2006 AMC 12B Problem 13

Question

Rhombus

ABCD

is similar to rhombus

BFDE.

The area of rhombus

ABCD

is

24,

and

\angle BAD = 60^\circ.

What is the area of rhombus

BFDE?

6

4\sqrt{3}

8

9

6\sqrt{3}

Since

\angle BAD = 60^\circ,

triangle

ABD

is equilateral, so

BD

equals a side

If the diagonals meet at

O,

then

\triangle ABO

is a

30^\circ

–

60^\circ

–

90^\circ

triangle; compare the long and short diagonals

Let the diagonals of

ABCD

meet at

O.

They bisect each other at right angles, and since

\angle BAD=60^\circ,

triangle

ABO

is a

30^\circ

–

60^\circ

–

90^\circ

triangle. Hence the half-diagonals satisfy

AO=\sqrt3\,BO.

The segment

BD

is the short diagonal of

ABCD

and the long diagonal of the similar rhombus

BFDE.

Thus the smaller-to-larger length ratio is

\frac{BD}{AC}=\frac{BO}{AO}=\frac{1}{\sqrt3}.

Areas scale by the square of this ratio, so the area of

BFDE

is

24\cdot\dfrac13=8.

Thus, the correct answer is C.

Problem 13 in other years