Art of Problem Solving

1998 AJHSME Problems/Problem 6: Difference between revisions

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==Solution 2==
==Solution 2==


We could take the area of each square.
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 10:08, 31 July 2011

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]

$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution 1

By inspection, you can notice that the triangle on the top row matches the hole in the bottom row.

This creates a $2\times3$ box, which has area $2\times3=\boxed{6}$

Solution 2

We could count the area contributed by each square on the $3 \times 3$ grid:

Top-left: $0$

Top: Triangle with area $\frac{1}{2}$

Top-right: $0$

Left: Square with area $1$

Center: Square with area $1$

Right: Square with area $1$

Bottom-left: Square with area $1$

Bottom: Triangle with area $\frac{1}{2}$

Bottom-right: Square with area $1$

Adding all of these together, we get $\boxed{6}$ or $\boxed{B}$

See also

1998 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AJHSME/AMC 8 Problems and Solutions