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Difference between revisions of "2020 AMC 8 Problems/Problem 18"

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~yofro
 
~yofro
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==Solution 3 (coordinate bashing)==
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Let the midpoint of segment <math>FE</math> be the origin. Evidently, point <math>D</math> is at <math>(-8, 0)</math> and <math>A</math> is at <math>(8, 0)</math>. Since points <math>C</math> and <math>B</math> share x-coordinates with <math>D</math> and <math>A</math>, respectively, we can just find the y-coordinate of <math>B</math> (which is just the width of the rectangle) and multiply this by <math>DA</math>, or <math>16</math>. Since the radius of the semicircle is <math>\frac{9+16+9}{2}</math>, or <math>17</math>, the equation of the circle that our semicircle is a part of is <math>x^2+y^2=289</math>. Since we know that the x-coordinate of <math>B</math> is <math>8</math>, we can plug this into our equation to obtain that <math>y=\pm15</math>. Since <math>y>0</math>, as the diagram suggests, we know that the y-coordinate of <math>B</math> is <math>15</math>. Therefore, our answer is <math>16\cdot 15</math>, or <math>\boxed{\textbf{(A) }240}</math>.
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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.
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- StarryNight7210
  
 
==See also==
 
==See also==

Revision as of 13:01, 18 November 2020

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


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

See also

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

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