1991 AJHSME Problems/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

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

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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1991 AJHSME Problems/Problem 10

Problem 10

Solution