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?

Answer 1

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

Answer 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.

Answer 3

I know the range is 14>x>4

Answer 4

and the third side can not be 5 or 9

Answer 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

Answer 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?

Answer 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

Answer 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.

Answer 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