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

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(→‎Solution 5 -SweetMango77)
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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>.  
 
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}{3}=\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>\textbf{(A) }240</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>\textbf{(A) }240</math>.
  
 
==See also==
 
==See also==

Revision as of 17:00, 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

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: $\textbf{(A) }240$.

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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