A 25 m long ladder rests against a vertical wall. Initially, the foot of the ladder is 7 m away from the wall. If the top of the ladder slides down 4 m, how far will the foot of the ladder move away from the wall?

Introduction / Context:
This problem involves a ladder sliding along a vertical wall while its foot moves along the ground. The ladder length remains constant, so the situation is governed by the Pythagoras theorem. You are given the initial position and then told that the top has slid down by 4 m. The goal is to determine the new position of the foot and how far it has moved from its original place.

Given Data / Assumptions:

• Length of the ladder L = 25 m (constant).
• Initially, the foot of the ladder is 7 m away from the wall.
• Initially, the height reached on the wall is unknown but can be found.
• The top of the ladder slides down by 4 m.
• The wall is vertical, the ground is horizontal, and the ladder is straight.

Concept / Approach:
The ladder, wall and ground form a right angled triangle where the ladder is the hypotenuse. Initially, we use Pythagoras theorem to find the height of the ladder on the wall. After the top slides down by 4 m, we again apply Pythagoras with the new height and the fixed ladder length to compute the new distance from the wall. The movement of the foot is the difference between the new base distance and the original 7 m.

Step-by-Step Solution:
Initial situation: base distance x1 = 7 m, ladder length L = 25 m. Let initial height be h1. By Pythagoras: h1² + 7² = 25². So h1² + 49 = 625 ⇒ h1² = 576 ⇒ h1 = 24 m. The top slides down 4 m, so new height h2 = 24 − 4 = 20 m. Let the new base distance be x2. Again, h2² + x2² = L². So 20² + x2² = 25² ⇒ 400 + x2² = 625 ⇒ x2² = 225. Thus x2 = 15 m. Movement of the foot = x2 − x1 = 15 − 7 = 8 m.

Verification / Alternative check:
Check both positions with the fixed ladder length: Initially, sqrt(7² + 24²) = sqrt(49 + 576) = sqrt(625) = 25 m. After movement, sqrt(15² + 20²) = sqrt(225 + 400) = sqrt(625) = 25 m again. This confirms that the ladder length remains consistent in both positions. The difference in base distances is exactly 8 m, which aligns with our algebraic result.

Why Other Options Are Wrong:

• 5 m, 6 m, 9 m, 10 m: These values do not satisfy the right triangle constraints when combined with the 4 m vertical movement and a fixed 25 m ladder. None gives both initial and final triangles with hypotenuse 25 m and correct heights.
• Only 8 m exactly maintains the constant ladder length and the stated vertical slide distance.

Common Pitfalls:
A frequent error is to treat the ladder movement as linear and assume the foot moves the same distance as the top. That is incorrect because the ladder is rotating. Another mistake is failing to recompute the new base distance using Pythagoras theorem after the top slides down. Always remember that the relationship between height and base for a fixed hypotenuse is non linear.

Final Answer:
The foot of the ladder will move away from the wall by 8 m.