Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AMC 12B Problems/Problem 16"
(→‎Solution 2)
 
(22 intermediate revisions by 13 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Regular hexagon <math>ABCDEF</math> has vertices <math>A</math> and <math>C</math> at <math>(0,0)</math> and <math>(7,1)</math>, respectively. What is its area?
 +
 +
<math>
 +
\mathrm{(A)}\ 20\sqrt {3}
 +
\qquad
 +
\mathrm{(B)}\ 22\sqrt {3}
 +
\qquad
 +
\mathrm{(C)}\ 25\sqrt {3}
 +
\qquad
 +
\mathrm{(D)}\ 27\sqrt {3}
 +
\qquad
 +
\mathrm{(E)}\ 50
 +
</math>
 +
 +
 +
<!-- ~mathbeast9 -->
  
 
== Solution ==
 
== Solution ==
 +
 +
To find the area of the regular hexagon, we only need to calculate the side length.
 +
a distance of <math>\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}</math> apart.  Half of this distance is the length of the longer leg of the right triangles.  Therefore, the side length of the hexagon is <math>\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}</math>.
 +
 +
The apothem is thus <math>\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}</math>, yielding an area of <math>\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3} \implies \mathrm{(C)}</math>.
 +
 +
== Solution 2 ==
 +
 +
Solution 2 has the exact same solution as Solution 1 denoted, but instead, we do not need to know the value of the apothem. We could just apply s, which is the side length in this problem, <math>\frac{5\sqrt{6}}{3}</math> into the hexagon area formula, <math>\frac{3  (5 \sqrt{2} )^2 \sqrt{3} }{2}=25\sqrt{3} \implies \mathrm{(C)}</math> <-- 5 root 6 over 3 instead of 5 root 2 btw
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}
 
{{AMC12 box|year=2006|ab=B|num-b=15|num-a=17}}
 +
{{MAA Notice}}

Latest revision as of 12:36, 2 July 2024

Problem

Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area?

$\mathrm{(A)}\ 20\sqrt {3} \qquad \mathrm{(B)}\ 22\sqrt {3} \qquad \mathrm{(C)}\ 25\sqrt {3} \qquad \mathrm{(D)}\ 27\sqrt {3} \qquad \mathrm{(E)}\ 50$


Solution

To find the area of the regular hexagon, we only need to calculate the side length. a distance of $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}$.

The apothem is thus $\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}$, yielding an area of $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3} \implies \mathrm{(C)}$.

Solution 2

Solution 2 has the exact same solution as Solution 1 denoted, but instead, we do not need to know the value of the apothem. We could just apply s, which is the side length in this problem, $\frac{5\sqrt{6}}{3}$ into the hexagon area formula, $\frac{3 (5 \sqrt{2} )^2 \sqrt{3} }{2}=25\sqrt{3} \implies \mathrm{(C)}$ <-- 5 root 6 over 3 instead of 5 root 2 btw

See also

2006 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_16&oldid=221509"