Difference between revisions of "2014 AMC 10A Problems/Problem 16"
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==Solution== | ==Solution== | ||
+ | ===Solution 1=== | ||
+ | Note that the region is a kite; hence its diagonals are perpendicular and it has area <math>\dfrac{ab}{2}</math> for diagonals of length <math>a</math> and <math>b</math>. Since <math>HF=1</math> as both <math>H</math> and <math>F</math> are midpoints of parallel sides of rectangle <math>GECD</math> and <math>CE=1</math>, we let <math>b=HF=1</math>. Now all we need to do is to find <math>a</math>. | ||
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+ | Let the other two vertices of the kite be <math>I</math> and <math>J</math> with <math>I</math> closer to <math>AD</math> than <math>J</math>. This gives us <math>a=IJ</math>. Now let <math>D=(0,0)</math>. We thus find that the equation of <math>\overleftrightarrow{AF}</math> is <math>4x+y=2</math> and that of <math>\overleftrightarrow{DH}</math> is <math>2x-y=0</math>. Solving this system gives us <math>x=\dfrac{1}{3}</math>, so the <math>x</math>-coordinate of <math>I</math> is <math>\dfrac{1}{3}</math>; in other words, <math>I</math> is <math>\dfrac{1}{3}</math> from <math>\overline{AD}</math>. By symmetry, <math>J</math> is also the same distance from <math>\overline{BC}</math>, so as <math>CD=1</math> we have <math>a=IJ=1-\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{1}{3}</math>. Hence the area of the kite is <math>\dfrac{ab}{2}=\dfrac{\frac{1}{3}\cdot1}{2}=\dfrac{1}{6}\implies\boxed{\textbf{(C)}\ \dfrac{1}{6}}</math>. | ||
==See Also== | ==See Also== |
Revision as of 07:05, 7 February 2014
Contents
[hide]Problem
In rectangle , , , and points , , and are midpoints of , , and , respectively. Point is the midpoint of . What is the area of the shaded region?
Solution
Solution 1
Note that the region is a kite; hence its diagonals are perpendicular and it has area for diagonals of length and . Since as both and are midpoints of parallel sides of rectangle and , we let . Now all we need to do is to find .
Let the other two vertices of the kite be and with closer to than . This gives us . Now let . We thus find that the equation of is and that of is . Solving this system gives us , so the -coordinate of is ; in other words, is from . By symmetry, is also the same distance from , so as we have . Hence the area of the kite is .
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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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