# Thread: Super Urgent Circles Help!

1. ## Super Urgent Circles Help!

In the figure above, CD is the diameter which meets the chord AB in E such that AE=BE=4cm. If CE = 3cm, find the radius of the circle.

2. Originally Posted by Ruler of Hell

In the figure above, CD is the diameter which meets the chord AB in E such that AE=BE=4cm. If CE = 3cm, find the radius of the circle.

Hello,

I've modified your drawing (see attachment)
You are dealing actually with a right triangle. The height divide the hypotenuse into 2 parts, called p and q. Then the product of these parts equals the square of the height (height theorem of a right triangle(?))

$\displaystyle h^2=p \cdot q$. Plug in the values you know: h = 4, q = 3

$\displaystyle 4^2=p \cdot 3 \Longleftrightarrow p = \frac{16}{3}$. Therefore the hypotenuse c = p + q:

$\displaystyle c = \frac{25}{3}$

EB

3. Hello, Ruler of Hell!

Another approach . . .
Code:
              * * *
*           *
*               *
*                 *

*         O         *
*         *         *
*       / | \       *
/   |   \R
*  /    E|  4  \  *
* - - - + - - - *A
*     |3    *
* * *
D

We have: .$\displaystyle OA = OD = R$ . and: .$\displaystyle OE = R - 3$

In right triangle $\displaystyle OEA\!:\;\;(R-3)^2 + 4^2 \:=\:R^2\quad\Rightarrow\quad R^2 - 6R + 9 + 16 \:=\:R^2$

Therefore: .$\displaystyle 6R \,=\,25\quad\Rightarrow\quad\boxed{R\,=\,\frac{25} {6}}$

4. @Soroban: After trying a lot, I also got the same answer.. Thanks a lot!

Thanks both of you people!

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# in a circle CD is the diameter which meets the chord AB in E such that AE=BE=4cm.If CE=3cm.Find the radius of the circle

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