2012 AMC 10A Problem 21

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}

Use midsegments in triangles

ABD

and

ABC

The quadrilateral is a rectangle with side lengths from

AD

and

BC

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.

Problem 21 in other years