Question

A 10-foot ladder leans against a wall so that it is 6 feet high at the top. The ladder is moved so that the base of the ladder travels toward the wall twice the distance that the top of the ladder moves up. How much higher is the top of the ladder now?

Answer 1

Answer:

The top of the ladder is now at 10 ft.

Step-by-step explanation:

At the start, we have a height H=6, a length L=10 and a base B, that has to be calculated by the Pythagorean theorem:

$B^2=L^2-H^2=10^2-6^2=100-36=64\\\\B=\sqrt{64}=8$

The base is moved twice the distance the height moves up.

We called this distance x, so we have:

$L^2=(H+x)^2+(B-2x)^2=H^2+2Hx+x^2+B^2-4Bx+4x^2\\\\L^2=(H^2+B^2)+5x^2+(2H-4B)x\\\\L^2=L^2+5x^2+(2H-4B)x\\\\0=5x^2+(2H-4B)x\\\\5x+(2H-4B)=0\\\\x=\dfrac{4B-2H}{5}=\dfrac{4*8-2*6}{5}=\dfrac{32-12}{5}=\dfrac{20}{5}=4$

The new height (H+x) is

$H'=H+x=6+4=10$

The base travels 2x=8, so the new base B' is 0.

This means that the ladder is all against the wall (L=H').