1968 AHSME Problems/Problem 35
Problem
draw(circle((0,0),10, 0, 180),black+linewidth(.75)); draw((-10,0)--(10,0),black+linewidth(.75)); draw((-sqrt(96),2)--(sqrt(96),2),black+linewidth(.75)); draw((-8,6)--(8,6),black+linewidth(.75)); draw((0,0)--(0,10),black+linewidth(.75)); draw((-8,6)--(-8,2),black+linewidth(.75)); draw((8,6)--(8,2),black+linewidth(.75)); dot((0,0)); MP("O",(0,0),S); MP("a",(5,0),S); MP("J",(0,10),N); MP("D",(sqrt(96),2),E); MP("C",(-sqrt(96),2),W); MP("F",(8,6),E); MP("E",(-8,6),W); MP("G",(0,2),NE); MP("H",(0,6),NE); MP("L",(-8,2),S); MP("M",(8,2),S); (Error making remote request. Unknown error_msg)
In this diagram the center of the circle is , the radius is
inches, chord
is parallel to chord
.
,
,
,
are collinear, and
is the midpoint of
. Let
(sq. in.) represent the area of trapezoid
and let
(sq. in.) represent the area of rectangle
Then, as
and
are translated upward so that
increases toward the value
, while
always equals
, the ratio
becomes arbitrarily close to:
Solution
Let , where
. Since the areas of rectangle
and trapezoid
are both half of rectangle
and trapezoid
, respectively, the ratios between their areas will remain the same, so let us consider rectangle
and trapezoid
. Draw radii
and
, both of which obviously have length
. By the Pythagorean theorem, the length of
is
, and the length of
is
. It follows that the area of rectangle
is
while the area of trapezoid
is
. Now, we want to find the limit, as
approaches
, of
. Note that this is equivalent to finding the same limit as
approaches
. Substituting
into
yields that trapezoid
has area
and that rectangle
has area
. Our answer thus becomes
See also
1968 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 34 |
Followed by Problem 35 | |
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