geometry help!!!!?

Question

FInd the length of the hypotenuse of the largest Pythagorean triple triangle in which 16 is the measure of a leg. Show the analysis that leads to your conclusion. (Hint: Begin by writing an equation using the Pythagorean Theorem, then use factoring to write a system of equations)

exactly how it was written by my geometry teacher..
answer, anyone??

Answer 1

Your question is imprecise. I assume you mean for all sides of the right triangle to have lengths that are integers.

h = length of hypotenuse
x = length of one leg
16 = length of other leg

The closer together in length the hypotenuse and longer leg of the triangle, the larger they can be in relation to the smaller leg.

First try h = x + 1 (since we are working in integers)

h² - x² = 16²
(x + 1)² - x² = 16²
x² + 2x + 1 - x² = 256
2x = 255
x = 255/2 Alas, x is not an integer.

Next try h = x + 2 (since we are working in integers)

(x + 2)² - x² = 16²
x² + 4x + 4 - x² = 256
4x = 252
x = 252/4 = 63 an integer.

h = x + 2 = 65

So the triangle has sides 65, 63, and 16.

Answer 2

Please note that my solution is the only one that actually follows the hint, "use factoring to write a system of equations".)

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You're given a leg, so you've got:

a² + 16² = c²

So

c² - a² = 16²

The left can be factored as (c - a)(c + a):

(c - a)(c + a) = 256

Now, 256 is 2^8, so there are 9 possible ways to find 2 numbers that multiply to give you 256: 1 x 256, 2 x 128, 4 x 64, 8 x 32, 16 x 16, 32 x 8, 64 x 4, 128 x 2, and 256 x 1.

This means that there are 9 possibilities you could try:

1) c - a = 1, c + a = 256
2) c - a = 2, c + a = 128
3) c - a = 4, c + a = 64
4) c - a = 8, c + a = 32
5) c - a = 16, c + a = 16
6) c - a = 32, c + a = 8
7) c - a = 64, c + a = 4
8) c - a = 128, c + a = 2
9) c - a = 256, c + a = 1

You could try each combination, but you know two extra things:

1) you want c (and therefore c + a, the first number) to be as big as possible, and
2) you know that a has to be positive...which means that c - a must be less than c + a...which means that you can discard 32x8, 64x4, and 128x2.

So, remembering we want the first number to be as big as possible, try 256 x 1 first:

c - a = 1
c + a = 256
2c = 257
c = 128.5

That won't work, because the definition of Pythagorean triples requires that all three numbers be integers. (See PS below.)

So try 128 x 2:

c + a = 128
c - a = 2
-------------
2c = 130
c = 65

Then c - a = 2, so 65 - a = 2, so a = 63.

So the answer is:

63, 16, 65.

As a check:

63² + 16² = 65²
3969 + 256 = 4225...check!!

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(PS: The numbers (127.5, 16, 128.5) also satisfy the Pythagorean theorem, but since Pythagorean triples must be integers by definition, you should probably not consider them the answer. Personally, I'd say 128.5 is the actual longest hypotenuse of a Pythagorean triple triangle that has 16 as a leg. I'd also say that (0.3, 0.4, and 0.5) is a Pythagorean triple, but I guess the definition doesn't agree with me.)

Answer 3

The Pythagorean equation is
a^2 + b^2 = c^2

If a=16, we want to find the largest possible value of b to get the largest possible value of c.
We have
b^2 = c^2 - 16
= (c+4)(c-4) But I can't see that this helps, so try something else.

"Pythagorean triple" suggests we want whole numbers only. Now if m and n are integers, then
m^2 - n^2, 2mn, m^2 + n^2 form a Pythagorean triple. The middle one could be 16 if we had m=8, n=1, and this would give
63, 16, 65.
Or if we had m=4, n=2, it would give 12, 16, 20 which isn't as big

Would we do any better trying to make m^2-n^2 equal to 16? Since it's equal to (m+n)(m-n), we have to keep m and n small enough that m+n doesn't exceed 16, and if it were 16 then m-n would have to be 1. That works if m=17/2 and n=15/2, and then we have fractions, which we don't want. But it does show that
16,255/2, 257/2 is a triple.
Next best is m-n=2, m+n=8, which gives m=5, n=3. This gives the triple
16, 30, 34. Too small.

Clearly the one we want is 16, 63, 65.

Answer 4

The 16-63-65 triangle is a large pythagorean triple.

Answer 5

Let n be the hypotenuse,

n^2 - 16^2 = (n+16)(n-16)

Now, you need to find maximum (n+16)(n-16) such that the product is a perfect square.

Let (n+16)(n-16) = m^2.

Since m ≤ n-1, the largest hypotenuse is 65 because
65^-16^ = 81(49) = 63^2

while 64 cannot be the leg.

Answer 6

h = hypotenuse and x = 2nd leg.
Then h^2 = x^2 +256
It appears that the largest triple is 16, 63, 65.

After that value of h and the and the longer leg become too close together for a triple to occur.

Answer 7

Hey, wait a minute. A is a leg and C is the hypotenuse, and if A^2 + 16^2 = C^2, we can get bigger and bigger hypotenuses C, simply by increasing A. There is no bounds for C. As A goes to infinity so too does C.

Unless your teacher wanted an integer answer for A and C, then the first person has it right.