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

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 $OD=OA=\frac{16}{2}= 8$. 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 Sidenote: 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 formula, we scale the figure down by $9$; this will give the height a length of $\sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$. Now, scaling back up by $9$, the height $DC$ is $9 \cdot \frac{5}{3} = 15$. The answer is then $15 \cdot 16 = \boxed{\textbf{(A) }240}$. -SweetMango77

## Solution 4 (Power of a Point Theorem)

Draw the other half of the circle as follows: $[asy] draw(arc((0,0),17,360,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("A",(8,0), 1.25*SE); dot("B",(8,15), 1.25*N); dot("C",(-8,15), 1.25*N); dot("D",(-8,0), 1.25*SW); dot("E",(17,0), 1.25*E); dot("F",(-17,0), 1.25*W); label("16",(0,0),N); label("9",(12.5,0),N); label("9",(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot("B'",(8,-15), 1.25*S); dot("C'",(-8,-15), 1.25*S); [/asy]$ By the Power of a Point Theorem, $FD\cdot DE = CD\cdot C'D$. By symmetry, $CD = C'D$. We see that $FD = 9$ and $DE = 16 + 9 = 25$. Substituting in these values, $9\cdot 25 = CD^2$, giving $CD^2 = 225$ and $CD = 15$. The area of the rectangle is therefore $15\cdot 16 = \boxed{\textbf{(A) }240}$.

## Solution 5 (Vertical Theorem) $[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("A",(8,0), 1.25*S); dot("B",(8,15), 1.25*N); dot("C",(-8,15), 1.25*N); dot("D",(-8,0), 1.25*S); dot("E",(17,0), 1.25*S); dot("F",(-17,0), 1.25*S); label("16",(0,0),S); label("9",(12.5,0),N); label("9",(-12.5,0),N); dot("G", (0,15), SE); dot("O", (0,0), NE); draw((0,0)--(0, 15)); draw((-7.5,15)--(0,0)); [/asy]$

According to the Pythagorean Theorem and the Vertical Theorem, we can find out that $OG=\sqrt{\left(\frac{2\times9+16}{2}\right)^2 - \left(\frac{16}{2}\right)^2}=15$. Therefore, the answer is $15\times16=\boxed{\textbf{(A) }240}$;

## Video Solution (🚀 Quick 🚀)

~Education, the Study of Everything

## Video Solution by North America Math Contest Go Go Go

~North America Math Contest Go Go Go

~savannahsolver

~Interstigation

## Video Solution by OmegaLearn

~ pi_is_3.14

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 