Question

Please help right triangles?

A 20-foot ladder leans against a wall so that it is 16 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? (Hint: Let 12−2x be the distance from the base of the ladder to the wall.)

Answer 1

Using substitution and the Pythagorean theorem, `x=16/5`.

Explanation:

When the 20ft ladder is 16ft up the wall, the distance of the base of the ladder is 12ft (it is a 3-4-5 right triangle). That's where the 12 in the hint "let 12-2x be the distance..." comes from.

In the new configuration, `a^2 +b^2=20^2`.

Let's say the base `a=12-2x` like the hint suggests.
Then the new height `b=16+x`.

Plug these `a` and `b` values into the Pythagorean equation above: `(12-2x)^2+(16+x)^2=20^2`.

Multiply these all out and get: `144-24x-24x+4x^2+256+16x+16x+x^2=400`.
which simplifies to `5x^2-16x=0`.

Factor out an `x`: `x(5x-16)=0`
We're only concerned with the `5x-16=0`; if `x=0`, it means the ladder didn't move.

So solve `5x-16=0` and get `x=16/5`