TRICKY GEOMETRY

For the purposes of this exercise, assume the co-ordinates of $B,R,I,M$ are $\left[0,0\right],\left[16,0\right],\left[16,18\right],\left[0,18\right]$ respectively. Given the premises of the question, we can deduce that $A = \left[6,18\right]$ and $H = \left[0,10\right]$ , and solving for the intersection of $(BA)$ and $(IH)$ gives $T = \left[4,12\right]$ .

So, given that

$2 \times \Delta(\overline{MATH}) = \Delta(\overline{ABM}) + \Delta(\overline{HIM}) - \Delta(\overline{HTB}) - \Delta(\overline{AIT})$

and

$\Delta(\overline{ABM}) = \frac{6\times18}{2} = 54 \\ \Delta(\overline{HIM}) = \frac{8\times16}{2} = 64 \\ \Delta(\overline{HTB}) = \frac{4\times10}{2} = 20 \\ \Delta(\overline{AIT}) = \frac{6\times10}{2} = 30,$

then

$\Delta(\overline{MATH}) = \frac{54+64-20-30}{2} = 34,$

as required.