• Nov 11th 2009, 09:06 AM
masters
The rectangle in the corner is 6 units by 12 units. Find the radius of the circle.
• Nov 11th 2009, 04:51 PM
Jester
Quote:

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?
• Nov 11th 2009, 08:44 PM
simplependulum
Quote:

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 (Rock)
• Nov 12th 2009, 04:00 AM
masters
Quote:

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

Quote:

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

Quote:

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 (Rock)

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.
• Nov 12th 2009, 05:54 AM
Wilmer
"If radius = 30 and rectangle is w by 2w, find w"
would have been "fun" (Nod)
• Nov 14th 2009, 09:36 PM
simplependulum
Quote:

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

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