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Difference between revisions of "1994 AHSME Problems/Problem 2"

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We can easily see the dimensions of each small rectangle. So the area of the last rectangle is <math>3\times 5=\boxed{\textbf{(B) }15}</math>.
 
We can easily see the dimensions of each small rectangle. So the area of the last rectangle is <math>3\times 5=\boxed{\textbf{(B) }15}</math>.
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
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==Solution 2==
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<asy>
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pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7);
 +
path CH=C--H;
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path BF=B--F;
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path FC=F--C;
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path DH=D--H;
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draw(A--B--C--D--cycle);
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draw(EE--F);
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draw(G--H);
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draw(CH,L=Label("$b$",position=MidPoint,align=(0,1)));
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draw(BF,L=Label("$y$",position=MidPoint,align=(1,0)));
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draw(FC,L=Label("$x$",position=MidPoint,align=(1,0)));
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draw(DH,L=Label("$a$",position=MidPoint,align=(0,1)));
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label("$6$", (1.5,6));
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label("$14$", (6.5,6));
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label("$35$", (6.5,2.5));
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</asy>
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 +
In case its not immediately obvious from inspection what the dimensions of the small rectangles are, we can work it out.  We know <math>ax</math> and <math>bx</math> and <math>by</math> and we want to know <math>ay</math>.  We can compute it as follows <math>ay = \frac{(ax)(by)}{bx} = \frac{ 6\cdot 35 }{14} = 15</math> and the answer is <math>\fbox{B}</math>
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==See Also==
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 +
{{AHSME box|year=1994|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 01:26, 28 May 2021

Problem

A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle? [asy] draw((0,0)--(10,0)--(10,7)--(0,7)--cycle); draw((0,5)--(10,5)); draw((3,0)--(3,7)); label("6", (1.5,6)); label("?", (1.5,2.5)); label("14", (6.5,6)); label("35", (6.5,2.5)); [/asy]

$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20 \qquad\textbf{(D)}\ 21 \qquad\textbf{(E)}\ 25$

Solution

[asy] pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7); path BG=shift(0,-0.5)*(B--G); path BF=shift(0.5,0)*(B--F); path FC=shift(0.5,0)*(F--C); path DH=shift(0,0.5)*(D--H); draw(A--B--C--D--cycle); draw(EE--F); draw(G--H); draw(BG,L=Label("$7$",position=MidPoint,align=(0,-1)),arrow=Arrows(),bar=Bars,red); draw(BF,L=Label("$5$",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(FC,L=Label("$2$",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(DH,L=Label("$3$",position=MidPoint,align=(0,1)),arrow=Arrows(),bar=Bars,red); label("$6$", (1.5,6)); label("$15$", (1.5,2.5),blue); label("$14$", (6.5,6)); label("$35$", (6.5,2.5)); [/asy]

We can easily see the dimensions of each small rectangle. So the area of the last rectangle is $3\times 5=\boxed{\textbf{(B) }15}$.

--Solution by TheMaskedMagician

Solution 2

[asy] pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7); path CH=C--H; path BF=B--F; path FC=F--C; path DH=D--H; draw(A--B--C--D--cycle); draw(EE--F); draw(G--H); draw(CH,L=Label("$b$",position=MidPoint,align=(0,1))); draw(BF,L=Label("$y$",position=MidPoint,align=(1,0))); draw(FC,L=Label("$x$",position=MidPoint,align=(1,0))); draw(DH,L=Label("$a$",position=MidPoint,align=(0,1))); label("$6$", (1.5,6)); label("$14$", (6.5,6)); label("$35$", (6.5,2.5)); [/asy]

In case its not immediately obvious from inspection what the dimensions of the small rectangles are, we can work it out. We know $ax$ and $bx$ and $by$ and we want to know $ay$. We can compute it as follows $ay = \frac{(ax)(by)}{bx} = \frac{ 6\cdot 35 }{14} = 15$ and the answer is $\fbox{B}$

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

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