Always looking for a shortcut (Pythagorean Theorem)

[MainFragger]

When I was in school many years ago, I remember learning about right triangles and Pythagorean Theorem. At one point the teacher pointed out that there is a bit of a simple shortcut to finding the hypoteneuse when side a and side b of the triangle are the same. Which is, you just use the number and the square root of two. In other words, if each side is 3, then its 3*sqrt(2)....if both sides are 4, then its 4*sqrt(2)..

Well, what happens when the sides are NOT equal? Initially, I assumed that whatever number you get, you divide by the sqrt(2) and you would be left with the difference between the two numbers you are using. Or the quotient. Or some variation of that...something that makes sense.. but when I use excel sheets to figure that out, well, lets just say that I haven't spotted it yet. I kind of toyed with the idea that if you subtract the highest possible square from the number, you would be left with a ratio that equates to how many line segments there are, versus how many are possible in the next triangle with matching sides. For example, with sides 3 and 4, the next triangle up would be 4 and 4, which is a total of 8 segments. So a triangle with 3 and 4 segments has 7/8 segments..

I thought maybe it would be more obvious if I pushed it so that some values were two numbers apart instead of one, but I still don't really see the pattern I am looking for.

I figure maybe someone more accomplished at math will see something that I might not notice.

The chart below (oops, just noticed a slight error in my first few cells, but don't worry, it doesn't affect the rest of the sheet) shows the sides in the first two columns, the length of the hypotenuse in the third column, and those values divided by the sqrt(2) in the fourth column).

1​	2​	FALSE​	#VALUE!​	1​	3​	3.162278​	2.236068​
2​	3​	3.605551​	2.5495097567964	2​	4​	4.472136​	3.162278​
3​	4​	5​	3.53553390593275	3​	5​	5.830952​	4.123106​
4​	5​	6.403124​	4.52769256906872	4​	6​	7.211103​	5.09902​
5​	6​	7.81025​	5.52268050859365	5​	7​	8.602325​	6.082763​
6​	7​	9.219544​	6.51920240520267	6​	8​	10​	7.071068​
7​	8​	10.63015​	7.51664818918648	7​	9​	11.40175​	8.062258​
8​	9​	12.04159​	8.51469318296323	8​	10​	12.80625​	9.055385​
9​	10​	13.45362​	9.51314879522026	9​	11​	14.21267​	10.04988​
10​	11​	14.86607​	10.5118980208144	10​	12​	15.6205​	11.04536​
11​	12​	16.27882​	11.5108644332214	11​	13​	17.02939​	12.04159​
12​	13​	17.69181​	12.5099960031968	12​	14​	18.43909​	13.0384​
13​	14​	19.10497​	13.5092560861063	13​	15​	19.84943​	14.03567​
14​	15​	20.51828​	14.508618128547	14​	16​	21.26029​	15.0333​
15​	16​	21.93171​	15.5080624192709	15​	17​	22.67157​	16.03122​
16​	17​	23.34524​	16.5075740192192	16​	18​	24.08319​	17.02939​
17​	18​	24.75884​	17.5071414000117	17​	19​	25.4951​	18.02776​
18​	19​	26.1725​	18.5067555233218	18​	20​	26.90725​	19.0263​
19​	20​	27.58623​	19.5064092031312	19​	21​	28.3196​	20.02498​
20​	21​	29​	20.50609665441	20​	22​	29.73214​	21.0238​

 

[lev888]

Hypotenuse length depends on both legs. If they are equal, yes, you get `asqrt(2)`. Why? Because:
`c = sqrt(a^2+b^2) = sqrt(a^2+a^2) = sqrt(2*a^2) = asqrt(2)`.
But in general you need both a and b to calculate c. So, I am not sure what you are trying to achieve.

 

[MainFragger]

When I was younger, Pythagoreans Theorem just seemed a bit overcomplicated to me in that you have to figure out an anwser by squaring numbers and adding them, and then unsquaring the answer. I've always been trying to find a quicker and more direct way to figure out the answer.

 

[Dr.Peterson]

When I was younger, Pythagoreans Theorem just seemed a bit overcomplicated to me in that you have to figure out an anwser by squaring numbers and adding them, and then unsquaring the answer. I've always been trying to find a quicker and more direct way to figure out the answer.

I imagine Pythagoras was pleased with how simple (and unexpected) his theorem was! I've seen far worse.

 

[Steven G]

If it turns out that a² +b² = c², then for any positive integer k it is also true that (ka)² + (kb)² = (kc)²

For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2. That is 3^2 + 4^2 = 5^2. If I multiply all three number by 2, then 6^2 + 8^2 = 36 + 64 = 100 = 10^2.

Now, if two sides of a right triangle are both 1, then the hypotenuse must be sqrt(1^2 + 1^2) = sqrt(1+1) = sqrt(2)

Now if you multiply the two legs of this triangle by k, then the sides will be k*1, k*1 and k*sqrt(2), ie k, k, k*sqrt(2)
This only worked because in a right triangle with legs of length 1 the hypotenuse must be sqrt(2).

Now if the legs of a right triangle were 1 and sqrt(3), then the hypotenuse must be 2. Now if you multiple the two legs by k, the hypotenuse will be 2*k. That is another rule your teacher could have given you!

The rule where the legs are the same is more friendly to remember so it is taught to students. Personally I think that remembering the theorem at the top is more valuable.