Judi leans a $25 \mathrm{~m}$ ladder against a vertical wall with the bottom of the ladder $7 \mathrm{~m}$ from the wall. (Please note that Judi is very strong - don't try this at home!) As she pulls the bottom of the ladder away from the wall, the top of the ladder slides $4 \mathrm{~m}$ down the wall. How far did she pull the bottom of the ladder from its original position?
(A) $4 \mathrm{~m}$
(B) $11 \mathrm{~m}$
(C) $2 \mathrm{~m}$
(D) $13 \mathrm{~m}$
(E) $8 \mathrm{~m}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-087.jpg?height=431&width=423&top_left_y=256&top_left_x=1293)

Part B: Each correct answer is worth 6.
Initially, the $25 \mathrm{~m}$ ladder has its bottom $7 \mathrm{~m}$ from the wall.

Suppose that its top is initially $h \mathrm{~m}$ above the ground.

By the Pythagorean Theorem, $h^{2}+7^{2}=25^{2}$ or $h^{2}+49=625$.

Thus, $h^{2}=625-49=576$, so $h=\sqrt{576}=24$, since $h>0$.

When the top slides $4 \mathrm{~m}$ down the wall, the top is now $24-4=20 \mathrm{~m}$ above the ground.

Suppose that the bottom ends up $d \mathrm{~m}$ from the wall.

Again by the Pythagorean Theorem, $20^{2}+d^{2}=25^{2}$, so $d^{2}=25^{2}-20^{2}=625-400=225$.

Since $d>0$, then $d=15$.

This means that the bottom is $15-7=8 \mathrm{~m}$ farther from the wall than in its original position.