2023 AMC 8 Problems/Problem 19

Solution 1

By AA~ similarity triangle we can find the ratio of the area of big: small —> $\frac{9}{4}$ then there are a relative $5$ for the $3$ trapezoids combines. For $1$ trapezoid it is a relative $53$ so now the ratio is $\frac{5}{\frac{3}{4}}$ which can simplify to $\boxed{\text(C) \frac{5}{12}}$

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat

Solution 2

Subtracting the larger equilateral triangle from the smaller one yields the sum of the three trapezoids. Since the ratio of the side lengths of the larger to the smaller one is $3:2$ , we can set the side lengths as $3$ and $2$ , respectively. So, the sum of the trapezoids is $\frac{9\sqrt{3}}{4}-\frac{4\sqrt{3}}{4}=\frac{5}{4}\sqrt{3}$ . We are also told that the three trapezoids are congruent, thus the area of each of them is $\frac{1}{3} \cdot \frac{5}{4}\sqrt{3}=\frac{5}{12}\sqrt{3}$ . Hence, the area is $\frac{\frac{5}{12}\sqrt{3}}{\sqrt{3}}=\boxed{\textbf{(C)}\ \frac{5}{12}}$ .