The rectangle in the corner is 6 units by 12 units. Find the radius of the circle.

2. Originally Posted by masters
The rectangle in the corner is 6 units by 12 units. Find the radius of the circle.
$\displaystyle R = 30$, No?

3. Originally Posted by Danny
$\displaystyle R = 30$, No?

$\displaystyle (R-6)^2 + (R-12)^2 = R^2$

$\displaystyle R^2 - 36R + 180 = 0$

$\displaystyle (R-30)(R-6) = 0$

but $\displaystyle R$ cannot be $\displaystyle 6$ so the answer is 30

4. Originally Posted by masters
The rectangle in the corner is 6 units by 12 units. Find the radius of the circle.
Originally Posted by Danny
$\displaystyle R = 30$, No?
Originally Posted by simplependulum

$\displaystyle (R-6)^2 + (R-12)^2 = R^2$

$\displaystyle R^2 - 36R + 180 = 0$

$\displaystyle (R-30)(R-6) = 0$

but $\displaystyle R$ cannot be $\displaystyle 6$ so the answer is 30
Very good, guys! I didn't want to stress your brains too much, so this one was not a really big challenge. I'll spice it up a little more in the future.

5. "If radius = 30 and rectangle is w by 2w, find w"
would have been "fun"

6. Originally Posted by Wilmer
"If radius = 30 and rectangle is w by 2w, find w"
would have been "fun"

In general

$\displaystyle (R - \omega)^2 + (R - 2\omega)^2 = R^2$

$\displaystyle R^2 - 2R( \omega + 2\omega ) + {\omega}^2( 1 + 2^2 ) = 0$

$\displaystyle R^2 - 6R(\omega) + 5{\omega}^2 = 0$

$\displaystyle ( R - 5 \omega )(R - \omega) = 0$

Therefore , the relationship between $\displaystyle R$ and $\displaystyle \omega$ is

$\displaystyle R = 5\omega$

or their ratio is 5