SOLUTION: A ladder leans against a house with it's base 15 feet from the house. When the ladder is pulled 9 feet further away from the house the upper end slides down 13 feet. How long is th

Algebra ->  Pythagorean-theorem -> SOLUTION: A ladder leans against a house with it's base 15 feet from the house. When the ladder is pulled 9 feet further away from the house the upper end slides down 13 feet. How long is th      Log On


   

Question 89044: A ladder leans against a house with it's base 15 feet from the house. When the ladder is pulled 9 feet further away from the house the upper end slides down 13 feet. How long is the ladder?
I understand how to figure Pythagoreans theorem with one variable, but how do you do it with two variables?
Found 2 solutions by malakumar_kos@yahoo.com, jim_thompson5910:
Answer by malakumar_kos@yahoo.com(315) About Me  (Show Source):
You can put this solution on YOUR website!

A ladder leans against a house with it's base 15 feet from the house. When the ladder is pulled 9 feet further away from the house the upper end slides down 13 feet. How long is the ladder?
I understand how to figure Pythagoreans theorem with one variable, but how do you do it with two variables?
Let ABC be a right angled triangle such that angle A = 90 degrees
BC will be the hypotenuse, AB willbe the distance at which the foot of the ladder stands from the house, AC is the height at which the top of the ladder
meets the building.Let the height of the building be = x ft
applying pythagorous theorem we get BC^2 = AB^2+AC^2 (AC=15 ft)
Therefore BC^2 = x^2+15^2...........eq'n (1)
When the ladder is pulled 9 ft away from the house the upper end slides
13 ftdown. The new triangle formed will be ADE such that DE is the
hypotenuse , AD = x-13 , and AE = 15+9 = 24 ft
In both the triangles the hypotenuse represents the length of the ladder which is a constant for a given ladder.
applying Pythagorous theorem for triangle ADE we get DE^2 = AD^2+AE^2
or DE^2 = (x-13)^2+(24)^2..........eq'n(2)
From eq'n(1) &eq'n(2) (x-13)^2+(24)^2 = x^2+15^2 (BC = DE)
x^2+169-26x+576 = x^2+225
169+576-225 = 26x
745-225 = 26x
520 = 26x
x = 520/26 = 20 ft
To find thelength of the ladder substitite the value of x in eq'n (1)
Therefore BC^2 = x^2+15^2
= 20^2+15^2
= 400+225
= 625
BC = sq rt of 625 = 25 ft
Therefore the length of the ladder = 25 ft



Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
We basically have this triangle set up:




a%5E2%2B15%5E2=c%5E2 Here is the original Pythagorean equation


Here is the original position of the ladder (blue) and the shifted ladder (green)
Notice that we subtract 13 from a to get a-13. So our new "a" is a-13. Also, we add 9 to our base to get 15%2B9. So our new "b" is 15%2B9=24

%28a-13%29%5E2%2B24%5E2=c%5E2 So if we shift the ladder, we get this equation.


Notice how the 2 equations are equal to c%5E2, so that means we can set the two equations equal to each other

%28a-13%29%5E2%2B24%5E2=a%5E2%2B15%5E2

%28a-13%29%5E2%2B576=a%5E2%2B225 Evaluate 24%5E2 and 15%5E2

a%5E2-26a%2B169%2B576=a%5E2%2B225 Foil %28a-13%29%5E2

a%5E2-26a%2B745=a%5E2-520 Combine like terms

-26a%2B745=-520 Subtract a%5E2 from both sides

-26a=-520 Subtract 745 from both sides

a=20 Divide both sides by -26

So the original height of the ladder is 20 feet, and it drops to 20-13=7 feet. Lets check our answer:

Check:

20%5E2%2B15%5E2=400%2B225=625


7%5E2%2B24%5E2=49%2B576=625

Since the 2 equations are equal, this means that we get the same hypotenuse. So the ladder is sqrt%28625%29=25 feet long. This also verifies our answer.