I wonder how quick you can work this out...
http://img297.imageshack.us/img297/2415/mathwq0.th.png
Printable View
I wonder how quick you can work this out...
http://img297.imageshack.us/img297/2415/mathwq0.th.png
Hello, Geometor!
Quote:
PQ and RS are parallel chords 6 cm apart.
PQ = 10 cm and RS = 14 cm.
(a) What is the shortest distance from the center to chord PQ?
(b) Find the radius of the circle.Code:* * *
* 5 T 5 *
P* - - - * - - - *Q
* | *
| x
* | *
* *O *
* | *
| 6-x
R*- - - - * - - - -*S
* 7 U 7 *
* *
* * *
Draw diameter TU through center O, perpendicular to PQ and RS.
. . Then: .TQ = 5, US = 7.
Draw radii OP = OQ = OR = OS = r.
Let x = TO, then 6 - x = OU.
In right triangle ORQ: .x² + 5² .= .r² .[1]
In right triangle OUS: .(6-x)² + 7² .= .r² .[2]
Equate [1] and [2]: .x² + 25 .= .36 - 12x + x² + 49 . → . 12x .= .60
Therefore: .(a) .x = 5
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
Substitute into [1]: .5² + 5² .= .r² . . → . . (b) .r = 5√2
alright, I see. Thanks!