2018 AMC 8 Problem 22

Question

Point

E

is the midpoint of side

\overline{CD}

in square

ABCD,

and

\overline{BE}

meets diagonal

\overline{AC}

at

F.

The area of quadrilateral

AFED

is

45.

What is the area of

ABCD?

100

108

120

135

144

Answer: B

Concepts:similarity square (geometry)area ratio

Difficulty rating: 1770

Small hint:

Let the square have side length

s

.

Big hint:

Find the small triangle cut from

\triangle ACD

, then subtract it from half the square.

Video solution:

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Written solution:

Let the square have side length

s,

and let

H

be the foot of the perpendicular from

F

to

\overline{BC}.

The right triangles

CAB

and

CFH

are similar, so

CH=FH.

Also, triangles

BFH

and

BEC

are similar, giving

\frac{FH}{EC}=\frac{BH}{BC}.

Since

EC=\frac{s}{2},

BC=s,

and

BH=s-CH=s-FH,

this becomes

\frac{2FH}{s}=1-\frac{FH}{s},

so

FH=CH=\frac{s}{3}.

Thus triangle

EFC

has base

EC=\frac{s}{2}

and height

CH=\frac{s}{3},

so its area is

\frac{s^2}{12}.

Therefore

\begin{aligned} [AFED]&=[\triangle ACD]-[\triangle EFC]\\ &=\dfrac{s^2}{2}-\dfrac{s^2}{12}\\ &=\dfrac{5s^2}{12}. \end{aligned}

Since this area is

45

, we get

s^2=108

, the area of the square.