1968 AHSME Problems/Problem 31
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 | |
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