2009 AMC 10A Problem 14

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 the rectangle have shorter side

x

and longer side

y,

and express the two square sides

The outer square has side

y + x

and the inner square has side

y - x,

with ratio

\sqrt{4} = 2

Let each rectangle have shorter side

x

and longer side

y.

The outer square has side length

y + x

and the inner square has side length

y - x.

Since the area ratio is

4,

the side ratio is

2,

so

y + x = 2(y - x),

which gives

y = 3x.

The ratio of longer to shorter side is

\dfrac{y}{x} = 3.

Thus, the correct answer is A.

Problem 14 in other years