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Difference between revisions of "1989 AHSME Problems/Problem 3"
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== Problem ==
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A square is cut into three rectangles along two lines parallel to a side, as shown. If the perimeter of each of the three rectangles is 24, then the area of the original square is
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<asy>
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draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);
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draw((3,0)--(3,9), dashed);
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draw((6,0)--(6,9), dashed);</asy>
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<math> \textrm{(A)}\ 24\qquad\textrm{(B)}\ 36\qquad\textrm{(C)}\ 64\qquad\textrm{(D)}\ 81\qquad\textrm{(E)}\ 96 </math>
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== Solution ==
 
Let <math>x</math> be the width of a rectangle so that the side of the square is <math>3x</math>. Since <math>2(3x)+2(x)=24</math> we have <math>x=3</math>. Thus the area of the square is <math>(3\cdot3)^2=81</math>.
 
Let <math>x</math> be the width of a rectangle so that the side of the square is <math>3x</math>. Since <math>2(3x)+2(x)=24</math> we have <math>x=3</math>. Thus the area of the square is <math>(3\cdot3)^2=81</math>.
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== See also ==
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{{AHSME box|year=1989|num-b=2|num-a=4}} 
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[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:40, 22 October 2014

Problem

A square is cut into three rectangles along two lines parallel to a side, as shown. If the perimeter of each of the three rectangles is 24, then the area of the original square is

[asy] draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(3,9), dashed); draw((6,0)--(6,9), dashed);[/asy]

$\textrm{(A)}\ 24\qquad\textrm{(B)}\ 36\qquad\textrm{(C)}\ 64\qquad\textrm{(D)}\ 81\qquad\textrm{(E)}\ 96$


Solution

Let $x$ be the width of a rectangle so that the side of the square is $3x$. Since $2(3x)+2(x)=24$ we have $x=3$. Thus the area of the square is $(3\cdot3)^2=81$.

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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

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