Elegant way to find the Perimeter of a right triangle | (step-by-step explanation) | #math #maths

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Elegant way to find the Perimeter of a right triangle | (step-by-step explanation) | #math #maths

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[yuusufliibaan]

❤❤❤ thanks 💯🙏 keep going my dear teacher ❤️

[gandelve]

The important extra information which is not emphasised is the requirement that sides must be positive integers. If sides can be any positive real number, there are an infinity of answers.

[DdDd-ssms]

With the given information there are endless solutions. When a nears 0, c nears 89+ . When a nears endles, c nears endles

[pratapkarishma]

We need not find the values of a and c seperately, as the question is 'What is the perimeter? ' Perimeter is a + b + c we have got the value of a + c = 7921, just add a (89) to this to get the perimeter. ( a + c ) + b = a + b + c = 7921 + 89 = 8010, which is the answer you got by finding the values of a and c.

[jakelabete]

This problem is incorrectly posed. If you move the point C either left or right the sides 'a' and 'c' will change and with them the perimeter. The problem is still solvable by making an additional assumption, which you actually do when you assign the values.

[marcellosangiorgio]

It is arbitrary to say that, if xy = zt, then x=z andy=t.
As a matter of fact, there are infinite triangles having a side = 89

[davek]

This solution only works if you assume all values are integers, which was not given as a condition.
Introduce fractions, and there are an infinite number of possible solutions.

[ybodoN]

For any odd number n greater than 1, there is a Pythagorean triple (n, m, m + 1) where m = ½ (n² − 1).
When n is a prime number, there is no other Pythagorean triple than this one and the perimeter is n² + n.

[raymondarata]

3-4-5, 5-12-13 and 7-24-25 are the three smallest Pythagorean triples where the the smallest side is listed first. There appears to be a pattern. That is c = b+1. The hypotenuse is one larger than the longer leg. Using a = 89, b, c = b+1, the Pythagorean Theorem and some algebra, you get b = 3960 and c = 3961. P = sum of three sides = 8010.

[Ctrl_Alt_Sup]

b=89 is a prime number
In fact for any prime number b, the second scenario always leads to a=0.
Also there is only one possible solution: c=(b²+1)/2 and a=(b²-1)/2
And a perimeter p = a+b+c = (b²-1)/2+b+(b²+1)/2 = (b²-1+2b+b²+1)/2 = (2b²+2b)/2 = b²+b
We check it with b=89, p=89²+89=7921+1=8010

We can deduce the following property...
For each prime number b, there exists a Pythagorean triplet (a, b, c) of non-zero natural integers satisfying the Pythagorean relation a²+b²=c² with c=(b²+1)/2 and a=(b² -1)/2

[hemendraparikh]

Something does not make sense. Can you not move the point C to the right keeping given side length at fixed 89 and thus change the sum of other two sides? By moving the point c anywhere on the line you would still keep the side length 'b' constant at 89 but change the perimeter of the triangle.

[longchen]

畢氏數(Pythagorean triple)有通解(General solutions) ：
(b, (b²-1/2), (b²+1)/2), 當b為奇數(odd)，或(2b, b²-1, b²+1)

[k.ervede]

You forgot to mention your condition that only whole numbers apply. If not, 7921 can also be divided by any other number less than 7921 to produce a fraction, e.g. 7921=(100)(79.21).
In that case c=89.605 and a=10.395.
The circumference is then 189.
This problem therefore gives an infinite number of answers.
(You also don't have to calculate a and c separately. If you know that (a+c) is a value, you can add the known value b.)

[kennethstevenson]

It looked like a 30, 60, 90 triangle so I took the given shortest side and formed three sides in the ratio of 1, 2, and root three. This produced sides of 89, 178, and 89 root 3. This checks with the Pythagorean Triple 7921 + 23763 = 31684.

[chrisbonney]

Surely there are many possible solutions

[walterbrown]

Your solution is only one of an infinite number of solutions. Side a could be 89, and we would have a right triangle with 2 45 degree angles. If I choose a value of 2 X 89 = 178 for c, then my right triangle would be a 30 60 90 right triangle. If you were one of my grade school math students, I would assign the following homework question for you: "How many angles and/or side lengths are required to uniquely specify any polygon ?"

[peterkovak]

Mathematical 'magic' was used here, because, in fact, as long as you don't have one more side or one more angle (except of the right one, of course), you have an infinite number of solutions.

[rogerdadd]

Surely there are many integer possibilities for a and c. You just need to push the point opposite the 89 length and a and c will change whilst 89 remains the same. I think this is a possible solution but not THE solution as it cannot be defined.

[Roy-tffe]

Nothing need be prime, and the given value can be an irrational square root (for example) and so long as the number whose square root is taken is factorable, you will have a solution for every possible combination of the factors (BASED ON the factors, not the factors directly). And the given one, of course, with the two unknown sides being a single unit apart (for purists who will be apoplectic realizing I mean "one times a number" to be considered a prime factorization). So if the known value is the square root of 255, 1*255, 3*85, 5*51, and 15*17 will all generate solutions.

(By the way, that last fact is why one uses a single pair of primes generating an encryption solution: using several gives the codebreaker several possible solutions.))

For example, from my last: 3*85. (85+3)/2 and (85-3)/2 are the two sides.

[dimuthdarshaka]

Solutions may not be full filled
Pythagoras values
Please check this sir.