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# DONT GET THIS. HELP PLZ

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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.

Guest Mar 23, 2017
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#1
+91237
+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

Find x using the quadratic formula

Find CD using the ythagorean theorum.

All done

The rule was copied from this site:

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

Melody  Mar 24, 2017
edited by Melody  Mar 24, 2017
#2
+79846
+2

We have, using the secant-tangent theorem, that

(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  *  4  = 64

AC  = 16

Therefore

sqrt (AC^2 + AD^2)  =  CD

sqrt  (16^2  + 8^2)  =   CD

sqrt ( 256 + 64)  =

sqrt ( 64 [ 4 + 1 ] )  =

8 sqrt (5)    ≈  17.889

CPhill  Mar 24, 2017
edited by CPhill  Jun 2, 2017
#3
+18777
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,

and AC is perperpendicular to AD,

then find CD.

$$\begin{array}{|rcll|} \hline AD^2 &=& AB \cdot AC & \text{secant-tangent theorem} \\ \mathbf{AC} & \mathbf{=} & \mathbf{ \frac{AD^2}{AB} }\\\\ CD^2 &=& AD^2 + AC^2 & \text{pythagoras} \\ CD^2 &=& AD^2 + AC^2 & | \quad AC= \frac{AD^2}{AB} \\ CD^2 &=& AD^2 + \left(\frac{AD^2}{AB}\right)^2 \\ CD^2 &=& AD^2 + \frac{AD^4}{AB^2} \\ CD^2 &=& AD^2\cdot \left( 1+\frac{AD^2}{AB^2} \right) \\ CD^2 &=& AD^2\cdot \left[1+\left(\frac{AD}{AB}\right)^2 \right] \\ \mathbf{CD} & \mathbf{=} & \mathbf{AD \cdot \sqrt{1+\left(\frac{AD}{AB}\right)^2 } }\\ \hline \end{array}$$

CD = ?

$$\begin{array}{|rcll|} \hline \mathbf{CD} & \mathbf{=} & \mathbf{AD \cdot \sqrt{1+\left(\frac{AD}{AB}\right)^2 } } \quad | \quad AB=4 \quad AD=8 \\ CD & = & 8 \cdot \sqrt{1+\left(\frac{8}{4}\right)^2 } \\ CD & = & 8 \cdot \sqrt{1+ 2^2 } \\ CD & = & 8 \cdot \sqrt{5 } \\ CD & = & 17.88854382 \\ \hline \end{array}$$

heureka  Mar 24, 2017

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