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Difference between revisions of "2012 AMC 10A Problems/Problem 21"

(Created page with "==Problem 21== Let points <math>A</math> = <math>(0 ,0 ,0)</math>, <math>B</math> = <math>(1, 0, 0)</math>, <math>C</math> = <math>(0, 2, 0)</math>, and <math>D</math> = <math>(...")
 
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==Problem 21==
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==Problem==
  
 
Let points <math>A</math> = <math>(0 ,0 ,0)</math>, <math>B</math> = <math>(1, 0, 0)</math>, <math>C</math> = <math>(0, 2, 0)</math>, and <math>D</math> = <math>(0, 0, 3)</math>. Points <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math> are midpoints of line segments <math>\overline{BD},\text{ }  \overline{AB}, \text{ } \overline {AC},</math> and <math>\overline{DC}</math> respectively. What is the area of <math>EFGH</math>?
 
Let points <math>A</math> = <math>(0 ,0 ,0)</math>, <math>B</math> = <math>(1, 0, 0)</math>, <math>C</math> = <math>(0, 2, 0)</math>, and <math>D</math> = <math>(0, 0, 3)</math>. Points <math>E</math>, <math>F</math>, <math>G</math>, and <math>H</math> are midpoints of line segments <math>\overline{BD},\text{ }  \overline{AB}, \text{ } \overline {AC},</math> and <math>\overline{DC}</math> respectively. What is the area of <math>EFGH</math>?
  
 
<math> \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} </math>
 
<math> \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} </math>
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==Solution ==
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Consider a tetrahedron with vertices at <math>A,B,C,D</math> on the <math>xyz</math>-plane. The length of <math>EF</math> is just one-half of <math>AD</math> because it is the midsegment of <math>\triangle ABD.</math> The same concept applies to the other side lengths. <math>AD=3</math> and <math>BC=\sqrt{1^2+2^2}=\sqrt{5}</math>. Then <math>EF=HG=\frac32</math> and <math>EH=FG=\frac{\sqrt{5}}{2}</math>. The line segments lie on perpendicular planes so quadrilateral <math>EFGH</math> is a rectangle. The area is
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<cmath>EF \cdot FG = \frac32 \cdot \frac{\sqrt{5}}{2} = \boxed{\textbf{(C)}\ \frac{3\sqrt{5}}{4}}</cmath>
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== See Also ==
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{{AMC10 box|year=2012|ab=A|num-b=20|num-a=22}}

Revision as of 01:43, 9 February 2012

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

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

See Also

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