SOLUTION: Hi, I really need help with the question below. It is from a homework handout so I do not have an ISBN. Thanks for your help.
A ladder 15 feet long leans against a wall. Th
Algebra ->
Pythagorean-theorem
-> SOLUTION: Hi, I really need help with the question below. It is from a homework handout so I do not have an ISBN. Thanks for your help.
A ladder 15 feet long leans against a wall. Th
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Question 190170: Hi, I really need help with the question below. It is from a homework handout so I do not have an ISBN. Thanks for your help.
A ladder 15 feet long leans against a wall. The top of the ladder reaches a height that is 3 feet more than twice the distance from the base of the ladder to the wall. Draw and label a diagram. How high up the wall does this ladder reach? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A ladder 15 feet long leans against a wall. The top of the ladder reaches a height that is 3 feet more than twice the distance from the base of the ladder to the wall. Draw and label a diagram. How high up the wall does this ladder reach?
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Draw the picture.
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The ladder is the hypotenuse of a right triangle.
Let the base of the right triangle be "x".
Then the height on the wall is "2x+3".
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Using Pythagoras you get:
x^2 + (2x+3)^2 = 15^2
x^2 + 4x^2 + 12x + 9 = 225
5x^2 + 12x - 216 = 0
Use the quadratic formula to get:
x = [-12 +- sqrt(12^2-4*5*-216)]/10
Positive solution:
x = [-12 + sqrt(4464)]/10
x = 5.38 ft
2x+3 = 13.96 ft. (height on the wall)
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Cheers,
Stan H.