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A circular table is pushed into a corner of the room, where two walls meet at a right angle.  A point P on the edge of the table (as shown below) has a distance of 6 from one wall, and a distance of 6 from the other wall.  Find the radius of the table.

 Aug 2, 2024
 #1
avatar+1926 
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Let's first graph this problem to have a visual on what we must do. 

 

First off, let's note we have a lot of right triangles to work with. This also means we almost certainly must use the pythaogrean thereom. 

So let's do it. First, we have the equation 

\((r -6)^2 + (r - 6)^2 = r^2 \\ 2r^2 - 24r + 72 = r^2 \\ r^2 - 24r + 72 = 0\)

 

Now, let's complete the square for r. 

\(r^2 - 24r = -72 \\ r^2 -24r + 144 = -72 + 144 \\ (r- 12)^2 = 72\)

 

since the distance cannot be negative, let's take the positive root for r. We have

\(r -12 = \sqrt{72} \\ r = \sqrt {72} + 12 \\ r = 6\sqrt 2 + 12 ≈ 20.49\)

 

So 20.49 is our answer. 

 

Thanks! :)

 Aug 2, 2024
edited by NotThatSmart  Aug 2, 2024

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