Since the point in the middle of both right angles is 6m, we can use this information to calculate the area of the triangle. Let's call the height of the triangle h, and the base of each right angle triangle as b.
Since the two right angle triangles are congruent, each has a height of h and a base of b/2. Using the Pythagorean theorem, we can find the value of h:
h^2 + (b/2)^2 = 6^2
h^2 + b^2/4 = 36
h^2 = 36 - b^2/4
Now, we can use the formula for the area of a triangle:
area = (1/2) x base x height
Substituting the values we have:
area = (1/2) x b x h
area = (1/2) x b x sqrt(36 - b^2/4)
To find the value of b that maximizes the area, we can differentiate the area formula with respect to b and set it equal to zero:
d(area)/db = (1/2) x (sqrt(36 - b^2/4) - (b^2/4) x (1/2) x (36 - b^2/4)^(-1/2)) = 0
Simplifying this equation, we get:
sqrt(36 - b^2/4) = b^2/4 x (36 - b^2/4)^(-1/2)
Squaring both sides, we get:
36 - b^2/4 = b^4/16 x (36 - b^2/4)^(-1)
b^4/16 - b^2/4 + 9 = 0
This is a quadratic equation in terms of b^2, so we can solve for b^2 using the quadratic formula:
b^2 = [-(1/2) +/- sqrt((1/2)^2 - 4 x (1/16) x 9)] / (2 x (1/16))
b^2 = [1 +/- sqrt(1 - 9/8)] / (1/8)
b^2 = [1 +/- sqrt(1/8)] x 8
b^2 = 8 +/- sqrt(2)
Since b^2 must be positive, we take the positive square root:
b^2 = 8 + sqrt(2)
Now, we can substitute this value of b into the area formula:
area = (1/2) x b x sqrt(36 - b^2/4)
area = (1/2) x sqrt(8 + sqrt(2)) x sqrt(36 - (8 + sqrt(2))/4)
area = (1/2) x sqrt(8 + sqrt(2)) x sqrt(34 - sqrt(2))
Therefore, the area of the triangle is approximately 47.21 square meters (rounded to two decimal places).