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Difference between revisions of "1970 AHSME Problems/Problem 24"

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\text{(E) } 12</math>
 
\text{(E) } 12</math>
  
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== Solution ==
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== Solution 1 ==
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<math>\fbox{B}</math>
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Let <math>ABCDEF</math> be our regular hexagon, with centre <math>O</math> - and join <math>AO, BO, CO, DO, EO, </math> and <math>FO</math>. Note that we form six equilateral triangles with sidelength <math>\frac{s}{2}</math>, where <math>s</math> is the sidelength of the triangle (since the perimeter of the two polygons are equal). If the area of the original equilateral triangle is <math>2</math>, then the area of each of the six smaller triangles is <math>1/4\times 2 = \frac{1}{2}</math> (by similarity area ratios).
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Thus, the area of the hexagon is <math>6\times \frac{1}{2}=3</math>, hence our answer is <math>\fbox{B}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 01:32, 29 June 2018

Problem

An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is $2$, then the area of the hexagon is

$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 6\quad \text{(E) } 12$

Solution 1

Let $ABCDEF$ be our regular hexagon, with centre $O$ - and join $AO, BO, CO, DO, EO,$ and $FO$. Note that we form six equilateral triangles with sidelength $\frac{s}{2}$, where $s$ is the sidelength of the triangle (since the perimeter of the two polygons are equal). If the area of the original equilateral triangle is $2$, then the area of each of the six smaller triangles is $1/4\times 2 = \frac{1}{2}$ (by similarity area ratios).

Thus, the area of the hexagon is $6\times \frac{1}{2}=3$, hence our answer is $\fbox{B}$

See also

1970 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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