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Difference between revisions of "1958 AHSME Problems/Problem 16"

(Created page with "== Problem == The area of a circle inscribed in a regular hexagon is <math> 100\pi</math>. The area of hexagon is: <math> \textbf{(A)}\ 600\qquad \textbf{(B)}\ 300\qquad \tex...")
 
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== Solution ==
 
== Solution ==
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<math>\fbox{}</math>
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We can split the hexagon into 6 equilateral triangles. If the area of the circle is <math>100pi</math>, then the radius is <math>10</math>. The radius is equal to the height of one of the equilateral triangles. Using <math>sin60 = \frac{\sqrt3}{2}</math>, we get the hypotenuse is <math>\frac{20\sqrt{3}}{3}</math>, which is also equal to the side length. Using the hexagon formula (or going back to the equilateral triangle formula <math>\cdot 6</math>), we get
 +
 
 +
<math>\frac{6}{4} \cdot s^2\sqrt{3} \Rightarrow</math> <math>\frac{6}{4} \cdot (\frac{20\sqrt3}{3})^2 \cdot \sqrt3 \Rightarrow</math> <math> \frac{6}{4} \cdot \frac{20\sqrt3}{3} \cdot \frac{20\sqrt3}{3} \cdot \sqrt3 \Rightarrow</math> <math>\boxed{200\sqrt3}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 00:38, 12 March 2017

Problem

The area of a circle inscribed in a regular hexagon is $100\pi$. The area of hexagon is:

$\textbf{(A)}\ 600\qquad \textbf{(B)}\ 300\qquad \textbf{(C)}\ 200\sqrt{2}\qquad \textbf{(D)}\ 200\sqrt{3}\qquad \textbf{(E)}\ 120\sqrt{5}$

Solution

We can split the hexagon into 6 equilateral triangles. If the area of the circle is $100pi$, then the radius is $10$. The radius is equal to the height of one of the equilateral triangles. Using $sin60 = \frac{\sqrt3}{2}$, we get the hypotenuse is $\frac{20\sqrt{3}}{3}$, which is also equal to the side length. Using the hexagon formula (or going back to the equilateral triangle formula $\cdot 6$), we get

$\frac{6}{4} \cdot s^2\sqrt{3} \Rightarrow$ $\frac{6}{4} \cdot (\frac{20\sqrt3}{3})^2 \cdot \sqrt3 \Rightarrow$ $\frac{6}{4} \cdot \frac{20\sqrt3}{3} \cdot \frac{20\sqrt3}{3} \cdot \sqrt3 \Rightarrow$ $\boxed{200\sqrt3}$

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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