Question

A 25-foot ladder is placed 7 feet from the wall, as shown.
25 ft.
7 ft.
If the top of the ladder slips down 4 feet, how many feet will the bottom slide out?

Answer 1

Answer:

The answer is 8 feet.

Step-by-step explanation:

Let x = height where the ladder hits the building

7 squared+x squared=25 squared

49+x squared=625

x squared=576

x=24 feet

Let x = distance from wall:

x^2 + 20^2 = 25^2

x^2 + 400 = 625

x^2 = 225

x = 15 feet

Answer 2

Pythagoras' theorem is a basic relationship between the three sides of a right triangle. The ladder slips out by 8ft from its initial position.

What is Pythagoras theorem?

The Pythagorean theorem, sometimes known as Pythagoras' theorem, is a basic relationship between the three sides of a right triangle in Euclidean geometry. The size of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides, according to this rule.

The length of the ladder is 25 ft, while the distance between the legs of the ladder and the wall is 7ft. Therefore, the initial height of the ladder on the wall can be written as,

$\rm (Hypotenuse)^2 = (Perpendicular)^2+(Base)^2$

$\rm 25^2 = (Perpendicular)^2 + 7^2\\\\625 =(Perpendicular)^2+ 49\\\\625-49 = (Perpendicular)^2\\\\(Perpendicular)^2 = 576\\\\(Perpendicular) = 24$

Now, If the top of the ladder slips down by 4 feet, then the top of the ladder is 20 ft from the base. Therefore, the distance between the bottom of the ladder and the wall can be written as,

$\rm (Hypotenuse)^2 = (Perpendicular)^2+(Base)^2$

$\rm 25^2 = 20^2 + (Base)^2\\\\625 =400+ (Base)^2\\\\625-400= (Perpendicular)^2\\\\(Perpendicular)^2 = 225\\\\(Perpendicular) = 15$

Further, it is given that the distance between the bottom of the ladder and the wall initially is 7ft while later it was 15ft. Thus, the ladder slips out by 8ft from its initial position.

Learn more about Pythagoras Theorem:

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