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Easy Question...I can't do it
Sun Sep 06, 2009 9:23 pm
Ok, My homework is "How many different isosceles triangles are possible if the the sides must have whole number lengths and the perimeter must be 93 inches?
I need All the different ways. please. I would do it myself, but for some reason, I can't.
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Sun Sep 06, 2009 9:52 pm
This took a little lateral thinking. I'm not an expert, so anyone out there is free to double-check my working.
Q. How many different isosceles triangles are possible if the the sides must have whole number lengths and the perimeter must be 93 inches? (Note that this question doesn't ask you to state
all the possible triangles
, as you said in your post, but
triangles are possible.)
Perimeter of 93 inches = 2 equal sides + 1 unequal side.
Let's call the equal sides x and the unequal side y. So:
93 = 2x + y
Let’s imagine that y = 1.
93 = 2x + 1
92 = 2x
46 = x
Let’s imagine that x = 1.
93 = 2(1) + y
93 = 2 + y
91 = y
So the range of possible y values is from 1 to 91 - that's 91 values. And the range of possible x values is from 1 to 46 - that's 46 possible values. Uh-oh! We can't have a ton of triangles without any x values. More trial and error required.
Let’s imagine that y = 2.
93 = 2x + 2
91 = 2x
x = not a whole number, so this is not a possible answer.
As you can see, y can only be an odd number. This slashes half of the possible triangles. Our range of possible y values is now odd numbers only from 1 to 91. That's 46 possible values. Both our x and our y are now giving us 46 possible triangles! That's great.
One final step is to exclude the values of y = 31, x = 31, because these would give us an equilateral, not isoceles, triangle.
So the final answer is 45 possible triangles.
(Edited to cover up my inability to count.)
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