Difference between revisions of "2012 AMC 10A Problems/Problem 15"
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<math> \textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}</math> | <math> \textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}</math> | ||
− | == Solution == | + | == Solution 1 == |
<center><asy> | <center><asy> | ||
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Since AC=2BC, <math>BC=\frac{1}{\sqrt{5}}</math>. <math>\triangle ABC</math> is a right triangle so the area is just <math>\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}</math> | Since AC=2BC, <math>BC=\frac{1}{\sqrt{5}}</math>. <math>\triangle ABC</math> is a right triangle so the area is just <math>\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}</math> | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | <center><asy> | ||
+ | unitsize(2cm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); | ||
+ | pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0); | ||
+ | draw(A--(2,0)); draw((0,-1)--F); draw(E--D); | ||
+ | draw(A--E); draw(B--D); draw((2,0)--F); | ||
+ | draw(A--F); draw(B--E); draw(C--G); | ||
+ | |||
+ | pair[] ps={A,B,C,D,E,F,G}; | ||
+ | dot(ps); | ||
+ | |||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,N); | ||
+ | label("$C$",C,W); | ||
+ | label("$D$",D,S); | ||
+ | label("$E$",E,S); | ||
+ | label("$F$",F,E); | ||
+ | label("$G$",G,N); | ||
+ | </asy></center> | ||
+ | |||
+ | Let <math>\text{E}</math> be the origin. Then, | ||
+ | <math>\text{D}=(1, 0)</math> | ||
+ | <math>\text{A}=(0, 2)</math> | ||
+ | <math>\text{B}=(1, 2)</math> | ||
+ | <math>\text{F}=(2, 1)</math> | ||
+ | |||
+ | <math>\bar{AB}</math> can be represented by the line <math>y=2x</math> | ||
+ | Also, <math>\bar{AF}</math> can be represented by the line <math>y=-\frac{1}{2}x+2</math> | ||
+ | |||
+ | Subtracting the second equation from the first gives us <math>\frac{5}{2}x-2=0</math>. | ||
+ | Thus, <math>x=\frac{4}{5}</math>. | ||
+ | Plugging this into the first equation gives us <math>y=\frac{8}{5}</math>. | ||
+ | |||
+ | Since <math>\text{C} (0.8, 1.6)</math>, <math>G</math> is <math>(0.8, 2)</math>. | ||
+ | |||
+ | <math>\bar{AB}=1</math> and <math>\bar{CG}=0.4</math>. | ||
+ | |||
+ | Thus, <math>[ABC]=\frac{1}{2} \cdot \bar{AB} \cdot \bar{CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}. The answer is then </math>(\text{B})$. | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2012|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2012|ab=A|num-b=14|num-a=16}} |
Revision as of 21:17, 17 February 2012
Contents
[hide]Problem
Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of ?
Solution 1
intersects at a right angle, so . The hypotenuse of right triangle BED is .
Since AC=2BC, . is a right triangle so the area is just
Solution 2
Let be the origin. Then,
can be represented by the line Also, can be represented by the line
Subtracting the second equation from the first gives us . Thus, . Plugging this into the first equation gives us .
Since , is .
and .
Thus, (\text{B})$.
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |