Difference between revisions of "1991 AJHSME Problems/Problem 10"
(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> | ||
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==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
Solution
Note the base AD=BC=4 and the height is the y-value of B and C which is 2 
