A catheter of a right triangle measures 30cm. The hypotenuse of the triangle measures 2cm more than double the measurement of the other catheter. What is the area of the triangle

Let the measurement of one catheter be x cm. According to the given information, the other catheter is 30 cm, and the hypotenuse is 2x+2 cm.

Using the Pythagorean theorem, we have:
x^2 + 30^2 = (2x + 2)^2
x^2 + 900 = 4x^2 + 8x + 4
3x^2 + 8x - 896 = 0

Solving the quadratic equation using the formula x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} where a = 3, b = 8, and c = -896, we find:
x = \frac{{-8 \pm \sqrt{{8^2 - 4*3*(-896)}}}}{2*3}
x = \frac{{-8 \pm \sqrt{{64 + 10752}}}}{6}
x = \frac{{-8 \pm \sqrt{{10816}}}}{6}
x = \frac{{-8 \pm 104}}{6}

Therefore, the possible values of x are:
1. x = \frac{{-8 + 104}}{6} = \frac{{96}}{6} = 16
2. x = \frac{{-8 - 104}}{6} = \frac{{-112}}{6} = -18.7 (discarded as it is not a valid measurement)

So, the measurement of one catheter is 16 cm, the other catheter is 30 cm, and the hypotenuse is 2(16) + 2 = 34 cm.

The area of the triangle is given by:
A = \frac{1}{2} \times \text{Base} \times \text{Height}
A = \frac{1}{2} \times 16 \times 30
A = 240 \, \text{cm}^2

\boxed{240 \, \text{cm}^2} is the area of the triangle.