A ladder 25 feet long is leaning against a wall that is perpendicular

25^2-7^2=576
it means that the height is equal to 24.

since the top of the ladder slips down 4 feet, then the height of the wall =24-4=20

the bottom =sqrt(25^2-20^)=sqrt(625-400)=15

ans is E , not C

New,total distance between foot ladder and the wall = 15
slip = 15-7 = 8 hence C

you can think of this like a triangle problem where the height and base change but the hypotenuse remains the same (25)

so at first the ladder's base is 7, with a hypotenuse of 25 you can use the Pythagorean theorem to figure out that the height is 24.

The new height is then 20, with a hypotenuse of 25, which is a version of the 3:4:5 common right triangle, so we know the new base is 15.

Its important to keep in mind what you are looking for (how far the ladder slips) otherwise you might incorrectly pick 15 at this point. However, the correct answer is 15-7 or 8 (C)

Ladder Before Movement
-------------
25^2 = 7^2 + x^2
625=49+x^2
x^2=24

Ladder After Movement
--------------------
25^2=20^2+x^2
625=400+x^2
225=x^2
x=15

Ladder moved 15-7 = 8 feet

C

The ladder will from a right angled triangle with the wall. The hypotenuse will be 25 feet and base will be 7 feet.
So the perpendicular =\sqrt{25^2-7^2} = \sqrt{576} = 24 feet

When the top of the ladder slips by 4 feet the perpendicular will be = (24- 4) feet = 20 feet
But the length of the ladder that is the hypotenuse will remain the same.

In this case the base will be =\sqrt{25^2 - 20 ^2} = 15 feet

So the distance from the base of the ladder to base of the wall will become = 15 ft
So the ladder slips by (15-7) ft = 8 ft (C)

The most difficult aspect of this is getting your head around what actually happens.
The ladder's length doesn't change, but it's distance in relation to the wall does when the ladder "slips down".

Imagine a ladder propped up against a wall and the base gets kicked out such that the ladder is still up against the wall, but the base is now further out.

This is the scenario.

Original ladder placement = \(25^2 = 7^2 + b^2\)

\(b^2 = 576; b = 24\)

Since the ladder slips down by \(4 feet, b - 4 = 20\)

\(a^2 + 20^2 = 25^2\)

\(a^2 = 225\)

\(a = 15\)

The bottom of the ladder is 15 feet from the wall.

\(15 - 7 = 8\)

The ladder slipped 8 feet. The answer is C.

With the Ladder up against the Perpendicular Wall, we have a Right Triangle:

Ladder Length = 25 = Hypotenuse

Ground = Bottom Leg = 7 feet

From the 90 degree Vertex to the Vertex where the Ladder meets the Wall = Leg 2 = X

x - 7 - 25 ------> this is the Pythagorean Triplet of: 7 - 24 -25


The Height from the Ground up to the Point where the ladder meets the wall is thus = 24 feet

When the ladder slips down -4 Feet, we know have a new Right Triangle:

Leg = 20 feet

Hypotenuse = ladder length = 25 feet

and new Distance on the Grounds = X = Leg

Using Pythagoras:

(X)^2 + (20)^2 = (25)^2

(X)^2 = (25)^2 - (20)^2

(X)^2 = (25 + 20) (25 - 20)

(X)^2 = (45)(5) -----> (9)*(25)

Sqrt( X'2) = Sqrt( 9 * 25 )

X = 15

The Ladder Slipped Down: 15 Feet - 7 feet = 8 feet further

-C-

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I understand the theory behind this question. However, does anyone have any tips on how to answer it in under 2 minutes?

For example, finding the square root of 576? Under exam conditions my mind will panic and freeze seeing that in front of me.

A ladder 25 feet long is leaning against a wall that is perpendicular to level ground. The bottom of the ladder is 7 feet from the base of the wall.
We have something like this:

Since the wall is perpendicular to the ground, we have a right triangle, which means we can apply the Pythagorean Theorem to write: x² + 7² = 25²
Simplify to get: x² + 49 = 625
Subtract 49 from both sides: x² = 576
Solve: x = 24

ASIDE: We could have avoided all of the calculations above had we recognized that 7 and 25 are two of the three values in the Pythagorean triple 7-24-25, which means the missing side must have length 24

So, the ladder, in its ORIGINAL position, extends to a height of 24 feet.

If the top of the ladder slips down 4 feet . . .
If the ladder slips down 4 feet, then the ladder's NEW height = 24 - 4 = 20
So, we have something like this:


Once again, we COULD apply the Pythagorean Theorem to find the value of y (I'll let you do that on your own)
However, we can save time by recognizing that 25 and 20 are two of the three values in the magnified version of the Pythagorean triple 3-4-5
That is, if we take the Pythagorean triple 3-4-5 and multiply each side length by 5, we get the equivalent Pythagorean triple 15-20-25
This means the missing side must have length 15
In other words, y = 15

. . . how many feet will the bottom of the ladder slip?
Originally, the bottom of the ladder was 7 feet from the wall.
Afterwards, the bottom of the ladder was 15 feet from the wall
So, the bottom slipped 8 feet

Answer: C

=)) as a non-english speaking learner, it's kind of difficult to understand the question lol

This is what i was thinking, I believe that i will need to memorize all the powers from 0 to 30 at least.

use the following pattern:
(n+1)^2 = (n)^2 + (n) + (n+1)
OR, by rearrenging same equation,
n^2 = (n+1)^2 - (n+1) - (n)

Hence its enough to memorize until 15^2, and now have 20^2 and 25^2 under your belt too, which are common an easy to remember. Will take alittle bit of time in exam but is better than nothing, in case you don´t want to memorize +10 extra squares.

E.g. 24^2 = 25^5 - 25 - 24 = 576