# 2012 AMC 10A Problems/Problem 15

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## Problem

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$?

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw((0,-2)--(1,-2)); draw(A--(0,-2)); draw(B--(1,-2)); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--(0,-2)); pair[] ps={A,B,C}; dot(ps); label("A",A,N); label("B",B,N); label("C",C,W); [/asy]$

$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$

## Solution

$[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); pair D=(1,-2), E=(0,-2); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D); draw(A--E); draw(B--D); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--E); pair[] ps={A,B,C,D,E}; dot(ps); label("A",A,N); label("B",B,N); label("C",C,W); label("D",D,S); label("E",E,S); label("1",(D--E),S); label("1",(A--B),N); label("2",(A--E),W); label("\sqrt{5}",(B--E),NW); [/asy]$

$AC$ intersects $BC$ at a right angle, so $\triangle ABC \sim \triangle BED$. The hypotenuse of right triangle BED is $\sqrt{1^2+2^2}=\sqrt{5}$.

$$\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC$$

$$\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}$$

Since AC=2BC, $BC=\frac{1}{\sqrt{5}}$. $\triangle ABC$ is a right triangle so the area is just $\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}$