Thread:

1. Chord length of a circle inscribed in a rectangle

The picture is of a rectangle whose base is greater than its height. A circle is drawn in the rectangle, which is tangent to the top, left, and bottom edges. the diagonal from the bottom left corner to the top right corner meets the circle in points E and F. Express the distance EF in terms of the base and height.

Not sure how to even start this off other than r=h/2

2. Let ABCD be the rectangle where AB is b and BC is h.
Draw a line PFQ parallel to AB.
PF is the diameter of the circle. PEF is a right angled triangle which is similar to BCD. Now compare equivalent sides to find the length of the chord EF in terms of h and b.

3. Originally Posted by sa-ri-ga-ma
Let ABCD be the rectangle where AB is b and BC is h.
Draw a line PFQ parallel to AB.
PF is the diameter of the circle.
...
I don't want to pick at you but in my opinion this is only correct if b = 2h. If you change the ratio of b and h then the line PF parallel to AB is not a diameter of the circle. See attachment.
Attached Thumbnails

4. I tried working it out to the best of my knowledge but I don't really know if it's correct:

I don't even know if that's correct or if I'm going in the right direction

5. Hello Mustard

Welcome to Math Help Forum!

I have an answer, based on the Tangent-Secant Theorem - and a whole lot of algebra. It is:
$EF = \sqrt{\frac{2bh^3}{b^2+h^2}}$
This checks out for two special cases:
When $b = h, EF = h$, which is so, because EF is then the diameter of the circle, which is of length $h$.

As $b \to \infty, EF \to 0$, which is also correct.

Perhaps you'd like to check my working.

To keep the algebra simpler, I let:
$h = 2r$

$b = r+s$

and, in the diagram you have drawn:
$DE = p$

$BF = q$

$BD = t$

$EF = x$

Then the lengths of the tangents from D and B to the circle, are $r$ and $s$ respectively. So, using the Tangent-Secant Theorem:
$p(p+x) = r^2$ ...(1)

$q(q+x) = s^2$ ...(2)

So, from (1):
$p^2+px-r^2 = 0$

$\Rightarrow p = \frac{-x+\sqrt{x^2+4r^2}}{2}$, taking the positive root

Similarly
$\Rightarrow q = \frac{-x+\sqrt{x^2+4s^2}}{2}$
But
$p+x+q = t$
Therefore
$\frac{-x+\sqrt{x^2+4r^2}}{2}+x+\frac{-x+\sqrt{x^2+4s^2}}{2}=t$

$\Rightarrow\sqrt{x^2+4r^2}+\sqrt{x^2+4s^2}=2t$

$\Rightarrow x^2+4r^2+x^2+4s^2+2\sqrt{(x^2+4r^2)(x^2+4s^2)}=4t^ 2$

and, then if we re-arrange, square both sides, and re-arrange again, this simplifies to:
$4t^2x^2=4(t^2-r^2-s^2)^2-16r^2s^2$ ...(3)
So we now have an expression for x that we can write in terms of $b$ and $h$.

Now:
$t^2 = b^2+h^2$ (Pythagoras' Theorem on $\triangle DBC$)
$= (r+s)^2+(2r)^2$

$= 5r^2 + 2rs+s^2$

So (3) becomes:
$4(b^2+h^2)x^2 = 4(4r^2+2rs)^2 - 16r^2s^2$
$=4(16r^4+16r^3s+4r^2s^2)-16r^2s^2$
$(b^2+h^2)x^2=16r^4+16r^3s$

$\Rightarrow x^2= \frac{16r^3(r+s)}{b^2+h^2}$

$\Rightarrow x =\sqrt{\frac{2bh^3}{b^2+h^2}}$

Grandad
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