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 1

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

Solution 2

$[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$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]$

We have $OC=17$ , as it is a radius, and $OD=8$ since it is half of $AD$ . This means that $CD=\sqrt{17^2-8^2}=15$ . So $16*15=\boxed{\textbf{(A)}240}$

~yofro

Solution 3 (coordinate bashing)

Let the midpoint of segment $FE$ be the origin. Evidently, point $D$ is at $(-8, 0)$ and $A$ is at $(8, 0)$ . Since points $C$ and $B$ share x-coordinates with $D$ and $A$ , respectively, we can just find the y-coordinate of $B$ (which is just the width of the rectangle) and multiply this by $DA$ , or $16$ . Since the radius of the semicircle is $\frac{9+16+9}{2}$ , or $17$ , the equation of the circle that our semicircle is a part of is $x^2+y^2=289$ . Since we know that the x-coordinate of $B$ is $8$ , we can plug this into our equation to obtain that $y=\pm15$ . Since $y>0$ , as the diagram suggests, we know that the y-coordinate of $B$ is $15$ . Therefore, our answer is $16\cdot 15$ , or $\boxed{\textbf{(A) }240}$ .

NOTE: The synthetic solution is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier solution.

- StarryNight7210

Solution 4

First, realize that $ABCD$ is not a square. Let $O$ be the midpoint of $FE$ . Since $FE=9+9+16=34$ , we have $OF=OE=\frac{34}{2}=17=OB$ because they are all radii. Since $O$ is also the midpoint of $AD$ , we have $OA=\frac{16}2=8$ . By the Pythagorean Theorem on $\triangle BAO$ , we find that $AB=15$ . The answer is then $16\cdot 15=\textbf{(A) }240$ .

-franzliszt

Solution 5 -SweetMango77

This is an example of a formula in the Introduction to Algebra book (a sidenote): with a semicircle: if the diameter is $1+n$ with the $1$ part at one side, and the $n$ part at the other side, then the height from the end of the $1$ side and the start of the $n$ side is $\sqrt{n}$ .

Using this, we can scale the image down by $9$ to get what we note: The other side will be $\frac{16+9}{9}=\frac{25}{9}=\left(\frac{5}{3}\right)^2$ . Then, the height of that part will be $\frac{5}{3}$ . But, we have to scale it back up by $9$ to get a height of $15$ . Multiplying by $16$ gives our desired answer: $\boxed{\textbf{(A) }240}$ .

Solution 6

A good diagram goes a long way. With a proper compass and ruler, this problem can be easily solved by a quick sketch. However, use this as a last resort. ==Solution 7== The other side will be $\frac{16+9}{9}=\frac{25} which we know it is a$ AB=15 $. or so$ 16\cdot 15=\textbf{(A) }240$. ~oceanxia

2020 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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

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