1998 AJHSME Problems/Problem 6

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 take the area of each square.

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 $\boxed{B}$

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1998 AJHSME Problems/Problem 6

Contents

Problem 6

Solution 1

Solution 2

See also