# 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 1

$[asy] unitsize(5mm); defaultpen(linewidth(.8pt)); dotfactor=2; for(int a=0; a<=10; ++a) for(int b=0; b<=10; ++b) { dot((a,b)); }; draw((4,0)--(0,5)--(3,4)--(10,10)--cycle); draw((0,0)--(10,0)--(10,10)--(3,4)--(0,5)--cycle); draw((10,4)--(0,4)--cycle); dot("A", (0,5), W); dot("B", (3,4), N); dot("C", (10,10), NE); dot("D", (0,4), W); dot("E", (10,4), E); dot("F", (0,0), SW); dot("G", (10,0), SE); dot("H", (4,0), S); [/asy]$

Divide the shape up as above. $$Area = [DEGF] + [ABD] + [BCE] - [AFH] - [CGH] = 4 \cdot 10 + \frac12 \cdot 1 \cdot 3 + \frac12 \cdot 7 \cdot 6 - \frac12 \cdot 5 \cdot 4 - \frac12 \cdot 6 \cdot 10 = 40 + \frac32 + 21 - 10 - 30 = \boxed{\textbf{(C)}\ 22\frac12}$$

## Solution 2

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 3

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

 2004 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 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