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

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== Problem 10 ==
 
== Problem 10 ==
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The area in square units of the region enclosed by parallelogram ABCD, with A(-1,0), B(0,2), C(4,2), D(3,0), is
 
  
−
<math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 15 } \qquad \mathrm{(E) \ 18 }  </math>
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The area in square units of the region enclosed by parallelogram <math>ABCD</math> is
 +
 
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<asy>
 +
unitsize(24);
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pair A,B,C,D;
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A=(-1,0); B=(0,2); C=(4,2); D=(3,0);
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draw(A--B--C--D); draw((0,-1)--(0,3)); draw((-2,0)--(6,0));
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draw((-.25,2.75)--(0,3)--(.25,2.75)); draw((5.75,.25)--(6,0)--(5.75,-.25));
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dot(origin); dot(A); dot(B); dot(C); dot(D); label("$y$",(0,3),N); label("$x$",(6,0),E);
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label("$(0,0)$",origin,SE); label("$D (3,0)$",D,SE); label("$C (4,2)$",C,NE);
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label("$A$",A,SW); label("$B$",B,NW);
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</asy>
 +
 
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<math>\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18</math>
  
 
==Solution==
 
==Solution==
−
Note the base AD=BC=4 and the height is the y-value of B and C which is 2 <math>\Rightarrow \text{Area}=2\cdot4=8 \Rightarrow \boxed{B}</math>
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 +
The base is <math>\overline{BC}=4</math>. The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is <math>4\cdot 2=8\Rightarrow \boxed{\mathrm{B}}</math>.
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==See Also==
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{{AJHSME box|year=1991|num-b=9|num-a=11}}
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[[Category:Introductory Geometry Problems]]
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{{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 (Problems • Answer Key • Resources)
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

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