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 12:33, 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 | |
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All AHSME Problems and Solutions |
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