# geometry/trig

Initially a fifty foot ladder rests against a wall. As I start to climb it, the ladder slides down, finally stopping such that the ladder touches the wall at a point 8 feet below where it originally touched the wall. During the slide, the base of the ladder slid 16 feet from its original position. How far is the top of the ladder from the ground, given that the wall is perpendicular to the ground?

(I can't figure out how to find the answer at all. Am I supposed to get a real number? All I know is that it is a right triangle and the hypotenuse is fifty. I mean, I know that the point on the wall after it slides would be x - 8? and the distance to the base would be y + 16? i think? I keep getting weird polynomials when I try to use trig. Please help !!)

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1. Let x = original side length (ladder-wall)
Let y = original side length (ladder-floor)

Well yeah. The new length of the side of the triangle (where ladder touches the wall) is x - 8. The new length of side of triangle (where ladder touches the floor) is y + 16. Of course, y is from the pythagorean expression of the original triangle:
50^2 = x^2 + y^2
y = sqrt(2500 - x^2)

Now that we have the side lengths of the new triangle, we can make a new pythagorean expression. The hypotenuse length is still 50 ft, since the length of the ladder itself did not change:
50^2 = (x - 8)^2 + (sqrt(2500 - x^2) + 16)^2

Solving for x:
2500 = x^2 - 16x + 64 + (2500 - x^2) + 32*sqrt(2500 - x^2) + 256
0 = -16x + 64 + 32*sqrt(2500 - x^2) + 256
0 = -16x + 320 + 32*sqrt(2500 - x^2)
16x - 320 = 32*sqrt(2500 - x^2)
0.5x - 10 = sqrt(2500 - x^2)
Square both sides:
0.25x^2 - 10x + 100 = 2500 - x^2
1.25x^2 - 10x - 2400 = 0
Divide everything by 1.25:
x^2 - 8x - 1920 = 0
(x - 48)(x + 40) = 0
x = -40 (extraneous root because dimensions cannot be negative)
x = 48 ft. (original side length, ladder-wall)

Now that you have this dimension, you can calculate for the lengths of the others.
Hope this helps~ u

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