+1839 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.
+1897 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! :)
