how do i do this question?
a ladder length 13m rests against a vertical wall with its foot on a horizontal floor at a distance of 5m from the wall. when the top of the ladder slips down a distance of x, the foot of the ladder moves out x. find the distance of x.
a^2 + b^2 = c^2
a^2 + 5^2 = 13^2 ==> 169-25=144=a^2
therefore a = 12
after adjustment;
(12-x)^2 +(5+x)^2 = 13^2
144 - 24x +x^2 + 25 + 10x +x^2 = 169
2x^2-14x=0
x(2x-14)=0
x = 7 or 0
Therefore the distance moved is 7m.
thanks!
but shouldn't it be 0 or -7?
oh, sorry, i get it
To solve the problem, we can use the Pythagorean theorem to find the length of the ladder (c) when it is in its original position.
The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.
In this case, the ladder represents the hypotenuse, so we can use the formula:
a^2 + b^2 = c^2
Where a is the distance between the foot of the ladder and the wall (5m) and c is the length of the ladder (13m).
Plugging in the values, we get:
5^2 + b^2 = 13^2
25 + b^2 = 169
Simplifying, we find:
b^2 = 144
Taking the square root of both sides, we get:
b = 12
So when the ladder is in its original position, the distance from the top of the ladder to the wall is 12m.
Now, let's consider what happens when the top of the ladder slips down a distance of x and the foot of the ladder moves out x.
In this case, we have a new right-angled triangle, with one side (a) now equal to 12 - x and the other side (b) equal to 5 + x. The length of the ladder (c) remains the same at 13m.
Now we can use the Pythagorean theorem again to find the new value of x.
(a- x)^2 + (b + x)^2 = c^2
Substituting the values:
(12 - x)^2 + (5 + x)^2 = 13^2
Expanding and simplifying, we get:
144 - 24x + x^2 + 25 + 10x + x^2 = 169
Combining like terms, we get:
2x^2 - 14x = 0
Factoring out x, we get:
x(2x - 14) = 0
This equation has two possible solutions: x = 0 and x = 7.
However, we need to consider the physical situation described in the problem. The problem states that when the top of the ladder slips down a distance of x, the foot of the ladder also moves out x.
If x is equal to 0, it means that the ladder did not slip at all, and both the top and the bottom of the ladder remain in their original positions. This is not a valid solution in this case, as it does not satisfy the condition described in the problem.
Therefore, the only valid solution is x = 7, which means that the top of the ladder slipped down 7m and the foot of the ladder moved outwards 7m.
So, the distance moved is 7m.
I apologize for the initial confusion in my previous response.