# The greatest possible perimeter of a right angled triangle with side 12

What is the greatest possible perimeter of a right angled triangle with integer lengths of one of the side has length 12?

A common right angled triangle for us is the triangle with sides (5,12,13).So the perimeter is 30.But I am not sure that it is the maximum or not.Please tell me the right answer with reason

• The perimeter of $(12,16,20)$ is $48$. Jul 1, 2017 at 6:19

The answer is $$84$$. It comes from the right triangle $$(12,35,37)$$.

You want the maximum of $$12+a+b$$ where $$a$$ and $$b$$ are such that $$12^2+a^2=b^2$$. This is equivalent to $$(b-a)(b+a)=144$$. Making the list of all possible ways of expressing $$144$$ as the product of two numbers both of which are even, you will find that the list is rather short. My answer comes from the fact that $$144=2\times72$$. For this decomposition, you get $$a=35$$ and $$b=37$$.

Let $a,b,c$ be the sides of the right-angled triangle. Assuming $12$ is not the hypotenuse $c$ (or else $a,b < 12$), we can assume that $12$ is one of the legs of the triangle, which we will name $b$. Since by the Pythagorean theorem, $$a^2+12^2=c^2,$$ and by the difference of squares we get $$(c+a)(c-a) = 144.$$

Since the sides are integers, $c+a$ and $c-a$ must be integers as well. We can systematically check if $c+a$ and $c-a$ are factors of $144$. Since we want to maximise $c+a$, and therefore $a,c$, let us check the factor pair $(1,144)$ first.

Since $c+a > c-a$, let us solve the system of equations:$$c+a = 144$$ $$c-a=1,$$ so therefore $2c = 145$, and $c$ is not an integer.

Can you use this method to try the next largest factor pair $(2,72)$? Will this factor pair work? From the lengths of $a,b,c$, can you now calculate the maximum perimeter of the triangle?

It is a known fact that the greatest triangle with integer sides and $12$ as one of the sides is $12,35,37$ hence the perimeter is $84$

As you have the triangle $(5,12,13)$, you already know that $12$ should not be the hypothenuse.

Assume that $x$ and $12$ are the two sides of the right angle. The hypothenuse is $\sqrt {x^2+12^2}$. You know that $\sqrt {x^2+12^2}>x$ so you can conclude that so $\sqrt {x^2+12^2}\geq x+1$ as the sides have an integral length. When you square and solve for $x$, you get $x\leq 71$. So there is a finite number of possibilities, and you can check them with a computer for example.

The answer turns out to be $x=35$ so the triangle is $(12, 35, 37)$ with perimeter $84$.

For finding largest perimeter we can use Ashis' Hypotenuse-Perimeter Formula According to the formula, if the given side is odd, $$\text{largest perimeter}=\text{Side}(\text{Side}+1).$$ If the given side is even, $$\text{largest perimeter}= \frac{\text{Side}(\text{Side}+2)}{2}.$$ This formula is 100% correct. Entering 12 as the side we get, \begin{align} \text{largest perimeter} &= \frac{\text{Side}(\text{Side}+2)}{2} \\ &=\frac{12(12+2)}{2} \\ &=6 \times 14=84 \\ \end{align}