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.

blackpanther Mar 22, 2024

#1**+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 \(23+2\sqrt{65}\).

hairyberry Mar 22, 2024

#1**+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