Question
in the diagram below of Circle of chord AB bisects chord CD at E. If AE equals 8 and BE equals 9 find the length of CD in simplest radical form
Answers
bobpursley
http://www.mathopenref.com/chordsintersecting.html
AB*BE=CE*DE
72=CE*DE but CE=DE=CD/2 (it was bisected)
72=(CD/2)^2
CD=you can do it from here.
AB*BE=CE*DE
72=CE*DE but CE=DE=CD/2 (it was bisected)
72=(CD/2)^2
CD=you can do it from here.