# 2020 AMC 8 Problems/Problem 18

## Problem

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 1

$[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("A",(8,0), 1.25*S); dot("B",(8,15), 1.25*N); dot("C",(-8,15), 1.25*N); dot("D",(-8,0), 1.25*S); dot("E",(17,0), 1.25*S); dot("F",(-17,0), 1.25*S); label("16",(0,0),N); label("9",(12.5,0),N); label("9",(-12.5,0),N); dot("O", (0,0), 1.25*S); draw((0,0)--(-8,15));[/asy]$

Let $O$ be the center of the semicircle. The diameter of the semicircle is $9+16+9=34$, so $OC = 17$. By symmetry, $O$ is in fact the midpoint of $DA$, so $OD=OA=\frac{16}{2}=$. By the Pythagorean theorem in right-angled triangle $ODC$ (or $OBA$), we have that $CD$ (or $AB$) is $\sqrt{17^2-8^2}=15$. Accordingly, the area of $ABCD$ is $16\cdot 15=\boxed{\textbf{(A) }240}$.

## Solution 2 (coordinate geometry)

Let the midpoint of segment $FE$ be the origin. Evidently, point $D=(-8,0)$ and $A=(8,0)$. Since points $C$ and $B$ share $x$-coordinates with $D$ and $A$ respectively, it suffices to find the $y$-coordinate of $B$ (which will be the height of the rectangle) and multiply this by $DA$ (which we know is $16$). The radius of the semicircle is $\frac{9+16+9}{2} = 17$, so the whole circle has equation $x^2+y^2=289$; as already stated, $B$ has the same $x$-coordinate as $A$, i.e. $8$, so substituting this into the equation shows that $y=\pm15$. Since $y>0$ at $B$, the y-coordinate of $B$ is $15$. Therefore, the answer is $16\cdot 15 = \boxed{\textbf{(A) }240}$.

(Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)

## Solution 3 -

We can use a result from the Art of Problem Solving Introduction to Algebra book: for a semicircle with diameter $(1+n)$, such that the $1$ part is on one side and the $n$ part is on the other side, the height from the end of the $1$ side (or the start of the $n$ side) is $\sqrt{n}$. To use this, we scale the figure down by $9$; then the height is $\sqrt{1+\frac{16}{9}} = \sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$. Now, scaling back up by $9$, the height $DC$ is $9 \cdot \frac{5}{3} = 15$. The answer is $15 \cdot 16 = \boxed{\textbf{(A) }240}$. -SweetMango77; edited by someone else (I don't know who)

## Video Solution

 2020 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions