Difference between revisions of "2004 AMC 8 Problems/Problem 14"

Revision as of 09:12, 19 January 2024

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

~isabelchen

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 by Problem 13	Followed by Problem 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

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

@@ Line 19: / Line 19: @@
 ==Solution 1==
-Let the bottom left corner be <math>(0,0)</math>. The points would then be <math>(4,0),(0,5),(3,4),</math> and <math>(10,10)</math>. Applying the [[Shoelace Theorem]],
-<cmath>\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}</cmath>
-==Solution 2==
-The figure contains <math>21</math> interior points and <math>5</math> boundary points. Using [[Pick's Theorem]], the area is <cmath>21+\frac{5}{2}-1=\boxed{\textbf{(C)}\ 22\frac12}</cmath>
-==Solution 3==
 <asy>
@@ Line 58: / Line 49: @@
 ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
+==Solution 2==
+Let the bottom left corner be <math>(0,0)</math>. The points would then be <math>(4,0),(0,5),(3,4),</math> and <math>(10,10)</math>. Applying the [[Shoelace Theorem]],
+<cmath>\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}</cmath>
+==Solution 3==
+The figure contains <math>21</math> interior points and <math>5</math> boundary points. Using [[Pick's Theorem]], the area is <cmath>21+\frac{5}{2}-1=\boxed{\textbf{(C)}\ 22\frac12}</cmath>
 ==See Also==
 {{AMC8 box|year=2004|num-b=13|num-a=15}}
 {{MAA Notice}}

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