geometry

A stick 5 cm long, a stick 9 cm long, and a third stick n cm long form a triangle. What is the sum of all possible whole number values of n?

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  1. another of these triangle problems?
    Have any ideas of your own on this one?

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    posted by Steve
  2. as a hint, suppose you joined the two sticks at the end.

    Now consider rotating the smaller one, from being end-to-end with the larger one, to overlapping it. What's the max and min possible length for the third side?

    Now list all the integers in that range and add 'em up.

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    posted by Steve
  3. I know the range is 14>x>4

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    posted by itrixie
  4. and the third side can not be 5 or 9

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    posted by itrixie
  5. so that leaves 6, 7, 8, 10, 11, 12,and 13. That added up all together is 67 but my teacher said that was incorrect

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    posted by itrixie
  6. 14+13+12+11+10+9+8+7+6+5+4=99
    13+12+11+10+9+8+7+6+5=81
    14+13+12+11+10+8+7+6=81
    Which one is it?

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    posted by itrixie
  7. Of course the 3rd side can be 5 or 9. That just makes an isosceles triangle. So, you want all the numbers n such that 4 < n < 14:
    5+6+7+8+9+10+11+12+13 = 81

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    posted by Steve
  8. 9<n<14
    Values of n: 10, 11, 12, 13.
    Sum = 10+11+12+13 = 46 cm.

    5<n<9
    Values of n: 6, 7, 8.
    Sum = 6+7+8 = 21 cm.

    n = 5 cm(Isosceles Triangle).

    n = 9 cm(Isosceles Triangle).

    Sum Total=5+6+7+8+9+10+11+12+13 = 81 cm.

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    posted by Henry
  9. I say use the Pythagorean theorem.
    a^2+b^2=C^2.
    5^2+9^2=C^2
    25+81=106
    THEN DO THE SQUARE ROOT OF 106 WHICH IS 10.30CM

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    posted by Makayla

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