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==
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==See Also==
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{{AMC10 box|year=2014|ab=A|num-b=14|num-a=15}}
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{{MAA Notice}}

Revision as of 22:17, 6 February 2014

Problem

Equilateral $\triangle ABC$ has side length $1$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$?

[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]

$\textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6$

Solution

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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

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