2010 AMC 12A Problem 3

Question

Rectangle

ABCD,

pictured below, shares

50\%

of its area with square

EFGH.

Square

EFGH

shares

20\%

of its area with rectangle

ABCD.

What is

\dfrac{AB}{AD}?

4

5

6

8

10

Let

s

be the side length of the square; the shaded overlap has width

s

and height

AD

Half the rectangle’s area equals the overlap, giving

AB=2s;

one fifth of the square’s area gives

AD=\tfrac{s}{5}

Let

s

be the side length of square

EFGH.

The shaded overlap has width

s

and height

AD,

so its area is

s\cdot AD.

Because the overlap is

50\%

of the rectangle,

s\cdot AD=\tfrac12\,AB\cdot AD,

so

AB=2s.

Because it is

20\%

of the square,

s\cdot AD=\tfrac15 s^2,

so

AD=\tfrac{s}{5}.

Therefore

\dfrac{AB}{AD}=\dfrac{2s}{\frac{s}{5}}=10.

Thus, E is the correct answer.

Problem 3 in other years