You can also do it with trig.
Let be the angle between the beam and the ground.
The length of the beam is
=length from ground to dumpster.
= length from dumpster to wall.
Now, if
Can you finish?.
So I have this one problem:
A wall is being supported by a beam passing over a dumpster 5 feet high and 4 feet from the wall. Find the length of the shortest possible beam.
I know that you need to use similar triangles, but besides that I don't really know what to do. Please Help!!!!!!!!!
This is what I have figured out:
from the ground to the beam (along the wall) is v, and from the dumpster to the beam is v-5
from the wall to the beam (along the ground) is u, and from the dumpster to the beam is u-4.
the beam is x.
I've got it all into one equation: x^2 = v^2 + (20 +4)^2
(that's 20 over v-5) (v-5)
I'm stuck now. PLEASE PLEASE PLEASE HELP!!!!!!!!!!!!!!!!!
Hello, jahichuanna!
This is a classic . . . The "shortest ladder" problem.
It is easier using Trigonometry . . .A wall is being supported by a beam passing over a dumpster 5 ft high and 4 ft from the wall.
Find the length of the shortest possible beam.Code:* C * | * | B * θ | * - - - * D * | 4 | * | | * |5 | * θ | | A * - - - - * - - - * E F 4
In right triangle
In right triangle
So we have: .
And that is the function we must minimize . . .