2012 AMC 10A Problems/Problem 15
Problem
Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of 
?
![[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw((0,-2)--(1,-2)); draw(A--(0,-2)); draw(B--(1,-2)); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--(0,-2)); pair[] ps={A,B,C}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); [/asy]](http://latex.artofproblemsolving.com/3/4/4/344ad68ce1c966777c56c860c4fc9b0da8bb06cb.png)
Solution
![[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); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--E); pair[] ps={A,B,C,D,E}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); label("$D$",D,S); label("$E$",E,S); label("$1$",(D--E),S); label("$1$",(A--B),N); label("$2$",(A--E),W); label("$\sqrt{5}$",(B--E),NW); [/asy]](http://latex.artofproblemsolving.com/9/6/a/96a4225d39bd0f0c94ba3c7de1676da3c06c84d4.png)
intersects 
at a right angle, so 
. The hypotenuse of right triangle BED is 
.
![\[\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC\]](http://latex.artofproblemsolving.com/0/1/5/0150847710c03762e544bcc9613888d3bb87c8ff.png)
![\[\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}\]](http://latex.artofproblemsolving.com/2/c/b/2cbb6282231d3ac6d83e3d38ecb4d13e4a76955b.png)
Since AC=2BC, 
. is a right triangle so the area is just 
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 | ||
