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.
Please help... It's super urgent super super urgently needed... PLEASE
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
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}} $