Difference between revisions of "1991 AJHSME Problems/Problem 10"

m (Solution)
 
Line 24: Line 24:
 
{{AJHSME box|year=1991|num-b=9|num-a=11}}
 
{{AJHSME box|year=1991|num-b=9|num-a=11}}
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:07, 5 July 2013

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 $4\cdot 2=8\Rightarrow \boxed{\mathrm{B}}$.

See Also

1991 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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. AMC logo.png