A4. A) 30 Consider the right triangle $CDE$ in the figure. We see that $CD=15-6=9$ and $DE=12$. By the Pythagorean theorem, it follows that $CE=\sqrt{9^{2}+12^{2}}=15$.

Triangle $EFG$ is similar to triangle $CDE$, so it holds that $EG: EF = CE: CD$. We see that $EG: 6 = 15: 9$, so $EG = \frac{6 \cdot 15}{9} = 10$. Similarly, we see that $AC: BC = CE: DE$, so $AC: 4 = 15: 12$. It follows that $AC = \frac{4 \cdot 15}{12} = 5$. The length of the ladder is therefore $AC + CE + EG = 5 + 15 + 10 = 30$.

![](https://cdn.mathpix.com/cropped/2024_04_17_2ddd9533af815a0f4acbg-1.jpg?height=520&width=417&top_left_y=2273&top_left_x=1411)
4. A box of $4 \mathrm{dm}$ by $15 \mathrm{dm}$ is pushed up against the wall. On top of it stands a second box, of $12 \mathrm{dm}$ by $6 \mathrm{dm}$. A ladder touches exactly the ground, the two boxes, and the wall. See the representation in the figure (not to scale). How many dm long is the ladder?
A) 30
B) $8 \sqrt{15}$
C) 31
D) $22 \sqrt{2}$
E) $18 \sqrt{3}$

![](https://cdn.mathpix.com/cropped/2024_04_17_cc2ba9e3b1603277d398g-2.jpg?height=331&width=325&top_left_y=163&top_left_x=1505)