2001 AMC 8 Problems/Problem 11

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]

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

Solutions

Solution 1

[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 $(3)(6)=18, \boxed{\text{C}}$

Solution 2

Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle. Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle.

$A_{trap} = A_{tri} + A_{rect}$

$A_{trap} = \frac{1}{2}bh + lw$

$A_{trap} = \frac{1}{2}\cdot 6 \cdot 2 + 6\cdot 2$

$A_{trap}  = 6 + 12 = 18 \rightarrow \boxed{C}$

See Also

2001 AMC 8 (ProblemsAnswer KeyResources)
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

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