1998 AMC 12 Problem 8

Question

A square with sides of length

1

is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find

x,

the length of the longer parallel side of each trapezoid.

\frac35

\frac23

\frac34

\frac56

\frac78

Answer: D

Concepts:trapezoid area equal areas

Difficulty rating: 1540

Small hint:

Each of the three regions has area

\frac{1}{3}

Big hint:

A trapezoid has height

\frac{1}{2}

and parallel sides

x

and

\frac{1}{2}

Solution:

Each trapezoid has area

\frac{1}{3}.

Its height is

\frac{1}{2},

and its parallel sides have lengths

x

and

\frac{1}{2}.

Therefore

\frac12\cdot\frac12\left(x+\frac12\right)=\frac13,

which gives

x=\frac{5}{6}.

Thus D is correct.