Difference between revisions of "1954 AHSME Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | Using the formula for the area of a hexagon given the circumradius: <math>\frac{3\sqrt{3}}{2\pi}=\frac{A}{10^2\pi}\implies 100\cdot3\sqrt{3}\pi=2A\pi\implies 50\cdot3\sqrt{3}=A\implies 150\sqrt{3} \boxed{(\textbf{A})}</math> | + | Using the formula for the area of a hexagon given the circumradius: <math>\frac{3\sqrt{3}}{2\pi}=\frac{A}{10^2\pi}\implies 100\cdot3\sqrt{3}\pi=2A\pi\implies 50\cdot3\sqrt{3}=A\implies 150\sqrt{3}\ \boxed{(\textbf{A})}</math> |
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| + | ==See Also== | ||
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| + | {{AHSME 50p box|year=1954|num-b=4|num-a=6}} | ||
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| + | {{MAA Notice}} | ||
Latest revision as of 19:36, 17 February 2020
Problem 5
A regular hexagon is inscribed in a circle of radius 
inches. Its area is:
Solution
Using the formula for the area of a hexagon given the circumradius: 
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
