This was an SAT question of the day, but I can't figure out how to come up with the right answer, I think you use Pythagoras Theorum but I'm not sure. Can anyone help with the process here?
A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out ___ ft.
Nevermind I figured it out!
Before the slipping occurs, the top of the ladder's distance above the floor is
sqrt(25^2 - 7^2) = 24 feet
If the top slips 4 feet, it then is 20 feet from the floor.
Now use the Pythagorean theorem again to get the distance of the base from the vertical wall.
sqrt (25^2 - 20^2) = __?__
15
To solve this problem, we can indeed use the Pythagorean theorem, which states that the square of the hypotenuse (in this case, the ladder) is equal to the sum of the squares of the other two sides (the vertical distance and the horizontal distance).
In this scenario, the ladder forms a right-angled triangle with the wall and the ground. Let's denote the vertical distance as "y" (the height of the ladder on the wall) and the horizontal distance as "x" (the distance the bottom of the ladder slides out).
Using the Pythagorean theorem, we have:
(25 - 4)^2 = 7^2 + x^2
Simplifying,
21^2 = 49 + x^2
441 = 49 + x^2
x^2 = 441 - 49
x^2 = 392
Taking the square root of both sides,
x = sqrt(392) or approximately 19.80 ft
Therefore, the bottom of the ladder will slide out approximately 19.80 feet when the top of the ladder slips down 4 feet.