A rectangular piece of paper $A E C D$ has dimensions $8 \mathrm{~cm}$ by $11 \mathrm{~cm}$. Corner $E$ is folded onto point $F$, which lies on $D C$, as shown. The perimeter of trapezoid $A B C D$ is closest to
(A) $33.3 \mathrm{~cm}$
(B) $30.3 \mathrm{~cm}$
(C) $30.0 \mathrm{~cm}$
(D) $41.3 \mathrm{~cm}$
(E) $35.6 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-130.jpg?height=301&width=358&top_left_y=793&top_left_x=1363)
Since $E$ is folded over onto $F$, then $B E$ is equal to $B F$ and $\angle B F C=\angle B E C=90^{\circ}$.

So $B E C F$ is a rectangle, and since $E C=8 \mathrm{~cm}$, then $B F=8 \mathrm{~cm}$ and so $B E=F C=8 \mathrm{~cm}$. Thus, $A B=A E-B E=11-8=3 \mathrm{~cm}$ and $B C=\sqrt{B F^{2}+F C^{2}}=\sqrt{64+64}=\sqrt{128} \approx 11.31 \mathrm{~cm}$ Therefore, the perimeter of the trapezoid is approximately $3+11.31+11+8=33.31 \mathrm{~cm}$, which is closest to $33.3 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-253.jpg?height=363&width=415&top_left_y=762&top_left_x=1424)

ANSWER: (A)