Question

In the figure below $WXYZ$ is a rectangle with $WX = 4$ and $WZ = 8.$ Point $M$ lies on $\overline{XY},$ point $A$ lies on $\overline{YZ},$ and $\angle WMA$ is a right angle. The areas of $\triangle WXM$ and $\triangle WAZ$ are equal. What is the area of $\triangle WMA?$

$13$

$14$

$15$

$16$

$17$

Solution:

Set $X = (0,0),$ $W = (0,4),$ $Y = (8,0),$ $Z = (8,4),$ with $M = (m, 0)$ on $\overline{XY}$ and $A = (8, a)$ on $\overline{YZ}.$

Since $\angle WMA = 90^\circ,$ $\overrightarrow{MW} \cdot \overrightarrow{MA} = (-m)(8-m) + 4a = 0,$ so $4a = m(8-m).$ The areas give $[\triangle WXM] = \tfrac12 \cdot 4 \cdot m = 2m$ and $[\triangle WAZ] = \tfrac12 \cdot 8 \cdot (4 - a) = 4(4 - a).$ Setting these equal yields $m = 8 - 2a.$

Substituting $a = \tfrac{8-m}{2}$ into $4a = m(8-m)$ gives $2(8-m) = m(8-m),$ so $m = 2$ and $a = 3.$ Then with $W = (0,4),$ $M = (2,0),$ $A = (8,3),$ $[\triangle WMA] = \tfrac12\,\bigl|2(3 - 4) + 8(4 - 0)\bigr| = \tfrac12 (30) = 15.$

Thus, the correct answer is C.

2024 AMC 12B Problem 7

Solution:

Problem 7 in Other Years