# 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

Let the bottom left corner be $(0,0)$. The points would then 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

The figure contains $21$ interior points and $5$ boundary points. Using Pick's Theorem, the area is $$21+\frac{5}{2}-1=\boxed{\textbf{(C)}\ 22\frac12}$$

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