2021 Fall AMC 12A Problems/Problem 21

Solution

First realize that $\triangle BCY \sim \triangle DAX.$ Thus, because $CY: XA = 2:3,$ we can say that $BY = 2s$ and $DX = 3s.$ From the Pythagorean Theorem, we have $AB =(2s)^2 + 4^2 = 4s^2 + 16$ and $CD = (3s)^2 + 3^2 = 9s^2 + 9.$ Because $AB = CD,$ from the problem statement, $4s^2 + 16 = 9s^2 + 9.$ Solving gives $s = \frac{\sqrt{7}}{\sqrt{5}}.$ To find the area of the trapezoid, we can compute the area of $\triangle ABC$ and add it to the area of $\triangle ACD.$ Thus the area of the trapezoid is $\frac{1}{2} \cdot 2 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 + \frac{1}{2} \cdot 3 \cdot \frac{\sqrt{7}}{\sqrt{5}} \cdot 6 = 15\frac{\sqrt{7}}{{5}} = 3\sqrt{35}.$ Thus the answer is $\boxed{\textbf {(C)} \: 3\sqrt{35}}.$

~NH14

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2021 Fall AMC 12A Problems/Problem 21

Solution