Difference between revisions of "1996 AJHSME Problems/Problem 22"
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==Solution 3== | ==Solution 3== | ||
− | We can find the total area of the grid, and subtract the triangles off of the area. The area of the grid is <math> | + | We can find the total area of the grid, and subtract the triangles off of the area. The area of the grid is <math>4\times 3=12</math> |
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==Solution 4== | ==Solution 4== |
Revision as of 15:30, 16 August 2020
Problem
The horizontal and vertical distances between adjacent points equal 1 unit. What is the area of triangle ?
Solution 1
takes up half of the 4x3 grid, so it has area of .
has height of and a base of , for an area of .
has height of and a base of , for an area of
Note that can be found by taking , and subtracting off and .
Thus, the area of , and the answer is .
There are other equivalent ways of dissecting the figure; right triangles and rectangle can also be used. You can also use and trapezoid .
Solution 2
Using the Shoelace Theorem, and labelling the points , we find the area is:
Area = , which is option .
Solution 3
We can find the total area of the grid, and subtract the triangles off of the area. The area of the grid is
Solution 4
Pick's theorem states , where B is the number of boundary points and I is the number of interior points. There are 3 boundary points and 0 interior points. Substituting the values in, we get that the answer is . ~ math_genius_11
See Also
1996 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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.