# 1985 AHSME Problems/Problem 13

## Problem

Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is $[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]$ $\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 5.5 \qquad \mathrm{(E) \ }6$

## Solution

### Solution 1

We see that the number of interior points is $5$ and the number of boundary points is $4$. Therefore, by Pick's Theorem, the area is $5+\frac{4}{2}-1=6, \boxed{\text{E}}$.

### Solution 2 $[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7)); draw((0,0)--(4,0)--(4,3)--(0,3)--cycle); label("A",(0,3),NW); label("B",(4,3),NE); label("C",(4,0),SE); label("D",(0,0),SW); label("E",(1,3),N); label("F",(4,1),E); label("G",(3,0),S); label("H",(0,1),W); [/asy]$ Draw in the perimeter of the rectangle and label the points as shown. We have $[ABCD]=(3)(4)=12$, $[AEH]=\frac{1}{2}(1)(2)=1$, $[EBF]=\frac{1}{2}(3)(2)=3$, $[FCG]=\frac{1}{2}(1)(1)=\frac{1}{2}$, and $[GDH]=\frac{1}{2}(3)(1)=\frac{3}{2}$. Therefore, $[EFGH]=[ABCD]-([AEH]+[EBF]+[FCG]+[GDH])$ $=12-(1+3+\frac{1}{2}+\frac{3}{2})=6, \boxed{\text{E}}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 