1996 AJHSME Problems/Problem 22
Problem
The horizontal and vertical distances between adjacent points equal 1 unit. The area of triangle is
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
See Also
1996 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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