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:

$12\dfrac12$

$25$

$50$

$75$

$87\dfrac12$

Answer: D

Concepts:equilateral triangle square (geometry)area percentage

Difficulty rating: 1800

Small Hint:

Convert the three given areas into the side lengths $AB,$ $BC,$ $CD$

Big Hint:

The decrease in $AD$ is absorbed entirely by the square’s side

Solution:

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.

1968 AMC 12 Problem 31

Answer: D

Difficulty rating: 1800

Small Hint:

Big Hint:

Solution:

Problem 31 in Other Years