Question

A ladder 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. If 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?

• A. $x\sqrt{5}-\sqrt{5x^2-(x+y{)}^2}$
• B. $x\sqrt{5}-\sqrt{5x^2-(x-y{)}^2}$
• C. $2x-\sqrt{5x^2-(x+y{)}^2}$
• D. $2x-\sqrt{5x^2-(x-y{)}^2}$

Answer 1

Answer: (C)

$2x-\sqrt{5x^2-(x+y{)}^2}$

Here AB is a ladder.
$AC=2x$, $BC=x$ and $BD=y$
From $△ABC,A{B}^2=B{C}^2+A{C}^2=x^2+(2x{)}^2=5x^2$
$\because AB=DE$
$\because DE=x\sqrt{5}$
From $△DCE,D{E}^2=D{C}^2+C{E}^2$
$\therefore 5x^2=(x+y{)}^2+C{E}^2$
$\therefore C{E}^2=5x^2-(x+y{)}^2$
or, $CE=\sqrt{5x^2-(x+y{)}^2}$
$\therefore AE=2x-\sqrt{5x^2-(x+y{)}^2}$