hlambach
  • hlambach
BC is tangent to circle A at B and to circle D at C (not drawn to scale). AB = 5, BC = 30, and DC = 3. Find AD to the nearest tenth. Rest in comments.
Mathematics
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SOLVED
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chestercat
  • chestercat
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hlambach
  • hlambach
I have NO idea how to solve this... Please help!
1 Attachment
Jack1
  • Jack1
all i can think of is pythag theorum
Jack1
  • Jack1
|dw:1370718164234:dw|

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More answers

Jack1
  • Jack1
|dw:1370718242350:dw|
Jack1
  • Jack1
ab and cd are parallel so ab - cd = bottom of ur new triangle
hlambach
  • hlambach
huum... how come? i
Jack1
  • Jack1
bc and d...to the middle ish of ab.... are parallel and equal
Jack1
  • Jack1
ab and cd are parallel because tangent make right angles with the radius to them
Jack1
  • Jack1
as their both rightangles to the same line = parallel
hlambach
  • hlambach
so the answer is ab-cd = ad?
Jack1
  • Jack1
not what ur thinking...
hlambach
  • hlambach
so 5-3 = AD?
hlambach
  • hlambach
taht doesn't make sense
Jack1
  • Jack1
ad = sqrt 2^2 + 30^2
hlambach
  • hlambach
whoops i forgot: A. 31 B. 30.1 C. 30.1 D.30.4
hlambach
  • hlambach
AD = √(2^2) + 30^2
Jack1
  • Jack1
|dw:1370718680940:dw|
Jack1
  • Jack1
SO SQRT 904 =??
Jack1
  • Jack1
|dw:1370718869362:dw| rightangles
hlambach
  • hlambach
ahhh! got it!!
hlambach
  • hlambach
pythagorean theorem!
hlambach
  • hlambach
thanks! i get it now!
Jack1
  • Jack1
always welcome

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