Difference between revisions of "1968 AHSME Problems/Problem 31"
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== Solution == | == Solution == | ||
− | The side of the square | + | 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>. |
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== See also == | == See also == |
Latest revision as of 08:12, 18 July 2024
Problem
In this diagram, not drawn to scale, Figures and
are equilateral triangular regions with respective areas of
and
square inches. Figure
is a square region with area
square inches. Let the length of segment
be decreased by
% of itself, while the lengths of
and
remain unchanged. The percent decrease in the area of the square is:
Solution
Given an equilateral triangle with side length , the area is given by
. Setting this equation equal to the area of triangle
,
, we find that
. Because triangle
is also equilateral, it is similar to trinagle
, and, because it has a quarter of the area of
, it has
of the side length. Thus, its sides have a length of
. Square
initially has an area of
, so it's side length starts at
. The initial length
, therefore, is
. Because
decreases by
%, it becomes
of its inital value, which is
. Because the sides of the triangles remain unchanged, this decrease of
must come from the side length of the square. Thus, the square's final side is
, which gives an area of
square inches.
to
is a decrease of
%. Therefore, our answer is
.
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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