You have a circle....a chord AB is 20 inches long....the distance between the midpoint of AB and the nearest point on the circle is 5
whats the radius?
did you look up what the intersecting chord theorem was? see here
EDIT: Dan's site seems a lot nicer than mine
can you tell us what the solution is now?
No you need to learn how to do research for yourself, you have been given
the name or the result you need. Now it seems improbable that you have not
covered this in class, so it will be in your notes or text book.
Now where I come from text books have indexes, so you could try looking it
up in the index. Also you could, as you appear to have access to the
internet, try typing it into Google. If you don't know what Google is then
follow this: link to Google.
Intersecting Chord Theorem
Let AB and CD be two chords in the same circle that intersect at a point
X. Then:
|AX|*|XB| = |CX|*|XD| ... (1)
(Where |UV| denotes the length of the line segment UV.)
Now look at the diagram I posted. Do you see two intersecting chords?
Have I marked their lengths?
So now label the points as in the statement of the theorem and then
note which lengths appear in the statement of the intersecting chord
theorem and so write (1) in terms of the lengths of these chords.
Now solve the resulting equation for R.
RonL
Hello, stones44!
No fancy theorems needed . . .
You have a circle and a chord AB is 20 inches long.
The distance between the midpoint of AB and the nearest point on the circle is 5.
What is the radius?Code:C * * * * |5 * * 10 | 10 * A*- - - - + - - - -*B * | * R * |R-5 * * * * * O * * * * * * * * * *
Let the radius be $\displaystyle R$.
We have a right triangle with sides: $\displaystyle R-5,\:10,\: R$
$\displaystyle R^2\:=\:(R-5)^2+10^2\quad\Rightarrow\quad R^2 \:=\:R^2 - 10R + 25 + 100$
. . $\displaystyle 10R \:=\:125\quad\Rightarrow\quad R\:=\:12.5$