1968 AHSME Problems/Problem 35
Problem
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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