Difference between revisions of "2001 AMC 8 Problems/Problem 11"

Revision as of 10:32, 28 May 2011

Problem

Points $A$ , $B$ , $C$ and $D$ have these coordinates: $A(3,2)$ , $B(3,-2)$ , $C(-3,-2)$ and $D(-3, 0)$ . The area of quadrilateral $ABCD$ is

$[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } [/asy]$

$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24$

Solution

$[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } { draw((3,2)--(3,-2)--(-3,-2)--(-3,0)--cycle,linewidth(1)); } label("$A$",(3,2),NE); label("$B$",(3,-2),SE); label("$C$",(-3,-2),SW); label("$D$",(-3,0),NW); [/asy]$

This quadrilateral is a trapezoid, because $AB\parallel CD$ but $BC$ is not parallel to $AD$ . The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are $AB$ and $CD$ , which have lengths $2$ and $4$ , respectively, so the length of the median is $\frac{2+4}{2}=3$ . $CB$ is perpendicular to the bases, so it is the height, and has length $6$ . Therefore, the area of the trapezoid is $(3)(6)=18, \boxed{\text{C}}$

2001 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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

Page

Toolbox

Search

Difference between revisions of "2001 AMC 8 Problems/Problem 11"

Revision as of 10:32, 28 May 2011

Problem

Solution

See Also