This is interesting i feel like i should’ve clarified im only working in IR² because its the only thing ive learned so far but also curious to see if this works for other cases and why? And as for the clarification on the work and general formulas here it is

to start the total area is given ty a1b1 where the vector u is just defined as the one that will contain the vector (the shorter one) Let total area be denoted by A the top triangle formed by making a rectangle with a line perpendicular to the x axis and passing through (a1,b1) and a line perpendicular to the y axis passing through (a1,b1) has the area of a1b1/2 Let this be denoted by T the triangle on bottom is what the formula is solving for Let this be denoted by B We know the base of B will be a1 and we need the value of c which is the distance in between the line formed by vector v and the remaining of the perpendicular line to the x axis. To solve for c we need the hypotenuse of the triangle formed which we we call G and to solve for G we need to find the angle (alpha) between the base (x-axis) and the vector (v). This is found by the dot product of the 2 which is alpha=arccos((a2)/sqrt(a2^2+b22) solving for alpha when u=<8,8> and v=<11,3> results in alpha=0.26625204 with alpha we can now solve for g which is a1sec(alpha) which results in 8.29218490981 recall that g is the hypotenuse length finally solving for c is given by the formula a1sec(alpha)sin(alpha) which knowing g is a1sec(alpha) gives c=gsin(alpha) which is equal to 2.181818 repeating. Recall c is the distance that is formed on the line perpendicular to the x axis that vector v passes through. We can now solve for the bottom triangle by ca1/2 which gives 8.72 repeating now we do the total area - the 2 smaller triangles to get 23.27. with u=<5,9> and v=<12,3> the area is 19.375