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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 $10$ from one wall, and a distance of $13$ from the other wall.  Find the radius of the table.

 

 Mar 22, 2024

Best Answer 

 #1
avatar+394 
+2

Draw a rectangle in the middle like so:

The width is r-13, while the length is r - 10, and the diagonal is r:

Therefore, use the pythagorean theorem.

\({(r-10)}^{2}+{(r-13)}^{2}={r}^{2}\)

\(r=\frac{46\pm \sqrt{1040}}{2}=23 \pm \sqrt{260} = 23\pm 2\sqrt{65}\)

We know because the distance of 13 to the wall, r > 13, therefore the only solution is \(23+2\sqrt{65}\).

 Mar 22, 2024
edited by hairyberry  Mar 22, 2024
 #1
avatar+394 
+2
Best Answer

Draw a rectangle in the middle like so:

The width is r-13, while the length is r - 10, and the diagonal is r:

Therefore, use the pythagorean theorem.

\({(r-10)}^{2}+{(r-13)}^{2}={r}^{2}\)

\(r=\frac{46\pm \sqrt{1040}}{2}=23 \pm \sqrt{260} = 23\pm 2\sqrt{65}\)

We know because the distance of 13 to the wall, r > 13, therefore the only solution is \(23+2\sqrt{65}\).

hairyberry Mar 22, 2024
edited by hairyberry  Mar 22, 2024

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