Difference between revisions of "1998 AJHSME Problems/Problem 6"
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==Solution 2== | ==Solution 2== | ||
| − | We could | + | We could count the area contributed by each square on the <math>3 \times 3</math> grid: |
Top-left: <math>0</math> | Top-left: <math>0</math> | ||
| + | |||
Top: Triangle with area <math>\frac{1}{2}</math> | Top: Triangle with area <math>\frac{1}{2}</math> | ||
| + | |||
Top-right: <math>0</math> | Top-right: <math>0</math> | ||
| + | |||
Left: Square with area <math>1</math> | Left: Square with area <math>1</math> | ||
| + | |||
Center: Square with area <math>1</math> | Center: Square with area <math>1</math> | ||
| + | |||
Right: Square with area <math>1</math> | Right: Square with area <math>1</math> | ||
| + | |||
Bottom-left: Square with area <math>1</math> | Bottom-left: Square with area <math>1</math> | ||
| + | |||
Bottom: Triangle with area <math>\frac{1}{2}</math> | Bottom: Triangle with area <math>\frac{1}{2}</math> | ||
| + | |||
Bottom-right: Square with area <math>1</math> | Bottom-right: Square with area <math>1</math> | ||
Adding all of these together, we get <math>\boxed{6}</math> or <math>\boxed{B}</math> | Adding all of these together, we get <math>\boxed{6}</math> or <math>\boxed{B}</math> | ||
| − | |||
== See also == | == See also == | ||
Revision as of 11:08, 31 July 2011
Contents
Problem 6
Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is
![[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]](http://latex.artofproblemsolving.com/e/5/1/e512d80ea585bdc760432f8f8d7ee955400e1d71.png)
Solution 1
By inspection, you can notice that the triangle on the top row matches the hole in the bottom row.
This creates a 
box, which has area 
Solution 2
We could count the area contributed by each square on the 
grid:
Top-left: 
Top: Triangle with area 
Top-right:
Left: Square with area 
Center: Square with area
Right: Square with area
Bottom-left: Square with area
Bottom: Triangle with area
Bottom-right: Square with area
Adding all of these together, we get 
or 
See also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
