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Difference between revisions of "1983 AHSME Problems/Problem 4"

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== Problem 4 ==
 
== Problem 4 ==
  
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Position <math>A,B,C,D,E,F</math> such that <math>AF</math> and <math>CD</math> are parallel, as are sides <math>AB</math> and <math>EF</math>,  
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[[File:pdfresizer.com-pdf-convert.png]]
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and sides <math>BC</math> and <math>ED</math>. Each side has length of <math>1</math> and it is given that <math>\measuredangle FAB = \measuredangle BCD = 60^\circ</math>.  
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In the adjoining plane figure, sides <math>AF</math> and <math>CD</math> are parallel, as are sides <math>AB</math> and <math>EF</math>,  
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and sides <math>BC</math> and <math>ED</math>. Each side has length <math>1</math>. Also, <math>\angle FAB = \angle BCD = 60^\circ</math>.  
 
The area of the figure is  
 
The area of the figure is  
  
 
<math>
 
<math>
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\text{(A)} \ \frac{\sqrt 3}{2} \qquad  
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\textbf{(A)} \ \frac{\sqrt 3}{2} \qquad  
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\text{(B)} \ 1 \qquad  
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\textbf{(B)} \ 1 \qquad  
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\text{(C)} \ \frac{3}{2} \qquad  
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\textbf{(C)} \ \frac{3}{2} \qquad  
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\text{(D)}\ \sqrt{3}\qquad
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\textbf{(D)}\ \sqrt{3}\qquad
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\text{(E)}\ </math>     
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\textbf{(E)}\ 2</math>     
  
 
[[1983 AHSME Problems/Problem 4|Solution]]
 
[[1983 AHSME Problems/Problem 4|Solution]]
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</asy>
 
</asy>
  
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Drawing the diagram as described, we create a convex hexagon with all side lengths equal to 1. In this case, it is a natural approach to divide the figure up into four equilateral triangles. The area, A, of one such equilateral triangle is <math>\frac{\sqrt{3}}{4}</math>, which gives a total of <math>4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}</math>, or <math>\boxed{D}</math>.
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By rotating the diagram and drawing the dotted lines, we see that the figure can be divided into four equilateral triangles, each of side length <math>1</math>. The area of one such equilateral triangle is <math>\frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}</math>, which gives a total of <math>4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}</math>, or <math>\boxed{D}</math>.
  
 
==See Also==
 
==See Also==
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{{MAA Notice}}
 
{{MAA Notice}}
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[[Category: Introductory Geometry Problems]]

Latest revision as of 05:36, 27 January 2019

Problem 4

Pdfresizer.com-pdf-convert.png

In the adjoining plane figure, sides $AF$ and $CD$ are parallel, as are sides $AB$ and $EF$, and sides $BC$ and $ED$. Each side has length $1$. Also, $\angle FAB = \angle BCD = 60^\circ$. The area of the figure is

$\textbf{(A)} \ \frac{\sqrt 3}{2} \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3}\qquad \textbf{(E)}\ 2$

Solution

Solution

[asy] pair A, B, C, D, E, F; A = (0, 1.732); B = (0.5, 0.866); C = (0,0); D = (1, 0); E = (1.5, 0.866); F = (1, 1.732); draw(A--B--C--D--E--F--A); label("$A$", A, NW); label("$B$", B, 3W); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E); label("$F$", F, NE); draw(B--D, dashed+linewidth(0.5)); draw(B--E, dashed+linewidth(0.5)); draw(B--F, dashed+linewidth(0.5)); [/asy]

By rotating the diagram and drawing the dotted lines, we see that the figure can be divided into four equilateral triangles, each of side length $1$. The area of one such equilateral triangle is $\frac{\sqrt{3}}{4} \cdot 1^2 = \frac{\sqrt{3}}{4}$, which gives a total of $4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}$, or $\boxed{D}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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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