Multiple choice

In a room, a man is standing on the ladder, which rests against a wall with its lower end at a distance X from the wall and its upper end at a height 2X above the floor. Due to the weight of the man, the lower end slides through a distance Y perpendicular to the edge between the floor and the wall, by how much does the upper end slide?

  1. X √5 - √5X2 - (X - Y)2

  2. X √5 + √5X2 - (X+ Y)2

  3. 2X - √ {5X2 - (X + Y)2

  4. 2X + √5X2 - (X - Y)2

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the Pythagorean theorem, the ladder length L satisfies L^2 = X^2 + (2X)^2 = 5X^2. After sliding, the new height h satisfies (X+Y)^2 + h^2 = 5X^2, so h = sqrt(5X^2 - (X+Y)^2). The vertical slide is 2X - h.

AI explanation

Using the Pythagorean theorem, the original length of the ladder is the square root of the sum of X squared and 2X squared, which simplifies to X times the square root of 5. After sliding, the new distance from the wall to the ladder's base is X plus Y. Applying the Pythagorean theorem again, the new height is the square root of the quantity 5X squared minus the quantity X plus Y squared. The vertical distance the upper end slides is the original height minus the new height, yielding 2X minus the square root of the quantity 5X squared minus the quantity X plus Y squared.

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