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Another word problem.
- Thread starter Rainbow
- Start date
D
Deleted member 4993
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The two sides of a right triangle differ by one. If the hypotenuse is 29 cm, then determine the lengths of the two sides.
This also has to be written in a quadratic equation and solved.
let one side be = x
other side = x + 1
Now use Pythagorean theorm.
Please share your work with us .
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The two > > > sides < < <[/b] of a right triangle differ by one. If the hypotenuse is 29 cm, then determine the lengths of the two > > >sides. < <
This also has to be written in a quadratic equation and solved.
Rainbow, the problem should have called these "legs," not "sides" because the problem necessarily meant for neither of them to be the hypotenuse.
A hypotenuse is also a side of a right triangle. Stating them as "legs" is clear.
I set up the quadratic equation using the pythagorean formula and found the triangle to be imaginary.
Did anyone else find this?
No.
Look at:
\(\displaystyle x^2 + (x + 1)^2 \ = \ 29^2\)
HallsofIvy
Elite Member
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- Jan 27, 2012
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Why are you assuming that the given "sides" are legs? There are two possibilities:Rainbow, the problem should have called these "legs," not "sides" because the problem necessarily meant for neither of them to be the hypotenuse.
A hypotenuse is also a side of a right triangle. Stating them as "legs" is clear.
1) The given sides are in fact legs so we must have \(\displaystyle x^2+ (x+ 1)^2= 29^2\), a simple quadratic equation.
2) One of the given sides is the hypotenuse. Since the hypotenuse is the longest side in a right triangle, it must be "x+1", so that x+ 1= 29, x= 28 is one leg and the other is given by \(\displaystyle 28^2+ y^2= 29^2\).
D
Deleted member 4993
Guest
2) One of the given sides is the hypotenuse. Since the hypotenuse is the longest side in a right triangle, it must be "x+1", so that x+ 1= 29, x= 28 is one leg and the other is given by \(\displaystyle 28^2+ y^2= 29^2\).
That's an irrational answer....:lol::lol::lol::lol:
I crack myself up......
Why are you assuming that the given "sides" are legs? There are two possibilities:
1) The given sides are in fact legs so we must have \(\displaystyle x^2+ (x+ 1)^2= 29^2\), a simple quadratic equation.
2) One of the given sides is the hypotenuse. Since the hypotenuse is the longest side in a right triangle, it must be "x+1", so that x+ 1= 29, x= 28 is one leg and the other is given by \(\displaystyle 28^2+ y^2= 29^2\).
It is a poorly-phrased question from the start, not to mention that it's faulty with its inconsistent use of units. \(\displaystyle \ \ \)
A proper problem poser would not state
"The two sides of a right triangle..." as there are three sides of a right triangle. \(\displaystyle \ \ \)
Having the word "the" in front of "two sides of a right triangle" changes the meaning.
What would make sense to state is "The two legs of a right triangle..."
(The original question was trying to have it both ways.)
Who knows what the poser really wanted!? The problem is very flawed.
HallsofIvy, in your line of thinking as to what the question means (or could mean),
an amendment to the problem that may work is:
"Two sides of a right triangle have lengths that differ by one centimeter.
If the hypotenuse is 29 cm., then determine the lengths of those two sides."