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Difference between revisions of "1983 AHSME Problems/Problem 4"
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== 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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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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The area of the figure is
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<math>
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\text{(A)} \ \frac{\sqrt 3}{2} \qquad
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\text{(B)} \ 1 \qquad
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\text{(C)} \ \frac{3}{2} \qquad
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\text{(D)}\ \sqrt{3}\qquad
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\text{(E)}\ </math>   
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[[1983 AHSME Problems/Problem 4|Solution]]
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==Solution==
 
<asy>
 
<asy>
 
pair A, B, C, D, E, F;
 
pair A, B, C, D, E, F;
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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>.
 
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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==See Also==
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{{AHSME box|year=1983|num-b=3|num-a=5}}
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{{MAA Notice}}

Revision as of 07:58, 2 September 2017

Problem 4

Position $A,B,C,D,E,F$ such that $AF$ and $CD$ are parallel, as are sides $AB$ and $EF$, and sides $BC$ and $ED$. Each side has length of $1$ and it is given that $\measuredangle FAB = \measuredangle BCD = 60^\circ$. The area of the figure is

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

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, A); label("B", (0.3, 0.866)); label("C", (-0.1, 0)); label("D", D, D); label("E", E, E); label("F", F, F); draw(B--D, dashed+linewidth(0.5)); draw(B--E, dashed+linewidth(0.5)); draw(B--F, dashed+linewidth(0.5)); [/asy]

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 $\frac{\sqrt{3}}{4}$, 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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