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Difference between revisions of "1968 AHSME Problems/Problem 31"

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

Latest revision as of 07:12, 18 July 2024

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 $\boxed{\textbf{(D)}}$.

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 30 		Followed by
Problem 32
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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