1992 AHSME Problems/Problem 24
Contents
Problem
Let be a parallelogram of area
with
and
. Locate
and
on segments
and
, respectively, with
. Let the line through
parallel to
intersect
at
. The area of quadrilateral
is
Solution 1
Use vectors. Place an origin at
, with
. We know that
, and also
, and now we can find the area of
by dividing it into two triangles and using cross-products (the expressions simplify using the fact that the cross-product distributes over addition, it is anticommutative, and a vector crossed with itself gives zero).
Solution 2
We note that is a parallelogram because
and
. Using the same reasoning,
is also a parallelogram.
Assume that the height of parallelogram with respect to base
is
. Then, the area of parallelogram
is
. The area of triangle
is
, which is half of the area of parallelogram
.
Likewise, the area of triangle is half the area of parallelogram
.
Thus,
The answer is \boxed{C. 5}
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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