2024 AMC 10B Problem 11

Set $X = (0,0),$ $Y = (8,0),$ $W = (0,4),$ $Z = (8,4),$ so $M = (m, 0)$ and $A = (8, a).$ The right angle means $\overrightarrow{MW} \cdot \overrightarrow{MA} = 0,$ which gives $-m(8 - m) + 4a = 0,$ that is $m(8 - m) = 4a.$ Equal areas $[WXM] = 2m$ and $[WAZ] = 4(4 - a)$ force $m = 8 - 2a,$ so $a = \tfrac{8 - m}{2}.$ Substitute back and $(8 - m)(2 - m) = 0.$ Taking $M \ne Y$ leaves $m = 2$ and $a = 3.$ Then $[WMA] = 32$ $- [WXM]$ $- [MYA]$ $- [AZW]$ $= 32 - 4 - 9 - 4$ $= 15.$ Thus, C is the correct answer.