Kim places two very long (and very heavy) ladders, each $15 \mathrm{~m}$ long, on a flat floor between two vertical and parallel walls. Each ladder leans against one of the walls. The two ladders touch the floor at exactly the same place. One ladder reaches $12 \mathrm{~m}$ up one wall and the other ladder reaches $9 \mathrm{~m}$ up the other wall. In metres, how far apart are the walls?

![](https://cdn.mathpix.com/cropped/2024_04_30_22929074a9efc78b4168g-1.jpg?height=399&width=721&top_left_y=874&top_left_x=759)
We label the diagram as shown.

![](https://cdn.mathpix.com/cropped/2024_04_30_75c1cf70fbe3735bfd92g-02.jpg?height=434&width=724&top_left_y=585&top_left_x=755)

Since $A F=D F=15 \mathrm{~m}$, then by the Pythagorean Theorem,

$$
B F=\sqrt{15^{2}-12^{2}}=\sqrt{81}=9 \mathrm{~m} \quad F C=\sqrt{15^{2}-9^{2}}=\sqrt{144}=12 \mathrm{~m}
$$

Therefore, $B C=B F+F C=9+12=21 \mathrm{~m}$, and so the walls are $21 \mathrm{~m}$ apart.

ANSWER: 21