Difference between revisions of "2007 AMC 8 Problems/Problem 8"
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==Video Solution by SpreadTheMathLove== | ==Video Solution by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=omFpSGMWhFc | https://www.youtube.com/watch?v=omFpSGMWhFc | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/Qdbpdc-Khg4 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=7|num-a=9}} | {{AMC8 box|year=2007|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:37, 28 October 2024
Contents
Problem
In trapezoid , is perpendicular to , , and . In addition, is on , and is parallel to . Find the area of .
Solution 1 (Area Formula for Triangles)
Clearly, is a square with side-length By segment subtraction, we have
The area of is ~Aplus95 (Solution)
~MRENTHUSIASM (Revision)
Solution 2 (Area Subtraction)
Clearly, is a square with side-length
Let the brackets denote areas. We apply area subtraction to find the area of ~MRENTHUSIASM
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
Video Solution by WhyMath
See Also
2007 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.