Difference between revisions of "2014 AMC 10A Problems/Problem 13"
(Created page with "==Problem== Equilateral <math>\triangle ABC</math> has side length <math>1</math>, and squares <math>ABDE</math>, <math>BCHI</math>, <math>CAFG</math> lie outside the triangle. ...") |
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<math> \textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6 </math> | <math> \textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6 </math> | ||
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| + | ==Solution== | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AMC10 box|year=2014|ab=A|num-b=14|num-a=15}} | ||
| + | {{MAA Notice}} | ||
Revision as of 23:17, 6 February 2014
Problem
Equilateral 
has side length 
, and squares 
, 
, 
lie outside the triangle. What is the area of hexagon 
?
![[asy] import graph; size(6cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,W); label("$F$",F,E); label("$G$",G,E); label("$H$",H,SE); label("$I$",I,SW); [/asy]](http://latex.artofproblemsolving.com/3/8/5/385e803d10dad89681d6fb0e49d8ece31a8ba00e.png)
Solution
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 15 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
