Find Radius of Smaller Circle

Printable View

July 15th 2009, 07:37 AM
magentarita

Find Radius of Smaller Circle

http://i239.photobucket.com/albums/f...oos/prisms.jpg
July 15th 2009, 09:08 AM
red_dog

Let $O_1, \ O_2$ be the centers of the circles. (The bigger circle has the center $O_1$ ), $A\in(O_1), \ B\in(O_2)$

Then $O_1A\perp AB, \ O_2B\perp AB$ .

Let $BC\parallel O_1O_2, \ C\in (O_1A)$

In the right triangle CAB apply Pitagora:

$BC^2=AC^2+AB^2\Rightarrow 400=(11-r)^2+19^2$

Now solve the quadratic. Remember that $r<11$
July 15th 2009, 09:17 AM
masters

1 Attachment(s)

Quote:

Originally Posted by magentarita

http://i239.photobucket.com/albums/f...oos/prisms.jpg

Hi magentarita,

Ok, here's what I think.

In my diagram I have drawn a line from the center of the larger gear to the tangent point of the smaller gear.

Use the Pythagorean Theorem to find its length.

$c^2=11^2+19^2$

$c=\sqrt{482}$

Find angle DBA using Arctan.

$\arctan \frac{19}{11}\approx 59.9314$

Angle BAC is also 59.9314 since alternate interior angles are congruent (BD and CA are parallel because we have two lines perpendicular to the same line)

Now look at triangle ABC. We know $c=\sqrt{482}$ , a = 20, and angle BAC = 59.9314.

Use the Law of Cosines.

$a^2=b^2+c^2-2bc \cos A$

$20^2=b^2 + (\sqrt{482})^2-2b(\sqrt{482}) \cos 59.9314$

This all boils down to

$b^2-22b+82=0$

Apply the quadratic formula to get your 2 results. One you have to throw away because it is bigger than the larger gear radius.

Now you have your answer.
July 15th 2009, 10:05 AM
earboth

1 Attachment(s)

Actually you are dealing with a right triangle. (See attachment)

1. $r = 11-x$

2. $x^2+19^2=20^2$

Thus the correct answer is b)
July 15th 2009, 06:33 PM
magentarita

ow...

I want to thank all of you for your reply.