Difference between revisions of "2012 AMC 10A Problems/Problem 15"
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<math>{AB}=1</math> and <math>{CG}=0.4</math>. | <math>{AB}=1</math> and <math>{CG}=0.4</math>. | ||
− | Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}. The answer is | + | Thus, <math>[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}</math>. The answer is <math>(\text{B})</math>. |
== 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 21:21, 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,
$\widebar{AB}$ (Error compiling LaTeX. Unknown error_msg) 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, . The answer is .
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 |