Question

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 $EFGH?$

$\sqrt{2}$

$\dfrac{2\sqrt{5}}{3}$

$\dfrac{3\sqrt{5}}{4}$

$\sqrt{3}$

$\dfrac{2\sqrt{7}}{3}$

Answer: C

Concepts:3D geometry midpoint distance formula

Difficulty rating: 1730

Solution:

Note that $EF = \dfrac{1}{2}AD$ since it is a midsegment of $\triangle ABD.$ Similarly, $HG = \dfrac{1}{2}AD$ and $FG = \dfrac{1}{2}BC.$

We also have that $\overline{EF}$ and $\overline{HG}$ are perpendicular to the $xy$ -plane, which means that they are perpendicular to $\overline{FG}$ and $\overline{EH}.$

This tells us that $EFGH$ is rectangle since $EF = HG.$ We have $EF = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}.$

We also have that $FG = \dfrac{1}{2} \sqrt{1^2 + 2^2} = \dfrac{\sqrt{5}}{2}.$

The area of $EFGH$ is then $EF \cdot FG = \dfrac{3}{2} \cdot \dfrac{\sqrt{5}}{2} = \dfrac{3\sqrt{5}}{4}.$

Thus, C is the correct answer.

2012 AMC 10A Problem 21

Answer: C

Difficulty rating: 1730

Solution:

Problem 21 in Other Years