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Difference between revisions of "1954 AHSME Problems/Problem 5"

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== Solution ==
 
== Solution ==
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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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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>

Revision as of 19:36, 17 February 2020

Problem 5

A regular hexagon is inscribed in a circle of radius $10$ inches. Its area is:

$\textbf{(A)}\ 150\sqrt{3} \text{ sq. in.} \qquad \textbf{(B)}\ \text{150 sq. in.} \qquad \textbf{(C)}\ 25\sqrt{3}\text{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.}$

Solution

Using the formula for the area of a hexagon given the circumradius: $\frac{3\sqrt{3}}{2\pi}=\frac{A}{10^2\pi}\implies 100\cdot3\sqrt{3}\pi=2A\pi\implies 50\cdot3\sqrt{3}=A\implies 150\sqrt{3}\ \boxed{(\textbf{A})}$

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