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

Revision as of 08:41, 20 November 2020

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

$[asy] draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot("$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 $OD=OA=\frac{16}{2}=$ . 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: 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, we scale the figure down by $9$ ; then the height is $\sqrt{1+\frac{16}{9}} = \sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$ . Now, scaling back up by $9$ , the height $DC$ is $9 \cdot \frac{5}{3} = 15$ . The answer is now $15 \cdot 16 = \boxed{\textbf{(A) }240}$ .

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.

@@ Line 1: / Line 1: @@
+==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>
@@ Line 7: / Line 8: @@
 ==Solution 1==
-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
-==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 <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>
+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 in fact the midpoint of <math>DA</math>, so <math>OD=OA=\frac{16}{2}= </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>.
-~yofro
-==Solution 3 (coordinate bashing)==
-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>.
-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 <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>.
--franzliszt
-==Solution 5 -SweetMango77==
+==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>.
-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>.
+(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.)
-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 3==
+We can use a result from the Art of Problem Solving <i>Introduction to Algebra</i> book: 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, we scale the figure down by <math>9</math>; then the height is <math>\sqrt{1+\frac{16}{9}} = \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 now <math>15 \cdot 16 = \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}{9}</math> which we know it is a <math>AB=15</math>. or so <math>16\cdot 15=\textbf{(A) }240</math>.
-~oceanxia
 {{AMC8 box|year=2020|num-b=17|num-a=19}}
 [[Category:Introductory Geometry Problems]]
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