19,125 views
36 votes
36 votes
Help please ASAP thank you

Help please ASAP thank you-example-1
User Xyand
by
3.1k points

1 Answer

13 votes
13 votes

Question 1

If we let
FE=x, then
DE=x-5.

Also, as
\overline{DF} bisects
\overline{BC}, this means
BE=EC=6.

Thus, by the intersecting chords theorem,


6(6)=x(x-5)\\\\36=x^2 - 5x\\\\x^2 - 5x-36=0\\\\(x-9)(x+4)=0\\\\x=-4, 9

However, as distance must be positive, we only consider the positive case, meaning FE=9

Question 2

If we let CE=x, then because AB bisects CD, CE=ED=x.

We also know that since FB=17, the radius of the circle is 17. So, this means that the diameter is 34, and as AE=2, thus means EB=32.

By the intersecting chords theorem,


2(32)=x^2\\\\64=x^2\\\\x=-8, 8

However, as distance must be positive, we only consider the positive case, meaning CE=8

User Onyeka
by
2.4k points