Difference between revisions of "1985 AHSME Problems/Problem 13"

Revision as of 12:00, 5 July 2013

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, $[EFGH]=[ABCD]-([AEH]+[EBF]+[FCG]+[GDH])$ $=12-(1+3+\frac{1}{2}+\frac{3}{2})=6, \boxed{\text{E}}$ .

1985 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 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

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

@@ Line 38: / Line 38: @@
 ==See Also==
 {{AHSME box|year=1985|num-b=12|num-a=14}}
+{{MAA Notice}}

Page

Toolbox

Search