Difference between revisions of "2012 AMC 10A Problems/Problem 15"
(→Solution 2) |
SuperSnivy (talk | contribs) m (→Solution 2) |
||
Line 96: | Line 96: | ||
Plugging this into the first equation gives us <math>y=\frac{8}{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> | + | Since <math>\text{C} (0.8, 1.6)</math>, <math>G</math> is <math>(0.8, 2)</math>, |
<math>{AB}=1</math> and <math>{CG}=0.4</math>. | <math>{AB}=1</math> and <math>{CG}=0.4</math>. |
Revision as of 11:41, 4 January 2013
Contents
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,
$\widebar{EB}$ (Error compiling LaTeX. ! Undefined control sequence.) 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 |