Let B, C, and D be points on a circle. Let BC and the tangent to the circle at D intersect at A. If AB = 4, AD = 8, and AC is perperpendicular to AD, then find CD.
I do not have time to do it all right now but it is easy enough anyway.
use the rule below:
Let BC= x
set up a quadric equation.
Find x using the quadratic formula
Find CD using the ythagorean theorum.
All done
The rule was copied from this site:
We have, using the secant-tangent theorem, that
AC * AB = AD^2
(BC + AB) * AB = AD^2
(BC + 4) * 4 = 8^2
4BC + 16 = 64 divide through by 4
BC + 4 = 16 subtract 4 from each side
BC = 12
So AC = (12 + 4) = 16
Using the Pythagorean Theorem to find CD, we have
AC * AB = AD^2
AC * 4 = 64
AC = 16
Therefore
AC^2 + AD^2 = CD^2
sqrt (AC^2 + AD^2) = CD
sqrt (16^2 + 8^2) = CD
sqrt ( 256 + 64) =
sqrt ( 64 [ 4 + 1 ] ) =
8 sqrt (5) ≈ 17.889
Let B, C, and D be points on a circle. Let BC and the tangent to the circle at D intersect at A.
If AB = 4,
AD = 8,
and AC is perperpendicular to AD,
then find CD.
AD2=AB⋅ACsecant-tangent theoremAC=AD2ABCD2=AD2+AC2pythagorasCD2=AD2+AC2|AC=AD2ABCD2=AD2+(AD2AB)2CD2=AD2+AD4AB2CD2=AD2⋅(1+AD2AB2)CD2=AD2⋅[1+(ADAB)2]CD=AD⋅√1+(ADAB)2
CD = ?
CD=AD⋅√1+(ADAB)2|AB=4AD=8CD=8⋅√1+(84)2CD=8⋅√1+22CD=8⋅√5CD=17.88854382