The lengths of the sides of a triangle are consecutive even integers.Find the length of the longest side if it is 14 units shorter than the perimeter

Since the sides are consecutive even integers, it has perimeter
P=x+(x+2)+(x+4)

If you are familiar with Pythagorean triangles, this alone gives away the solution.

By Pythagoras:

x^{2}+(x+2)^{2}=(x+4)^{2}

Solve for x. You will notice the length of the hypoteneuse is 14 shorter than the perimeter.
It works out that way.

If you are familiar with Pythagorean triangles, this alone gives away the solution.


Looks like we have no reason to assume that the solution is a right angle triangle and ignore that the perimeter is longer than the longest side by a certain number.
Wouldn't it be just this (here a is the shortest side)?

a + (a+2) + (a+4) = (a+4) + 14 => a = 6 and so on.

I was thinking this as well... the triangle is not necessarily a right angled triangle.

Or, if we want to go faster, we do:

Let the longest side be x.

P - 14 = x
P = (x-4) + (x-2) + x

[(x-4) + (x-2) + x] - 14 = x

Yes, my bad. False assumption.

Force of habit. I am so used to these problems involving right triangles.

Coincidentally, I had just gone over a problem that stated "one side of a right triangle is 7 units more than the short side, and the hypoteneuse is 8 more than the short side. What are the side lengths?".