Difference between revisions of "2014 AMC 10A Problems/Problem 13"
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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> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
==Solution 1== | ==Solution 1== | ||
Line 53: | Line 54: | ||
==Solution 2== | ==Solution 2== | ||
− | + | As seen in the previous solution, segment <math>GH</math> is <math>\sqrt{3}</math>. Think of the picture as one large equilateral triangle, <math>\triangle{JKL}</math> with the sides of <math>2\sqrt{3}+1</math>, by extending <math>EF</math>, <math>GH</math>, and <math>DI</math> to points <math>J</math>, <math>K</math>, and <math>L</math>, respectively. This makes the area of <math>\triangle{JKL}</math> <math>\dfrac{\sqrt{3}}{4}(2\sqrt{3}+1)^2=\dfrac{12+13\sqrt{3}}{4}</math>. | |
− | + | <asy> | |
+ | import graph; | ||
+ | size(10cm); | ||
+ | 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; | ||
+ | |||
+ | pair J = rotate(60,I)*D; | ||
+ | pair K = rotate(60,E)*F; | ||
+ | pair L = rotate(60,G)*H; | ||
+ | |||
+ | 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); | ||
+ | |||
+ | draw(I--J--D); | ||
+ | draw(E--K--F); | ||
+ | draw(G--L--H); | ||
+ | |||
+ | 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); | ||
+ | label("$J$",J,SW); | ||
+ | label("$K$",K,N); | ||
+ | label("$L$",L,SE); | ||
+ | </asy> | ||
+ | |||
+ | Triangles <math>\triangle{DIJ}</math>, <math>\triangle{EFK}</math>, and <math>\triangle{GHL}</math> have sides of <math>\sqrt{3}</math>, so their total area is <math>3(\dfrac{\sqrt{3}}{4}(\sqrt{3})^2)=\dfrac{9\sqrt{3}}{4}</math>. | ||
+ | |||
+ | Now, you subtract their total area from the area of <math>\triangle{JKL}</math>: | ||
+ | |||
+ | <math>\dfrac{12+13\sqrt{3}}{4}-\dfrac{9\sqrt{3}}{4}=\dfrac{12+13\sqrt{3}-9\sqrt{3}}{4}=\dfrac{12+4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | We will use, <math>\frac{1}{2}ab\sin x</math> to find the area of the following triangles. Since <math>\angle A=360</math>, <math>\angle EAF=360-90-90-60=120</math>. | ||
+ | |||
+ | The Area of <math>\triangle AEF</math> is <math>\frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(120)</math>. Noting, <math>\sin (2x) = 2\sin (x)\cos (x)</math>, | ||
+ | |||
+ | Area of <math>\triangle AEF = \frac{1}{2} \cdot 1 \cdot 1 \cdot 2 \cdot \sin(60) \cdot \cos(60) = \dfrac{\sqrt{3}}{4}</math>, | ||
+ | |||
+ | Area of <math>\triangle ABC = \frac{1}{2} \cdot 1 \cdot 1 \cdot \sin(60) = \dfrac{\sqrt{3}}{4}</math>, | ||
+ | |||
+ | Area of square ABDE = 1, | ||
+ | |||
+ | Therefore the composite area of the entire figure is, <cmath>3 \cdot [\triangle AEF] + [\triangle ABC] + 3 \cdot [ABDE] = 3 \dfrac{\sqrt{3}}{4} + \dfrac{\sqrt{3}}{4} + 3 \cdot 1 = 4 \dfrac{\sqrt{3}}{4} + 3 = \sqrt{3} + 3 \implies\boxed{\textbf{(C)}\ 3+\sqrt3}</cmath> | ||
+ | |||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | We know that the area is equal to 3*EAF+3*ACGF+ABC. We also know that ACGF and the rest of the squares' area is equal to 1. Therefore the answer is 3*EAF+ABC+3. The only one with "+3" or "3+" is C, our answer. Very unreliable. | ||
+ | -Reality Writes | ||
+ | |||
+ | ==Solution 5== | ||
+ | <math>\angle{AEF} = 180- \angle{BAC} = 120</math> | ||
+ | |||
+ | The area of the obtuse triangle is <math>\frac{1}{2}\sin{120} = \frac{\sqrt{3}}{4}</math> | ||
+ | |||
+ | The total area is <math>3\left(1 + \frac{\sqrt{3}}{4}\right) + \frac{\sqrt{3}}{4} = \sqrt{3} + 3</math> | ||
+ | |||
+ | ~mathboy282 | ||
− | |||
==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 21:45, 26 September 2021
Problem
Equilateral has side length , and squares , , lie outside the triangle. What is the area of hexagon ?
Solution 1
The area of the equilateral triangle is . The area of the three squares is .
Since , .
Dropping an altitude from to allows to create a triangle since is isosceles. This means that the height of is and half the length of is . Therefore, the area of each isosceles triangle is . Multiplying by yields for all three isosceles triangles.
Therefore, the total area is .
Solution 2
As seen in the previous solution, segment is . Think of the picture as one large equilateral triangle, with the sides of , by extending , , and to points , , and , respectively. This makes the area of .
Triangles , , and have sides of , so their total area is .
Now, you subtract their total area from the area of :
Solution 3
We will use, to find the area of the following triangles. Since , .
The Area of is . Noting, ,
Area of ,
Area of ,
Area of square ABDE = 1,
Therefore the composite area of the entire figure is,
Solution 4
We know that the area is equal to 3*EAF+3*ACGF+ABC. We also know that ACGF and the rest of the squares' area is equal to 1. Therefore the answer is 3*EAF+ABC+3. The only one with "+3" or "3+" is C, our answer. Very unreliable. -Reality Writes
Solution 5
The area of the obtuse triangle is
The total area is
~mathboy282
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 |
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