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 "2014 AMC 10A Problems/Problem 13"

(→‎See Also: dates)
(→‎Solution)
Line 44: Line 44:
  
 
==Solution==
 
==Solution==
−
The area of the equilateral triangle is simply <math>\dfrac{\sqrt{3}}{4}</math>. The area of the three squares is <math>3\times 1=3</math>.
+
The area of the equilateral triangle is <math>\dfrac{\sqrt{3}}{4}</math>. The area of the three squares is <math>3\times 1=3</math>.
  
 
Since <math>\angle C=360</math>, <math>\angle GCH=360-90-90-60=120</math>.
 
Since <math>\angle C=360</math>, <math>\angle GCH=360-90-90-60=120</math>.
Line 51: Line 51:
  
 
Therefore, the total area is <math>3+\dfrac{\sqrt{3}}{4}+\dfrac{3\sqrt{3}}{4}=3+\dfrac{4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}</math>.
 
Therefore, the total area is <math>3+\dfrac{\sqrt{3}}{4}+\dfrac{3\sqrt{3}}{4}=3+\dfrac{4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}</math>.
 +
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2014|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2014|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:30, 9 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

The area of the equilateral triangle is $\dfrac{\sqrt{3}}{4}$. The area of the three squares is $3\times 1=3$.

Since $\angle C=360$, $\angle GCH=360-90-90-60=120$.

Dropping an altitude from $C$ to $GH$ allows to create a $30-60-90$ triangle since $\triangle GCH$ is isosceles. This means that the height of $\triangle GCH$ is $\dfrac{1}{2}$ and half the length of $GH$ is $\dfrac{\sqrt{3}}{2}$. Therefore, the area of each isosceles triangle is $\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{4}$. Multiplying by $3$ yields $\dfrac{3\sqrt{3}}{4}$ for all three isosceles triangles.

Therefore, the total area is $3+\dfrac{\sqrt{3}}{4}+\dfrac{3\sqrt{3}}{4}=3+\dfrac{4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}$.

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_13&oldid=59504"