Difference between revisions of "2001 AMC 8 Problems/Problem 11"
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− | ==Problem== | + | == Problem == |
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Points <math>A</math>, <math>B</math>, <math>C</math> and <math>D</math> have these coordinates: <math>A(3,2)</math>, <math>B(3,-2)</math>, <math>C(-3,-2)</math> and <math>D(-3, 0)</math>. The area of quadrilateral <math>ABCD</math> is | Points <math>A</math>, <math>B</math>, <math>C</math> and <math>D</math> have these coordinates: <math>A(3,2)</math>, <math>B(3,-2)</math>, <math>C(-3,-2)</math> and <math>D(-3, 0)</math>. The area of quadrilateral <math>ABCD</math> is | ||
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<math>\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24</math> | <math>\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24</math> | ||
− | + | == Solutions == | |
− | + | === Solution 1 === | |
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− | ==Solution 1== | ||
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<asy> | <asy> | ||
for (int i = -4; i <= 4; ++i) | for (int i = -4; i <= 4; ++i) | ||
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This quadrilateral is a trapezoid, because <math> AB\parallel CD </math> but <math> BC </math> is not parallel to <math> AD </math>. The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are <math> AB </math> and <math> CD </math>, which have lengths <math> 2 </math> and <math> 4 </math>, respectively, so the length of the median is <math> \frac{2+4}{2}=3 </math>. <math> CB </math> is perpendicular to the bases, so it is the height, and has length <math> 6 </math>. Therefore, the area of the trapezoid is <math> (3)(6)=18, \boxed{\text{C}} </math> | This quadrilateral is a trapezoid, because <math> AB\parallel CD </math> but <math> BC </math> is not parallel to <math> AD </math>. The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are <math> AB </math> and <math> CD </math>, which have lengths <math> 2 </math> and <math> 4 </math>, respectively, so the length of the median is <math> \frac{2+4}{2}=3 </math>. <math> CB </math> is perpendicular to the bases, so it is the height, and has length <math> 6 </math>. Therefore, the area of the trapezoid is <math> (3)(6)=18, \boxed{\text{C}} </math> | ||
− | ==Solution 2== | + | === Solution 2 === |
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Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle. Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle. | Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle. Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle. | ||
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<math>A_{trap} = 6 + 12 = 18 \rightarrow \boxed{C}</math> | <math>A_{trap} = 6 + 12 = 18 \rightarrow \boxed{C}</math> | ||
− | ==See Also== | + | == See Also == |
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{{AMC8 box|year=2001|num-b=10|num-a=12}} | {{AMC8 box|year=2001|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:31, 19 October 2020
Problem
Points , , and have these coordinates: , , and . The area of quadrilateral is
Solutions
Solution 1
This quadrilateral is a trapezoid, because but is not parallel to . The area of a trapezoid is the product of its height and its median, where the median is the average of the side lengths of the bases. The two bases are and , which have lengths and , respectively, so the length of the median is . is perpendicular to the bases, so it is the height, and has length . Therefore, the area of the trapezoid is
Solution 2
Using the diagram above, the figure can be divided along the x-axis into two familiar regions that do not overlap: a right triangle and a rectangle. Since the areas do not overlap, the area of the entire trapezoid is the sum of the area of the triangle and the area of the rectangle.
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.