2020 AMC 8 Problems/Problem 18

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 (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 $DE = 25$ . Substituting in these values, $9\cdot 25 = CD^2$ , giving $CD^2 = 225$ and $CD = 15$ . The area of the rectangle is therefore $15\cdot 16 = \boxed{\textbf{(A) }240}$ .

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=5Qo4pG3Uk_U

~North America Math Contest Go Go Go

Video Solution by WhyMath

https://youtu.be/l9wZS3qGSCg

~savannahsolver

Video Solution

https://youtu.be/VnOecUiP-SA

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=852

~Interstigation

Video Solution by OmegaLearn

https://youtu.be/7SwJdAEOeAg?t=23

~ pi_is_3.14

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

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

Page

Toolbox

Search