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Difference between revisions of "2001 AMC 8 Problems/Problem 11"
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==Problem==
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== 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>
  
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== Solution #1 ==
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==Solution 1==
 
  
 
<asy>
 
<asy>
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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>
  
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==Solution 2==
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== 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>
  
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==See Also==
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==Solution 3==
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Using the diagram in solution 1 we can use the Pick’s Theorem to solve this problem.There are twelve lattice points inside the quadrilateral and fourteen lattice points on the quadrilateral. Applying the Pick’s Theorem, 12 + <math>\frac{14}{2}</math> - 1 = 18 <math>\rightarrow \boxed{C}</math>
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==Video Solution==
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https://youtu.be/5gldUJaZZCg  Soo, DRMS, NM
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== See Also ==
 
{{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 21:57, 9 January 2025

Problem

Points $A$, $B$, $C$ and $D$ have these coordinates: $A(3,2)$, $B(3,-2)$, $C(-3,-2)$ and $D(-3, 0)$. The area of quadrilateral $ABCD$ is

[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } [/asy]

$\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24$

Solution #1

[asy] for (int i = -4; i <= 4; ++i) { for (int j = -4; j <= 4; ++j) { dot((i,j)); } } draw((0,-4)--(0,4),linewidth(1)); draw((-4,0)--(4,0),linewidth(1)); for (int i = -4; i <= 4; ++i) { draw((i,-1/3)--(i,1/3),linewidth(0.5)); draw((-1/3,i)--(1/3,i),linewidth(0.5)); } { draw((3,2)--(3,-2)--(-3,-2)--(-3,0)--cycle,linewidth(1)); } label("$A$",(3,2),NE); label("$B$",(3,-2),SE); label("$C$",(-3,-2),SW); label("$D$",(-3,0),NW); [/asy]


This quadrilateral is a trapezoid, because $AB\parallel CD$ but $BC$ is not parallel to $AD$. 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 $AB$ and $CD$, which have lengths $2$ and $4$, respectively, so the length of the median is $\frac{2+4}{2}=3$. $CB$ is perpendicular to the bases, so it is the height, and has length $6$. Therefore, the area of the trapezoid is $(3)(6)=18, \boxed{\text{C}}$

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.

$A_{trap} = A_{tri} + A_{rect}$

$A_{trap} = \frac{1}{2}bh + lw$

$A_{trap} = \frac{1}{2}\cdot 6 \cdot 2 + 6\cdot 2$

$A_{trap} = 6 + 12 = 18 \rightarrow \boxed{C}$

Solution 3

Using the diagram in solution 1 we can use the Pick’s Theorem to solve this problem.There are twelve lattice points inside the quadrilateral and fourteen lattice points on the quadrilateral. Applying the Pick’s Theorem, 12 + $\frac{14}{2}$ - 1 = 18 $\rightarrow \boxed{C}$

Video Solution

https://youtu.be/5gldUJaZZCg Soo, DRMS, NM


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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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