Art of Problem Solving

1991 AJHSME Problems/Problem 10: Difference between revisions

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New page: == 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 <math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 8 } \qquad \m...
 
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== Problem 10 ==
== 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
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>
<math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 15 } \qquad \mathrm{(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>
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>

Revision as of 18:00, 8 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

Note the base AD=BC=4 and the height is the y-value of B and C which is 2 $\Rightarrow \text{Area}=2\cdot4=8 \Rightarrow \boxed{B}$