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Question 1005225: How do I solve?? 13 footladder latter is placed so that it reaches to a point on the wall that is 2 feet higher than twice the distance from the base of the wall to the base of the ladder. How far from the wall is the base of the ladder? How high does the ladder reach???please help me..
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! this forms a triangle.
the ladder is the hypotenuse of the triangle with a length of 13.
the base of the triangle is equal to x, or any other variable name you want to use.
the height of the triangle is equal to 2x + 2.
this represent 2 more than twice the distance from the wall to the ladder.
use pythagorus to find the value of x.
pythagorus says:
a^2 + b^2 = c^2
set a = x
set b = 2x + 2
set c = 13
you get:
a^2 + b^2 = c^2 becomes:
x^2 + (2x+2)^2 = 13^2
simplify to get x^2 + 4x^2 + 4x + 4x + 4 = 169
simplify further to get x^2 + 4x^2 + 8x + 4 = 169
combine like terms to get 5x^2 + 8x + 4 = 169
subtract 169 from both sides of the equation to get 5x^2 + 8x - 165 = 0
factor this quadratic any way you can and you will get:
(5x + 33) * (x - 5) = 0
solve for x and you will get:
x = -33/5 and x = 5
since x has to be greater than 0, your solution is x = 5.
the distance from the wall to the base of the ladder is 5 feet.
the distance from the top of the ladder to the ground is 2*5+2 = 12 feet.
since 12^2 + 5^2 = 13^2, the measurements look good.
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