Difference between revisions of "2016 AMC 10B Problems/Problem 23"
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draw(X--Y); | draw(X--Y); | ||
− | label(" | + | label("$A$",A,SW); |
− | label(" | + | label("$B$",B,SE); |
− | label(" | + | label("$C$",C,ESE); |
− | label(" | + | label("$D$",D,NE); |
− | label(" | + | label("$E$",E,NW); |
− | label(" | + | label("$F$",F,WSW); |
− | label(" | + | label("$W$",W,ENE); |
− | label(" | + | label("$X$",X,ESE); |
− | label(" | + | label("$Y$",Y,WSW); |
− | label(" | + | label("$Z$",Z,WNW); |
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draw(F--C--B--E--D--A); | draw(F--C--B--E--D--A); | ||
− | label(" | + | label("$A$",A,SW); |
− | label(" | + | label("$B$",B,SE); |
− | label(" | + | label("$C$",C,ESE); |
− | label(" | + | label("$D$",D,NE); |
− | label(" | + | label("$E$",E,NW); |
− | label(" | + | label("$F$",F,WSW); |
− | label(" | + | label("$W$",W,ENE); |
− | label(" | + | label("$X$",X,ESE); |
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− | label(" | + | label("$Z$",Z,WNW); |
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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>. |
Revision as of 20:22, 25 February 2016
Contents
Problem
In regular hexagon , points , , , and are chosen on sides , , , and respectively, so lines , , , and are parallel and equally spaced. What is the ratio of the area of hexagon to the area of hexagon ?
Solution 1
We draw a diagram to make our work easier:
Assume that is of length . Therefore, the area of is . To find the area of , we draw , and find the area of the trapezoids and .
From this, we know that . We also know that the combined heights of the trapezoids is , since and are equally spaced, and the height of each of the trapezoids is . From this, we know and are each of the way from to and , respectively. We know that these are both equal to .
We find the area of each of the trapezoids, which both happen to be , and the combined area is .
We find that is equal to .
At this point, you can answer and move on with your test.
Solution 2
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 u have draw these lines, it's just a matter of counting triangles. There are small triangles in hexagon , and small triangles in the whole hexagon.
There are $22<math> small triangles in hexagon <math>ZWCXYF<math>, and <math>9 \text{ small triangles} \cdot 6 \text{ triangles}= 54<math> small triangles in the whole hexagon <math>ABCDEF<math>.
Thus, the answer is <math>\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$ (Error compiling LaTeX. ! Missing $ inserted.).
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 |
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