Isosceles Triangle Geometry Problem

  • Thread starter Bashyboy
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  • #1
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Homework Statement


The measure of the sides of an isosceles triangle are represented by x + 5, 3x +13, and 4x + 11. What are the measures of the sides? Two answers are possible.

Homework Equations





The Attempt at a Solution


Well, I set up three different triangles, to account for the different placements of the sides; but I only came up with one solution. Where is the second solution coming from?
 

Answers and Replies

  • #2
Dick
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Two of those sides must be equal, right? Pick two of them, set them equal and find x. There are actually three different ways to do that. You'll find one value of x doesn't work.
 
  • #3
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Well, I did do that, and I got -8, -6, and 2.
 
  • #4
Dick
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Well, I did do that, and I got -8, -6, and 2.

Are those the x values? How did you get -8 and -6? I did get x=2 for one pair.
 
  • #5
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Oh! terribly sorry: I meant to write -4 and -3. And I found them by creating three different triangles, each case having its side equal to on of the others. For instance, I said a = x + 5,
b = 3x + 13, and c = 4x + 11; then I arbitrarily assigned a and b to be the equivalent sides and set them equal to each other and solved for x. Then, in the second case, I said a = x + 5, but this time I set b = 4x + 11; and once again, I said that a and b were the sides of the isosceles triangle that were equal and consequently set them equal to each other and solved for x. I followed the same procedure for the third case.
 
  • #6
Dick
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Oh! terribly sorry: I meant to write -4 and -3. And I found them by creating three different triangles, each case having its side equal to on of the others. For instance, I said a = x + 5,
b = 3x + 13, and c = 4x + 11; then I arbitrarily assigned a and b to be the equivalent sides and set them equal to each other and solved for x. Then, in the second case, I said a = x + 5, but this time I set b = 4x + 11; and once again, I said that a and b were the sides of the isosceles triangle that were equal and consequently set them equal to each other and solved for x. I followed the same procedure for the third case.

And which one of those gave you x=(-3)? I'm ok with x=2 and x=(-4).
 
  • #7
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Blimey, I erred once again; the -3 should be a -2, and I found it by setting x + 5 = 4x + 11
 
  • #8
Dick
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Blimey, I erred once again; the -3 should be a -2, and I found it by setting x + 5 = 4x + 11

Ok, so you've got x=2, -2 or -4. Which of those correspond to real triangles? Check what the side lengths are in each case.
 
  • #9
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Oh, okay, I see: I never plugged the values back into each expression; as soon as I saw the negative value, I completely dismissed it, thinking that you can't have a negative measurement. So, the two answers should be 2 and -2? and this corresponds to two possible triangles?
 
  • #10
Dick
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Oh, okay, I see: I never plugged the values back into each expression; as soon as I saw the negative value, I completely dismissed it, thinking that you can't have a negative measurement. So, the two answers should be 2 and -2? and this corresponds to two possible triangles?

Yes, x=(-4) gives you a side with negative length. Can't have that.
 
  • #11
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Okay, thank you so very much for your time and help.
 

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