1968 AMC 12 Problem 31

Question

In this diagram, not drawn to scale, figures

\mathrm{I}

and

\mathrm{III}

are equilateral triangular regions with respective areas of

32\sqrt3

and

8\sqrt3

square inches. Figure

\mathrm{II}

is a square region with area

32

square inches. Let the length of segment

AD

be decreased by

12\dfrac12\%

of itself, while the lengths of

AB

and

CD

remain unchanged. The percent decrease in the area of the square is:

12\dfrac12

25

50

75

87\dfrac12

Convert the three given areas into the side lengths

AB,

BC,

CD

The decrease in

AD

is absorbed entirely by the square’s side

From

(\frac{\sqrt3}{4})s^2=32\sqrt3,

AB=8\sqrt2.

The square has

BC=4\sqrt2,

and the smaller equilateral triangle also has

CD=4\sqrt2.

Thus

AD=16\sqrt2.

Decreasing it by

\frac{1}{8}

removes

2\sqrt2

from

BC,

whose new length is

2\sqrt2.

The square’s area falls from

32

to

8,

a

75\%

decrease.

Therefore, the correct answer is D.

Problem 31 in other years