1991 AJHSME Problems/Problem 10

Problem 10

The area in square units of the region enclosed by parallelogram $ABCD$ is

$[asy] unitsize(24); pair A,B,C,D; A=(-1,0); B=(0,2); C=(4,2); D=(3,0); draw(A--B--C--D); draw((0,-1)--(0,3)); draw((-2,0)--(6,0)); draw((-.25,2.75)--(0,3)--(.25,2.75)); draw((5.75,.25)--(6,0)--(5.75,-.25)); dot(origin); dot(A); dot(B); dot(C); dot(D); label("$y$",(0,3),N); label("$x$",(6,0),E); label("$(0,0)$",origin,SE); label("$D (3,0)$",D,SE); label("$C (4,2)$",C,NE); label("$A$",A,SW); label("$B$",B,NW); [/asy]$

$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$

Solution

The base is $\overline{BC}=4$ . The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is $4\cdot 2=8\Rightarrow \boxed{\mathrm{B}}$ .

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1991 AJHSME Problems/Problem 10

Problem 10

Solution

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