2020 AMC 8 Problems/Problem 18

Rectangle $ABCD$ is inscribed in a semicircle with diameter $\overline{FE},$ as shown in the figure. Let $DA=16,$ and let $FD=AE=9.$ What is the area of $ABCD?$

$[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$A$",(8,0), 1.25*S); dot("$B$",(8,15), 1.25*N); dot("$C$",(-8,15), 1.25*N); dot("$D$",(-8,0), 1.25*S); dot("$E$",(17,0), 1.25*S); dot("$F$",(-17,0), 1.25*S); label("$16$",(0,0),N); label("$9$",(12.5,0),N); label("$9$",(-12.5,0),N); [/asy]$ $\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272$

Solution

First, realize $ABCD$ is not a square. It can easily be seen that the diameter of the semicircle is $9+16+9=34$ , so the radius is $\frac{34}{2}=17$ . Express the area of Rectangle $ABCD$ as $16h$ , where $h=AB$ . Notice that by the Pythagorean theorem $8^2+h^{2}=17^{2}\implies h=15$ . Then, the area of Rectangle $ABCD$ is equal to $16\cdot 15=\boxed{\textbf{(A) }240}$ . ~icematrix

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2020 AMC 8 Problems/Problem 18

Solution

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