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Difference between revisions of "2016 AMC 10B Problems/Problem 23"

(→‎Solution 2)
 
(28 intermediate revisions by 15 users not shown)
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<math>\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}</math>
 
<math>\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}</math>
−
 
  
 
==Solution 1==
 
==Solution 1==
Line 16: Line 15:
 
E=(0,sqrt(3));
 
E=(0,sqrt(3));
 
F=(-1/2,sqrt(3)/2);
 
F=(-1/2,sqrt(3)/2);
−
W=(4/3,2sqrt(3)/3);
+
X=(4/3,2sqrt(3)/3);
−
X=(4/3,sqrt(3)/3);
+
W=(4/3,sqrt(3)/3);
−
Y=(-1/3,sqrt(3)/3);
+
Z=(-1/3,sqrt(3)/3);
−
Z=(-1/3,2sqrt(3)/3);
+
Y=(-1/3,2sqrt(3)/3);
 
draw(A--B--C--D--E--F--cycle);
 
draw(A--B--C--D--E--F--cycle);
 
draw(W--Z);
 
draw(W--Z);
Line 36: Line 35:
 
</asy>
 
</asy>
  
−
Assume that <math>AB</math> is of length <math>1</math>.  Therefore, the area of <math>ABCDEF</math> is <math>\frac{3\sqrt 3}2</math>.  To find the area of <math>WCXYFZ</math>, we draw <math>\overline{CF}</math>, and find the area of the trapezoids <math>WCFZ</math> and <math>CXYF</math>.  
+
Assume that <math>AB</math> is of length <math>1</math>.  Therefore, the area of <math>ABCDEF</math> is <math>\frac{3\sqrt 3}2</math>.  To find the area of <math>WCXYFZ</math>, we draw <math>\overline{CF}</math>, and find the area of the trapezoids <math>WCFZ</math> and <math>CXYF</math>.
  
 
<asy>
 
<asy>
Line 69: Line 68:
 
From this, we know that <math>CF=2</math>.  We also know that the combined heights of the trapezoids is <math>\frac{\sqrt 3}3</math>, since <math>\overline{ZW}</math> and <math>\overline{YX}</math> are equally spaced, and the height of each of the trapezoids is <math>\frac{\sqrt 3}6</math>.  From this, we know <math>\overline{ZW}</math> and <math>\overline{YX}</math> are each <math>\frac 13</math> of the way from <math>\overline{CF}</math> to <math>\overline{DE}</math> and <math>\overline{AB}</math>, respectively.  We know that these are both equal to <math>\frac 53</math>.
 
From this, we know that <math>CF=2</math>.  We also know that the combined heights of the trapezoids is <math>\frac{\sqrt 3}3</math>, since <math>\overline{ZW}</math> and <math>\overline{YX}</math> are equally spaced, and the height of each of the trapezoids is <math>\frac{\sqrt 3}6</math>.  From this, we know <math>\overline{ZW}</math> and <math>\overline{YX}</math> are each <math>\frac 13</math> of the way from <math>\overline{CF}</math> to <math>\overline{DE}</math> and <math>\overline{AB}</math>, respectively.  We know that these are both equal to <math>\frac 53</math>.
  
−
We find the area of each of the trapezoids, which both happen to be <math>\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}</math>, and the combined area is <math>\frac{11\sqrt 3}{18}^{*}</math>.
+
We find the area of each of the trapezoids, which both happen to be <math>\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}</math>, and the combined area is <math>\frac{11\sqrt 3}{18}</math>.
  
 
We find that <math>\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}</math> is equal to <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}</math>.
 
We find that <math>\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}</math> is equal to <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}</math>.
  
 +
==Solution 2==
 +
 +
<asy>
 +
pair A,B,C,D,E,F,W,X,Y,Z;
 +
A=(0,0);
 +
B=(1,0);
 +
C=(3/2,sqrt(3)/2);
 +
D=(1,sqrt(3));
 +
E=(0,sqrt(3));
 +
F=(-1/2,sqrt(3)/2);
 +
W=(4/3,2sqrt(3)/3);
 +
X=(4/3,sqrt(3)/3);
 +
Y=(-1/3,sqrt(3)/3);
 +
Z=(-1/3,2sqrt(3)/3);
 +
draw(A--B--C--D--E--F--cycle);
 +
draw(W--Z);
 +
draw(X--Y);
 +
draw(F--C--B--E--D--A);
 +
 +
label("$A$",A,SW);
 +
label("$B$",B,SE);
 +
label("$C$",C,ESE);
 +
label("$D$",D,NE);
 +
label("$E$",E,NW);
 +
label("$F$",F,WSW);
 +
label("$W$",W,ENE);
 +
label("$X$",X,ESE);
 +
label("$Y$",Y,WSW);
 +
label("$Z$",Z,WNW);
 +
</asy>
  
−
<math>^*</math> At this point, you can answer <math>\textbf{(C)}</math> and move on with your test.
+
First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are <math>22</math> small triangles in hexagon <math>ZWCXYF</math>, and <math>9 \cdot 6 = 54</math> small triangles in the whole hexagon.  
  
−
==Solution 2==
+
Thus, the answer is <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}</math>.
  
 +
==Solution 3 (Similar Triangles)==
 
<asy>
 
<asy>
−
pair A,B,C,D,E,F,W,X,Y,Z,S,K,R,U,H,I,O,P,Q;
+
pair A,B,C,D,E,F,W,X,Y,Z;
 
A=(0,0);
 
A=(0,0);
 
B=(1,0);
 
B=(1,0);
Line 90: Line 120:
 
Y=(-1/3,sqrt(3)/3);
 
Y=(-1/3,sqrt(3)/3);
 
Z=(-1/3,2sqrt(3)/3);
 
Z=(-1/3,2sqrt(3)/3);
−
S=(-1/6,sqrt(3)/6);
+
pair G = (0.5, sqrt(3)*3/2);
−
H=(-1/6, 5sqrt(3)/6);
 
−
P=(7/6, 5sqrt(3)/6);
 
−
U=(7/6,sqrt(3)/6);
 
−
K=(1/3, 0);
 
−
R=(2/3, 0);
 
−
I=(1/3,sqrt(3));
 
−
O=(2/3,sqrt(3));
 
−
Q=(1/2, sqrt(3)/2);
 
−
 
 
 
draw(A--B--C--D--E--F--cycle);
 
draw(A--B--C--D--E--F--cycle);
 
draw(W--Z);
 
draw(W--Z);
 
draw(X--Y);
 
draw(X--Y);
−
draw(F--C--B--E--D--A);
+
draw(E--G--D);
−
draw(S--U);
+
draw(F--C);
−
draw(K--R);
 
−
draw(Z--K);
 
−
draw(H--R);
 
−
draw(I--U);
 
−
draw(O--X);
 
−
draw(H--P);
 
−
draw(I--Y);
 
−
draw(O--S);
 
−
draw(P--K);
 
−
draw(W--R);
 
  
 
label("$A$",A,SW);
 
label("$A$",A,SW);
Line 126: Line 137:
 
label("$Y$",Y,WSW);
 
label("$Y$",Y,WSW);
 
label("$Z$",Z,WNW);
 
label("$Z$",Z,WNW);
−
label("$S$",S,WSW);
+
label("$G$",G,N);
−
label("$K$",K,SSW);
 
−
label("$R$",R,SSE);
 
−
label("$U$",U,ESE);
 
−
label("$H$",H,WNW);
 
−
label("$I$",I,NNW);
 
−
label("$O$",O,NNE);
 
−
label("$P$",P,ENE);
 
−
label("$Q$",Q,N);
 
 
</asy>
 
</asy>
 +
Extend <math>\overline{EF}</math> and <math>\overline{CD}</math> to meet at point <math>G</math>, as shown in the diagram. Then <math>\triangle GZW \sim \triangle GFC</math>. Then <math>[GZW] = \left(\dfrac53\right)^2[GED]</math> and <math>[GFC] = 2^2[GED]</math>. Subtracting <math>[GED]</math>, we find that <math>[EDWZ] = \dfrac{16}{9}[GED]</math> and <math>[EDCF] = 3[GED]</math>. Subtracting again, we find that <cmath>[ZWCF] = [EDCF] - [EDWZ] = \dfrac{11}{9}[GED].</cmath>Finally, <cmath>\dfrac{[WCXYFZ]}{[ABCDEF]} = \dfrac{[ZWCF]}{[EDCF]} = \dfrac{\dfrac{11}{9}[GED]}{3[GED]} = \textbf{(C) } \dfrac{11}{27}.</cmath>
 +
 +
 +
==Solution 4 (Extending Lines)==
 +
 +
Refer to the diagram from Solution 1.
 +
 +
Let us start by connecting points <math>F</math> to <math>C</math> to create a new line segment <math>FC</math>. We drop a perpendicular line segment from <math>E</math> to side <math>FC</math> at point <math>P</math>. Since <math>\angle FED = 120</math>, and <math>\angle PED = 90</math>, we know that <math>\angle FEP = 30.</math>
 +
 +
Therefore, <math>\triangle EFP</math> is a 30-60-90 triangle. Assume that <math>EP = \sqrt{3}</math>. Therefore, we know that <math>FP = 1</math>, <math>EF = 2</math>.
 +
 +
Let us draw <math>PQ</math> such that <math>PQ</math> is perpendicular to <math>ZW</math>. Since the distance between the parallel lines are equal, we know that <math>PQ</math> is half of the distance between the parallel lines. Hence, <math>PQ</math> will be one-third the length of <math>EP</math>.
  
−
First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have draw these lines, it's just a matter of counting triangles. There are <math>22</math> small triangles in hexagon <math>ZWCXYF</math>, and <math>9 \cdot 6 = 54</math> small triangles in the whole hexagon.
+
From this, we also know that <math>EQ = EP - PQ = EP - \frac{1}{3} EP = \frac{2}{3} EP</math>.  
  
−
There are <math>22</math> small triangles in hexagon <math>ZWCXYF</math>, and <math>9 \text{ small triangles} \cdot 6 \text{ triangles}= 54</math> small triangles in the whole hexagon <math>ABCDEF</math>.
+
Hence, we know that <math>EQ = \frac{2}{3} EP = \frac{2\sqrt{3}}{3}.</math> Since <math>\angle EQZ = 90, FEP = 30</math> we know that <math>\triangle ZEQ</math> is also a 30-60-90 triangle. From this, we know that <math>ZQ = \frac{2}{3}</math>.  
  
−
Thus, the answer is <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}</math>.
+
Let us drop another perpendicular line segment from <math>D</math> to point <math>T</math> such that <math>DT</math> is perpendicular to <math>ZW</math>.
 +
 
 +
Since <math>ED = QT</math> and since <math>ED</math> is the side length of the regular hexagon, we know that <math>ED = EF = QT = 2.</math>
 +
 
 +
By symmetry, we also know that <math>WT = \frac{2}{3}</math>.
 +
 
 +
Therefore, we can find the length of <math>ZW = ZQ + QT + QW = \frac{2}{3} + 2 + \frac{2}{3}</math>. Hence, we know that <math>ZW = \frac{10}{3}.</math>
 +
 
 +
Now, we can find the area of trapezoid <math>EDZW = \frac{ED + ZW}{2} \cdot EQ = \frac{2 + \frac{10}{3}}{2} \cdot \frac{2\sqrt{3}}{3} = \frac{16\sqrt{3}}{9}.</math>
 +
 
 +
By symmetry, we know that the area of trapezoid <math>YZAB = \frac{16\sqrt{3}}{9}.</math>
 +
 
 +
Using the area of a hexagon formula, we get that the area of regular hexagon <math>ABCDEF = \frac{3\sqrt{3}}{2} \cdot 2^2 = 6\sqrt{3}.</math>
 +
 
 +
Hence, the area of hexagon <math>WCXYFZ = 6\sqrt{3} - 2\left(\frac{16\sqrt{3}}{9}\right) = \frac{22\sqrt{3}}{9}.</math>
 +
 
 +
Hence, the ratio of the area of hexagon <math>WCXYFZ</math> to the area of hexagon <math>ABCDEF</math> is <math>\dfrac{\frac{22\sqrt{3}}{9}}{6\sqrt{3}} = \boxed{\frac{11}{27} \implies C}.</math>
 +
 
 +
~yk2007 :D
 +
 
 +
==Solution 5==
 +
 
 +
We will do this by area subtraction. Drawing in the required lines, we drop an altitude from A to WZ and an altitude from E to XY. You get 2 30-60-90 triangles, and given that the side length is <math>x</math>, AZ is <math>2x/3</math>, which means that WZ is <math>5x/3</math>, adding in AB. Using the 30-60-90 triangle again, you get that the area of trapezoid <math>ABWZ</math> is <math>\frac{4x}{3} \cdot</math> <math>\frac{x\sqrt3}{3}</math>. Repeat on the other side to get that the area of both of these trapezoids combined are <math>\frac{8x^2\sqrt3}{9}</math>. Finding the area of the hexagon, dividing, and subtracting, gets you C.
 +
 
 +
-dragoon (LATEX help)
 +
 
 +
== Video Solution by Pi Academy ==
 +
https://youtu.be/N2eca474ljo?si=PLw0R0-KGp1zAnuQ
 +
 
 +
~ Pi Academy
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/GrCtzL0S-Uo?t=638
 +
 
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2016|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:53, 8 October 2024

Problem

In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?

$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$

Solution 1

We draw a diagram to make our work easier: [asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); X=(4/3,2sqrt(3)/3); W=(4/3,sqrt(3)/3); Z=(-1/3,sqrt(3)/3); Y=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW); [/asy]

Assume that $AB$ is of length $1$. Therefore, the area of $ABCDEF$ is $\frac{3\sqrt 3}2$. To find the area of $WCXYFZ$, we draw $\overline{CF}$, and find the area of the trapezoids $WCFZ$ and $CXYF$.

[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW); [/asy]

From this, we know that $CF=2$. We also know that the combined heights of the trapezoids is $\frac{\sqrt 3}3$, since $\overline{ZW}$ and $\overline{YX}$ are equally spaced, and the height of each of the trapezoids is $\frac{\sqrt 3}6$. From this, we know $\overline{ZW}$ and $\overline{YX}$ are each $\frac 13$ of the way from $\overline{CF}$ to $\overline{DE}$ and $\overline{AB}$, respectively. We know that these are both equal to $\frac 53$.

We find the area of each of the trapezoids, which both happen to be $\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}$, and the combined area is $\frac{11\sqrt 3}{18}$.

We find that $\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}$ is equal to $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$.

Solution 2

[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW); [/asy]

First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \cdot 6 = 54$ small triangles in the whole hexagon.

Thus, the answer is $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$.

Solution 3 (Similar Triangles)

[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW); label("$G$",G,N); [/asy] Extend $\overline{EF}$ and $\overline{CD}$ to meet at point $G$, as shown in the diagram. Then $\triangle GZW \sim \triangle GFC$. Then $[GZW] = \left(\dfrac53\right)^2[GED]$ and $[GFC] = 2^2[GED]$. Subtracting $[GED]$, we find that $[EDWZ] = \dfrac{16}{9}[GED]$ and $[EDCF] = 3[GED]$. Subtracting again, we find that \[[ZWCF] = [EDCF] - [EDWZ] = \dfrac{11}{9}[GED].\]Finally, \[\dfrac{[WCXYFZ]}{[ABCDEF]} = \dfrac{[ZWCF]}{[EDCF]} = \dfrac{\dfrac{11}{9}[GED]}{3[GED]} = \textbf{(C) } \dfrac{11}{27}.\]


Solution 4 (Extending Lines)

Refer to the diagram from Solution 1.

Let us start by connecting points $F$ to $C$ to create a new line segment $FC$. We drop a perpendicular line segment from $E$ to side $FC$ at point $P$. Since $\angle FED = 120$, and $\angle PED = 90$, we know that $\angle FEP = 30.$

Therefore, $\triangle EFP$ is a 30-60-90 triangle. Assume that $EP = \sqrt{3}$. Therefore, we know that $FP = 1$, $EF = 2$.

Let us draw $PQ$ such that $PQ$ is perpendicular to $ZW$. Since the distance between the parallel lines are equal, we know that $PQ$ is half of the distance between the parallel lines. Hence, $PQ$ will be one-third the length of $EP$.

From this, we also know that $EQ = EP - PQ = EP - \frac{1}{3} EP = \frac{2}{3} EP$.

Hence, we know that $EQ = \frac{2}{3} EP = \frac{2\sqrt{3}}{3}.$ Since $\angle EQZ = 90, FEP = 30$ we know that $\triangle ZEQ$ is also a 30-60-90 triangle. From this, we know that $ZQ = \frac{2}{3}$.

Let us drop another perpendicular line segment from $D$ to point $T$ such that $DT$ is perpendicular to $ZW$.

Since $ED = QT$ and since $ED$ is the side length of the regular hexagon, we know that $ED = EF = QT = 2.$

By symmetry, we also know that $WT = \frac{2}{3}$.

Therefore, we can find the length of $ZW = ZQ + QT + QW = \frac{2}{3} + 2 + \frac{2}{3}$. Hence, we know that $ZW = \frac{10}{3}.$

Now, we can find the area of trapezoid $EDZW = \frac{ED + ZW}{2} \cdot EQ = \frac{2 + \frac{10}{3}}{2} \cdot \frac{2\sqrt{3}}{3} = \frac{16\sqrt{3}}{9}.$

By symmetry, we know that the area of trapezoid $YZAB = \frac{16\sqrt{3}}{9}.$

Using the area of a hexagon formula, we get that the area of regular hexagon $ABCDEF = \frac{3\sqrt{3}}{2} \cdot 2^2 = 6\sqrt{3}.$

Hence, the area of hexagon $WCXYFZ = 6\sqrt{3} - 2\left(\frac{16\sqrt{3}}{9}\right) = \frac{22\sqrt{3}}{9}.$

Hence, the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$ is $\dfrac{\frac{22\sqrt{3}}{9}}{6\sqrt{3}} = \boxed{\frac{11}{27} \implies C}.$

~yk2007 :D

Solution 5

We will do this by area subtraction. Drawing in the required lines, we drop an altitude from A to WZ and an altitude from E to XY. You get 2 30-60-90 triangles, and given that the side length is $x$, AZ is $2x/3$, which means that WZ is $5x/3$, adding in AB. Using the 30-60-90 triangle again, you get that the area of trapezoid $ABWZ$ is $\frac{4x}{3} \cdot$ $\frac{x\sqrt3}{3}$. Repeat on the other side to get that the area of both of these trapezoids combined are $\frac{8x^2\sqrt3}{9}$. Finding the area of the hexagon, dividing, and subtracting, gets you C.

-dragoon (LATEX help)

Video Solution by Pi Academy

https://youtu.be/N2eca474ljo?si=PLw0R0-KGp1zAnuQ

~ Pi Academy

Video Solution by OmegaLearn

https://youtu.be/GrCtzL0S-Uo?t=638

~ pi_is_3.14

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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

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