# 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 1

$[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7)); draw((0,0)--(4,0)--(4,3)--(0,3)--cycle); label("A",(0,3),NW); label("B",(4,3),NE); label("C",(4,0),SE); label("D",(0,0),SW); label("E",(1,3),N); label("F",(4,1),E); label("G",(3,0),S); label("H",(0,1),W); [/asy]$ We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle $ABCD$ is $(3)(4) = 12$, while the areas of the triangles $AEH$, $EBF$, $FCG$, and $GDH$ are, respectively, \begin{align*}&\text{area of } \triangle AEH = \frac{1}{2}(1)(2) = 1, \\ &\text{area of } \triangle EBF = \frac{1}{2}(3)(2) = 3, \\ &\text{area of } \triangle FCG = \frac{1}{2}(1)(1) = \frac{1}{2}, \text{ and} \\ &\text{area of } \triangle GDH = \frac{1}{2}(3)(1) = \frac{3}{2}.\end{align*} Hence the area of the given quadrilateral $EFGH$ is $12-\left(1+3+\frac{1}{2}+\frac{3}{2}\right) = \boxed{\text{(E)} \ 6}$.

## Solution 2

The number of lattice points (i.e. pegs on the board) strictly inside the quadrilateral is $5$ and the number of lattice points on its boundary is $4$. Therefore, by Pick's theorem, its area is $5+\frac{4}{2}-1 = \boxed{\text{(E)} \ 6}$.

## See Also

 1985 AHSME (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 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 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

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