In the figure, ABCD is a 2 by 2 square, E is the midpoint of AD, and F is on BE. If CF is perpendicular to BE, then the area of quadrilateral CDEF is

1997 AMC 12 Problem 9

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In the figure,

ABCD

is a

2

2

square,

E

is the midpoint of

\overline{AD},

and

F

is on

\overline{BE}.

\overline{CF}

is perpendicular to

\overline{BE},

then the area of quadrilateral

CDEF

2

3-\frac{\sqrt3}{2}

\frac{11}{5}

\sqrt5

\frac94

Place

B=(0,0),

C=(2,0),

A=(0,2),

and

E=(1,2)

Find

F

as the intersection of

BE

with the line through

C

perpendicular to

BE

With the coordinates in the first hint,

BE

has equation

y=2x,

and the perpendicular through

C

y=-\frac{(x-2)}{2}.

Their intersection is

F=(\frac{2}{5},\frac{4}{5}).

The square has area

4,

while triangles

AEF

and

BCF

have areas

\frac{3}{5}

and

\frac{4}{5},

respectively. Hence

[CDEF]=4-\frac{3}{5}

{}-\frac{4}{5}=\frac{11}{5},

so C is correct.