Difference between revisions of "1992 AHSME Problems/Problem 24"
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− | Assume that the height of parallelogram <math>ABFG</math> | + | Assume that the height of parallelogram <math>ABFG</math> with respect to base <math>AB</math> is <math>x</math>. Then, the area of parallelogram <math>ABFG</math> is <math>AB * x</math>. The area of triangle <math>EFG</math> is <math>1/2 * AB * x</math>, which is half of the area of parallelogram <math>ABFG</math>. |
Revision as of 18:11, 4 September 2022
Contents
[hide]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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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