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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.

 Mar 23, 2017
 #1
avatar+118703 
+2

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   laugh

 

The rule was copied from this site:

http://topdrawer.aamt.edu.au/Geometric-reasoning/Big-ideas/Circle-geometry/Angle-and-chord-properties

 

 Mar 24, 2017
edited by Melody  Mar 24, 2017
 #2
avatar+130477 
+2

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

 

 

 

cool cool cool

 Mar 24, 2017
edited by CPhill  Jun 2, 2017
 #3
avatar+26397 
0

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=ABACsecant-tangent theoremAC=AD2ABCD2=AD2+AC2pythagorasCD2=AD2+AC2|AC=AD2ABCD2=AD2+(AD2AB)2CD2=AD2+AD4AB2CD2=AD2(1+AD2AB2)CD2=AD2[1+(ADAB)2]CD=AD1+(ADAB)2

 

CD = ?

CD=AD1+(ADAB)2|AB=4AD=8CD=81+(84)2CD=81+22CD=85CD=17.88854382

 

 

laugh

 Mar 24, 2017

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