1968 AHSME Problems/Problem 31

Problem

$[asy] draw((0,0)--(10,20*sqrt(3)/2)--(20,0)--cycle,black+linewidth(.75)); draw((20,0)--(20,12)--(32,12)--(32,0)--cycle,black+linewidth(.75)); draw((32,0)--(37,10*sqrt(3)/2)--(42,0)--cycle,black+linewidth(.75)); MP("I",(10,0),N);MP("II",(26,0),N);MP("III",(37,0),N); MP("A",(0,0),S);MP("B",(20,0),S);MP("C",(32,0),S);MP("D",(42,0),S); [/asy]$

In this diagram, not drawn to scale, Figures $I$ and $III$ are equilateral triangular regions with respective areas of $32\sqrt{3}$ and $8\sqrt{3}$ square inches. Figure $II$ is a square region with area $32$ square inches. Let the length of segment $AD$ be decreased by $12\tfrac{1}{2}$ % of itself, while the lengths of $AB$ and $CD$ remain unchanged. The percent decrease in the area of the square is:

$\text{(A)}\ 12\tfrac{1}{2}\qquad\text{(B)}\ 25\qquad\text{(C)}\ 50\qquad\text{(D)}\ 75\qquad\text{(E)}\ 87\tfrac{1}{2}$

Solution

Given an equilateral triangle with side length $s$, the area is given by $\frac{s^2\sqrt{3}}{4}$. Setting this equation equal to the area of triangle $I$, $32\sqrt{3}$, we find that $s=8\sqrt{2}$. Because triangle $III$ is also equilateral, it is similar to trinagle $I$, and, because it has a quarter of the area of $I$, it has $\sqrt{\frac{1}{4}}=\frac{1}{2}$ of the side length. Thus, its sides have a length of $4\sqrt{2}$. Square $II$ initially has an area of $32$, so it's side length starts at $\sqrt{32}=4\sqrt{2}$. The initial length $AD$, therefore, is $16\sqrt{2}$. Because $AD$ decreases by $12.5$ %, it becomes $\frac{7}{8}$ of its inital value, which is $14\sqrt{2}$. Because the sides of the triangles remain unchanged, this decrease of $2\sqrt{2}$ must come from the side length of the square. Thus, the square's final side is $2\sqrt{2}$, which gives an area of $8$ square inches. $32$ to $8$ is a decrease of $75$ %. Therefore, our answer is $\boxed{\textbf{(D)}}$.