A 50ft ladder rests against a wall so that the top of the ladder is 48ft. from the ground. As you start climbing the ladder slips and the top of the ladder drops 8ft. How far does the bottom of the ladder slide away from the wall? (From its original position.)
Well, this ladder seems to be playing a little game of "Hide and Slide!" Let's see how it goes.
The ladder starts with a length of 50ft and the top is 48ft from the ground. But oh dear, it slips down 8ft! Must've hit a banana peel or something.
Now, the top of the ladder is at 48ft - 8ft = 40ft from the ground. So, the bottom of the ladder must have slid away from the wall, let's call that distance "x."
We can use some good old Pythagoras here. According to him, the ladder, the ground distance (x), and the vertical distance (40ft) form a right triangle.
So, we have x^2 + 40^2 = 50^2. Let's do the math.
x^2 + 1600 = 2500
Subtracting 1600 from both sides, we're left with x^2 = 900.
Taking the square root of both sides, we get x = 30ft.
And there you have it! The bottom of the ladder slid 30ft away from the wall. Watch out for those slippery ladders next time!
To solve this problem, we can use the Pythagorean Theorem. According to the theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's call the distance the bottom of the ladder slides away from the wall "x".
The original position of the ladder forms a right-angled triangle, with the ladder being the hypotenuse, the distance the bottom of the ladder slides as one side, and the distance from the ground to the top of the ladder as the other side.
Using the Pythagorean Theorem, we can express this relationship as:
x^2 + 48^2 = 50^2
Simplifying the equation, we have:
x^2 + 2304 = 2500
Subtracting 2304 from both sides, we get:
x^2 = 196
Taking the square root of both sides, we have:
x = ±14
Since we are looking for the distance the bottom of the ladder slides away from the wall, we take the positive value of x:
x = 14
Therefore, the bottom of the ladder slides 14ft away from the wall from its original position.
To solve this problem, we can use the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the lengths of the two legs (sides adjacent to the right angle) is equal to the square of the length of the hypotenuse.
In this case, the ladder forms the hypotenuse of a right triangle with the wall. Let's call the distance the bottom of the ladder slides away from the wall "x" (in feet).
Using the Pythagorean theorem:
x^2 + (48 - 8)^2 = 50^2
Simplify the equation:
x^2 + 40^2 = 50^2
x^2 + 1600 = 2500
x^2 = 2500 - 1600
x^2 = 900
Taking the square root of both sides:
x = √900
x = 30
Therefore, the bottom of the ladder slides 30 feet away from the wall from its original position.
Original distance from wall --- x ft
x^2 + 48^2 = 50^2
x^2 = 2500 - 2304
x^2 = 196
x = √196 = 13 ft
New height above the ground = 40 ft
ladder is still 50 ft long
let the distance from the wall be y ft
repeat my calculations using the new values
(Looks like a serious accident is about to happen , a 50 ft ladder ??? )