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

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==Solution==
 
==Solution==
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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>.

Revision as of 12:19, 9 February 2009

Problem 10

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

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

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

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