2009 AMC 12A Problem 8

Question

Four congruent rectangles are placed as shown. The area of the outer square is

4

times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?

3

\sqrt{10}

2 + \sqrt{2}

2\sqrt{3}

4

Let each rectangle have shorter side

x

and longer side

y

The outer square has side

x + y,

the inner square has side

y - x,

and their side ratio is

\sqrt{4} = 2

Let the rectangles have shorter side

x

and longer side

y.

The outer square has side

x + y

and the inner square has side

y - x.

Since the outer area is

4

times the inner area, the side ratio is

\sqrt{4} = 2,

so

x + y = 2(y - x).

This gives

y = 3x,

so the ratio of longer to shorter side is

3.

Thus, the correct answer is A.

Problem 8 in other years