A 25 foot ladder leans against a vertical wall. The foot of the ladder is 7 feet from the base of the wall. If the top of the ladder slips 4 feet down the wall, how far will the foot of the ladder slide?
One way to do this is to use Pythagoras' Theorem twice. Sort of a before and after.
This would be much better if I could just draw the pictures. But I can't, so here goes...
Visualize the ladder as the hypotenuse of a right triangle.
Then the hypotenuse is 25 and the ground side is 7.
By the theorem a2 + b2 = c2, then a2 (the wall) = c2 (the ladder) – b2 (the ground)
a2 = 252 – 72
a2 = 625 – 49 = 576
a = sqrt(576) = 24 (the height of the wall side)
When the ladder slips 4 feet down the wall, the new "a" becomes 20
The hypotenuse stays the same 25 because it's a ladder
So now we can say b2 (the ground) = 252 (the ladder) – 202 (the wall)
b2 = 625 – 400 = 225
b = sqrt(225) = 15 (the length of the ground side)
When the top of the ladder slid down the side of the wall 4 feet,
the bottom of the ladder slid out until it was 15 feet from the wall.
The ladder was already 7 feet from the wall, so the bottom of the ladder slid 8 feet.
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