2012 AMC 10A Problems/Problem 21

Problem

Let points $A$ = $(0 ,0 ,0)$ , $B$ = $(1, 0, 0)$ , $C$ = $(0, 2, 0)$ , and $D$ = $(0, 0, 3)$ . Points $E$ , $F$ , $G$ , and $H$ are midpoints of line segments $\overline{BD},\text{ } \overline{AB}, \text{ } \overline {AC},$ and $\overline{DC}$ respectively. What is the area of $EFGH$ ?

$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{3}\qquad\textbf{(C)}\ \frac{3\sqrt{5}}{4}\qquad\textbf{(D)}\ \sqrt{3}\qquad\textbf{(E)}\ \frac{2\sqrt{7}}{3}$

Solution

Consider a tetrahedron with vertices at $A,B,C,D$ on the $xyz$ -plane. The length of $EF$ is just one-half of $AD$ because it is the midsegment of $\triangle ABD.$ The same concept applies to the other side lengths. $AD=3$ and $BC=\sqrt{1^2+2^2}=\sqrt{5}$ . Then $EF=HG=\frac32$ and $EH=FG=\frac{\sqrt{5}}{2}$ . The line segments lie on perpendicular planes so quadrilateral $EFGH$ is a rectangle. The area is

$EF \cdot FG = \frac32 \cdot \frac{\sqrt{5}}{2} = \boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}$

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2012 AMC 10A Problems/Problem 21

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