Something i've been wondering about in an expressable format

[Curd]

something I've been wondering about in an expressable format...

I've had thoughts about this before but this is the first time that I've been presented with such an ideal situation to express it (i couldn't figure out how to ask this question before, and still am not able to fully tie it into the other problems I've noticed it in).


anyway, i was looking at this problem, and i noticed that both me and the book came up with the same hypotenuse, but we came up with wildly different answers for what the sides measurements were.


i went with the idea that i might as well have both sides even since i have nothing telling me they're not. the book obviously took a different path.

so, i was thinking, are there not an infinite number of possible combinations for the lengths of the two unknown sides?

and therefore is not the accurate answer a^2 + b^2 = 169?

also, how exactly do you calculate all the possible answers? and how do you find the range in which those answers will fall?

and also there's the issue of this same type of situation effecting the answers to other problems.

you may notice that they wrote their solution equation differently from the way i wrote mine and of course received a particular restricted answer because of it. which means that there are any number of ways to write this equation, so long as those ways fit into a certain restricted pattern, and that each of those way will get you a different answer that generates the same hypotenuse.


and on top of that, how does this issue effect the calculation of other problems? i recall thinking some of these problems were rather arbitrarily solved... perhaps i will come up with a good example of this in the future.

anyway, what is the most productive manner by which to view this issue?

 

[jbunniii]

so, i was thinking, are there not an infinite number of possible combinations for the lengths of the two unknown sides?

and therefore is not the accurate answer a^2 + b^2 = 169?

There are infinitely many side lengths (a,b) satisfying this equation.

However, you did not take into account the additional constraint from the problem statement: the sum of the side lengths must be 17. Yours sum to 18.38, so while this yields a right triangle with the desired hypotenuse, sqrt(169) = 13, it doesn't satisfy the other condition.

 

[Curd]

ah! thanks for pointing that out.

 

[Curd]

okay, tell me if this is an okay way to think of problems like this


they say the sum of the sides not the hypotenuse is 17.

so, i labelled each side x and got x + x = 17

so that means that x = 17 - x

which of course doesn't make a lot of sense and would make sense were I to use more than one variable (why don't they teach us how to use more than one variable for problems like this anyway?)

but, it allows me to make sense of why one side is x and the other is 17 - x according to the solution in the book.

and it works.

what do you think?

 

[HallsofIvy]

No, it is not necessarily true that the two sides are equal. If you call the two sides x and y, then you have x+ y= 17 so that y= 17- x. By the Pythagorean theorem, with the hypotenuse equal to 13, you also have $$x^2+ y^2= 169$$. Putting y= 17- x into that, $$x^2+ (17- x)^2= 169$$. Multiply that out then solve the quadratic equation for x.

 

[Curd]

No, it is not necessarily true that the two sides are equal. If you call the two sides x and y, then you have x+ y= 17 so that y= 17- x. By the Pythagorean theorem, with the hypotenuse equal to 13, you also have $$x^2+ y^2= 169$$. Putting y= 17- x into that, $$x^2+ (17- x)^2= 169$$. Multiply that out then solve the quadratic equation for x.

i know I'm supposed to use 17-x and i know that both sides are not necessarily equal.

I was trying to come up with a way of thinking of the two sides that made sense algebraically.

that's why i had x+x=17. i don't see how i could us...

i guess i could also think of the sides as x and y and so have x+y=17 and therefore y=17-x and therefore i could plug 17-x in for y in the equation so that the equation only has one variable x, correct?

so, i would have x^2 + y^2 = c^2

and therefore x^2 + (17-x)^2 = c^2



is this a better way of thinking of it?


figuring out how to put these things in algebraic terms is my most difficult problem.

 

[Mark44]

they say the sum of the sides not the hypotenuse is 17.

so, i labelled each side x and got x + x = 17

If you label each side x, you are tacitly assuming that the two legs are equal, which is not given in the problem.

so that means that x = 17 - x

which of course doesn't make a lot of sense and would make sense were I to use more than one variable (why don't they teach us how to use more than one variable for problems like this anyway?)

but, it allows me to make sense of why one side is x and the other is 17 - x according to the solution in the book.

and it works.

what do you think?

i know I'm supposed to use 17-x and i know that both sides are not necessarily equal.

I was trying to come up with a way of thinking of the two sides that made sense algebraically.

that's why i had x+x=17. i don't see how i could us...

i guess i could also think of the sides as x and y and so have x+y=17 and therefore y=17-x and therefore i could plug 17-x in for y in the equation so that the equation only has one variable x, correct?

so, i would have x^2 + y^2 = c^2

and therefore x^2 + (17-x)^2 = c^2

Use the fact that the hypotenuse is 17, so your equation is x² + (17 - x)² = 17 = 169.

is this a better way of thinking of it?


figuring out how to put these things in algebraic terms is my most difficult problem.

 

[Curd]

so, the new way of thinking (for me it's new) is the good way?

 

[Mark44]

If it allows you to correctly translate the problem statement into one or more equations, it's a good way.