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

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==Problem==
 
Rectangle <math>ABCD</math> is inscribed in a semicircle with diameter <math>\overline{FE},</math> as shown in the figure. Let <math>DA=16,</math> and let <math>FD=AE=9.</math> What is the area of <math>ABCD?</math>
 
Rectangle <math>ABCD</math> is inscribed in a semicircle with diameter <math>\overline{FE},</math> as shown in the figure. Let <math>DA=16,</math> and let <math>FD=AE=9.</math> What is the area of <math>ABCD?</math>
  
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<math>\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272</math>
 
<math>\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272</math>
  
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==Solution 1==
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==Solution 1 (Pythagorean Theorem)==
−
First, realize <math>ABCD</math> is not a square. It can easily be seen that the diameter of the semicircle is <math>9+16+9=34</math>, so the radius is <math>\frac{34}{2}=17</math>. Express the area of Rectangle <math>ABCD</math> as <math>16h</math>, where <math>h=AB</math>. Notice that by the Pythagorean theorem <math>8^2+h^{2}=17^{2}\implies h=15</math>. Then, the area of Rectangle <math>ABCD</math> is equal to <math>16\cdot 15=\boxed{\textbf{(A) }240}</math>. ~icematrix
+
<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 <math>O</math> be the center of the semicircle. The diameter of the semicircle is <math>9+16+9=34</math>, so <math>OC = 17</math>. By symmetry, <math>O</math> is the midpoint of <math>DA</math>, so <math>OD=OA=\frac{16}{2}= 8</math>. By the Pythagorean theorem in right-angled triangle <math>ODC</math> (or <math>OBA</math>), we have that <math>CD</math> (or <math>AB</math>) is <math>\sqrt{17^2-8^2}=15</math>. Accordingly, the area of <math>ABCD</math> is <math>16\cdot 15=\boxed{\textbf{(A) }240}</math>.
 +
 
 +
==Solution 2 (Coordinate Geometry)==
 +
Let the midpoint of segment <math>FE</math> be the origin. Evidently, point <math>D=(-8,0)</math> and <math>A=(8,0)</math>. Since points <math>C</math> and <math>B</math> share <math>x</math>-coordinates with <math>D</math> and <math>A</math> respectively, it suffices to find the <math>y</math>-coordinate of <math>B</math> (which will be the height of the rectangle) and multiply this by <math>DA</math> (which we know is <math>16</math>). The radius of the semicircle is <math>\frac{9+16+9}{2} = 17</math>, so the whole circle has equation <math>x^2+y^2=289</math>; as already stated, <math>B</math> has the same <math>x</math>-coordinate as <math>A</math>, i.e. <math>8</math>, so substituting this into the equation shows that <math>y=\pm15</math>. Since <math>y>0</math> at <math>B</math>, the y-coordinate of <math>B</math> is <math>15</math>. Therefore, the answer is <math>16\cdot 15 = \boxed{\textbf{(A) }240}</math>.
 +
 
 +
(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 2==
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==Solution 3==
 +
We can use a result from the Art of Problem Solving <i>Introduction to Algebra</i> book Sidenote: for a semicircle with diameter <math>(1+n)</math>, such that the <math>1</math> part is on one side and the <math>n</math> part is on the other side, the height from the end of the <math>1</math> side (or the start of the <math>n</math> side) is <math>\sqrt{n}</math>. To use this formula, we scale the figure down by <math>9</math>; this will give the height a length of <math>\sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}</math>. Now, scaling back up by <math>9</math>, the height <math>DC</math> is <math>9 \cdot \frac{5}{3} = 15</math>. The answer is then <math>15 \cdot 16 = \boxed{\textbf{(A) }240}</math>.
 +
-[[User:Sweetmango77|SweetMango77]]
  
−
<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>
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==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]], <math>FD\cdot DE = CD\cdot C'D</math>. By symmetry, <math>CD = C'D</math>. We see that <math>FD = 9</math> and <math>DE = 16 + 9 = 25</math>. Substituting in these values, <math>9\cdot 25 = CD^2</math>, giving <math>CD^2 = 225</math> and <math>CD = 15</math>.
 +
The area of the rectangle is therefore <math>15\cdot 16 = \boxed{\textbf{(A) }240}</math>.
 +
 
 +
==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 <math> OG=\sqrt{\left(\frac{2\times9+16}{2}\right)^2 - \left(\frac{16}{2}\right)^2}=15 </math>. Therefore, the answer is <math> 15\times16=\boxed{\textbf{(A) }240} </math>;
 +
 
 +
~[[User:Bloggish|Bloggish]]
 +
 
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=bHNrBwwUCMI
 +
 
 +
~NiuniuMaths
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=_0fveg1r3WFmwmpw&t=3334
 +
 
 +
~Math-X
 +
 
 +
==Video Solution (🚀 Quick 🚀)==
 +
https://youtu.be/pA6Smw91Bc0
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution by North America Math Contest Go Go Go==
 +
https://www.youtube.com/watch?v=5Qo4pG3Uk_U
  
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We have <math>OC=17</math>, as it is a radius, and <math>OD=8</math> since it is half of <math>AD</math>. This means that <math>CD=\sqrt{17^2-8^2}=15</math>. So <math>16*15=\boxed{\textbf{(A)}240}</math>
+
~North America Math Contest Go Go Go
  
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~yofro
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==Video Solution by WhyMath==
 +
https://youtu.be/l9wZS3qGSCg
  
−
==Solution 3 (coordinate bashing)==
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~savannahsolver
  
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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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==Video Solution==
 +
https://youtu.be/VnOecUiP-SA
  
−
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.
+
==Video Solution by Interstigation==
 +
https://youtu.be/YnwkBZTv5Fw?t=852
  
−
- StarryNight7210
+
~Interstigation
  
−
==Solution 4==
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==Video Solution by OmegaLearn==
 +
https://youtu.be/7SwJdAEOeAg?t=23
  
−
First, realize that <math>ABCD</math> is not a square. Let <math>O</math> be the midpoint of <math>FE</math>. Since <math>FE=9+9+16=34</math>, we have <math>OF=OE=\frac{34}{2}=17=OB</math> because they are all radii. Since <math>O</math> is also the midpoint of <math>AD</math>, we have <math>OA=\frac{16}2=8</math>. By the Pythagorean Theorem on <math>\triangle BAO</math>, we find that <math>AB=15</math>. The answer is then <math>16\cdot 15=\textbf{(A) }240</math>.
+
~ pi_is_3.14
  
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-franzliszt
+
==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=S_CnKCuOA_w
  
−
==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 <math>1+n</math> with the <math>1</math> part at one side, and the <math>n</math> part at the other side, then the height from the end of the <math>1</math> side and the start of the <math>n</math> side is <math>\sqrt{n}</math>.
 
  
−
Using this, we can scale the image down by <math>9</math> to get what we note: The other side will be <math>\frac{16+9}{9}=\frac{25}{9}=\left(\frac{5}{3}\right)^2</math>. Then, the height of that part will be <math>\frac{5}{3}</math>. But, we have to scale it back up by <math>9</math> to get a height of <math>15</math>. Multiplying by <math>16</math> gives our desired answer: <math>\boxed{\textbf{(A) }240}</math>.
 
  
−
==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 <math>\frac{16+9}{9}=\frac{25} which we know it is a </math>AB=15<math>. or so </math>16\cdot 15=\textbf{(A) }240$.
 
−
~oceanxia
 
 
{{AMC8 box|year=2020|num-b=17|num-a=19}}
 
{{AMC8 box|year=2020|num-b=17|num-a=19}}
−
 
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
>B)
 +
hey ya'lll

Latest revision as of 15:35, 26 January 2024

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 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 $15\times16=\boxed{\textbf{(A) }240}$;

~Bloggish

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=bHNrBwwUCMI

~NiuniuMaths

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=_0fveg1r3WFmwmpw&t=3334

~Math-X

Video Solution (🚀 Quick 🚀)

https://youtu.be/pA6Smw91Bc0

~Education, the Study of Everything

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=S_CnKCuOA_w



2020 AMC 8 (Problems • Answer Key • Resources)
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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. AMC logo.png

>B) hey ya'lll

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