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,
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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