.ck12.org
b)
Solution:Again, we can use Theorem 9-13. Set up an equation and solve forx.
a) 8· 24 = ( 3 x+ 1 )· 12
192 = 36 x+ 12
180 = 36 x
5 =x
b) 32· 21 = (x− 9 )(x− 13 )
672 =x^2 − 22 x+ 117
0 =x^2 − 22 x− 555
0 = (x− 37 )(x+ 15 )
x= 37 ,− 15
However,x 6 =−15 because length cannot be negative, sox=37.
Segments from Secants
In addition to forming an angle outside of a circle, the circle can divide the secants into segments that are proportional
with each other.
If we draw in the intersecting chords, we will have two similar triangles.