Difference between revisions of "1992 AHSME Problems/Problem 24"
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== Solution 2 == | == Solution 2 == | ||
− | By noting the following: <cmath> | + | By noting the following: <cmath>\begin{align*}|EFHG| &= |\triangle EGH| + |\triangle EFH| \ &= \frac{1}{2}|AEHD|+\frac{1}{2}|EHCB| \ &= \frac{1}{2}[|AEHD|+|EHCB|] \ &= \frac{1}{2}|ABCD| \ &= 5 \end{align*}</cmath> we see that the answer is <math>\fbox{C}</math>. |
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== See also == | == See also == | ||
{{AHSME box|year=1992|num-b=23|num-a=25}} | {{AHSME box|year=1992|num-b=23|num-a=25}} |
Revision as of 08:08, 28 June 2018
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
By noting the following: we see that the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.