Difference between revisions of "2004 AMC 8 Problems/Problem 14"
Isabelchen (talk | contribs) |
Isabelchen (talk | contribs) m (→Solution 1) |
||
Line 46: | Line 46: | ||
Divide the shape up as above. | Divide the shape up as above. | ||
− | <cmath>Area = [ | + | <cmath>Area = [DEGF] + [ABD] + [BCE] - [AFH] - [CGH] = 4 \cdot 10 + \frac12 \cdot 1 \cdot 3 + \frac12 \cdot 7 \cdot 6 - \frac12 \cdot 5 \cdot 4 - \frac12 \cdot 6 \cdot 10 = 40 + \frac32 + 21 - 10 - 30 = \boxed{\textbf{(C)}\ 22\frac12}</cmath> |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Latest revision as of 09:13, 19 January 2024
Problem
What is the area enclosed by the geoboard quadrilateral below?
Solution 1
Divide the shape up as above.
Solution 2
Let the bottom left corner be . The points would then be and . Applying the Shoelace Theorem,
Solution 3
The figure contains interior points and boundary points. Using Pick's Theorem, the area is
See Also
2004 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.