Difference between revisions of "1970 AHSME Problems/Problem 24"
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\text{(E) } 12</math> | \text{(E) } 12</math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | <math>\fbox{B}</math> | + | 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). |
| + | |||
| + | 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 == | ||
Revision as of 02:32, 29 June 2018
Problem
An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 
, then the area of the hexagon is
Solution 1
Let 
be our regular hexagon, with centre 
- and join 
and 
. Note that we form six equilateral triangles with sidelength 
, where 
is the sidelength of the triangle (since the perimeter of the two polygons are equal.) If the area of the original equilateral triangle is , then the area of each of the six smaller triangles is 
(by similarity area ratios).
Thus, the area of the hexagon is 
, hence our answer is
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
