2004 AMC 8 Problems/Problem 14

Problem

What is the area enclosed by the geoboard quadrilateral below?

[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=2;  for(int a=0; a<=10; ++a) for(int b=0; b<=10; ++b)  {   dot((a,b));  };  draw((4,0)--(0,5)--(3,4)--(10,10)--cycle); [/asy]

$\textbf{(A)}\ 15\qquad \textbf{(B)}\ 18\frac12 \qquad \textbf{(C)}\ 22\frac12 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 41$

Solution

Assign points to each of the four vertices and use the shoelace theorem to find the area. Letting the bottom left corner be $(0,0)$, counting the boxes, the points would be $(4,0),(0,5),(3,4),$ and $(10,10)$. Applying the Shoelace Theorem,

\[\text{Area} = \frac12 \begin{vmatrix} 4 & 0 \\ 0 & 5 \\ 3 & 4 \\ 10 & 10 \end{vmatrix} = \frac12 |(20+30)-(15+40+40)| = \frac12 |50-95| = \boxed{\textbf{(C)}\ 22\frac12}\]

Solution 2

Apply Pick's Theorem on the figure, and you will get \[\boxed{\textbf{(C)}\ 22\frac12}\]

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AJHSME/AMC 8 Problems and Solutions

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