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?

      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