Difference between revisions of "1979 AHSME Problems/Problem 10"
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==Solution== | ==Solution== | ||
| + | Solution by e_power_pi_times_i | ||
| + | |||
| + | Notice that quadrilateral <math>Q_1Q_2Q_3Q_4</math> consists of <math>3</math> equilateral triangles with side length <math>2</math>. Thus the area of the quadrilateral is <math>3\cdot(\frac{2^2\cdot\sqrt{3}}{4}) = 3\cdot\sqrt{3} = \boxed{\textbf{(D) } 3\sqrt{3}}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 11:33, 6 January 2017
Problem 10
If 
is a regular hexagon whose apothem (distance from the center to midpoint of a side) is 
,
and 
is the midpoint of side 
for 
, then the area of quadrilateral 
is
Solution
Solution by e_power_pi_times_i
Notice that quadrilateral consists of 
equilateral triangles with side length . Thus the area of the quadrilateral is 
.
See also
| 1979 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
