in the diagram, AE is tangent to the circle at point a and secant DE intersects the circle at point C and D. the iines intersect

Question

Rama09 [41]

2 years ago

9

in the diagram, AE is tangent to the circle at point a and secant DE intersects the circle at point C and D. the iines intersect

outside of the circle at point E

Mathematics

1 answer:

AlexFokin [52] · Answer 1 · 2023-08-22T17:32:46+03:00

AlexFokin [52]2 years ago

8 0

Recall the secant-tangent theorem, and you have
EA^2 = EC*CD
12^2 = 8*(x+10)
and now ED = EC+CD = 8+x+10

I suspect a typo somewhere in the murk above