Solving for the area of a right triangle

Here's how I did it:

The three side lengths are $x, 2x+4,$ and $2x+6$ . I could use the Pythagorean theorem, but I'm feeling lazy, so I'll check out multiples of common Pythagorean triples:

The most common triple is $(3,4,5)$ , but that doesn't look useful here, as the longer leg is much closer in length to the hypotenuse. So I move on to the second most common triple: $(5,12,13)$ . Let's see...if I let $x=5$ then the difference between the longer leg and twice the shorter leg is $2$ , while the difference between the hypotenuse and the longer leg is $3$ . Off by a factor of $2$ .

So let $x=10$ and everything works. We're looking at a $(10,24,26)$ right triangle and its area is $120$ .

Let the smallest side be $x$ m in length. Then the hypotenuse has the length $2x+6$ m and the third side has the length $2x+4$ m. Therefore $(2x+6)^2=(2x+4)^2+x^2\implies x^2-8x-20=0$ . Hence $x=10, 2x+4=24$ and the area of the triangle is $\dfrac{1}{2}\times 10\times 24=\boxed {120}$ m $^2$ .