Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2001 AMC 8 Problems/Problem 11"

(created solution)
 
(→‎Solution 1)
 
(5 intermediate revisions by 5 users not shown)
Line 1: Line 1:
−
==Problem==
+
== Problem ==
−
 
 
 
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
  
Line 23: Line 22:
 
<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>
  
−
 
+
== Solution #1 ==
−
 
 
−
 
 
−
 
 
−
 
 
−
==Solution==
 
  
 
<asy>
 
<asy>
Line 59: Line 53:
 
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>
  
−
==See Also==
+
== 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.
 +
 
 +
<math>A_{trap} = A_{tri} + A_{rect}</math>
 +
 
 +
<math>A_{trap} = \frac{1}{2}bh + lw</math>
 +
 
 +
<math>A_{trap} = \frac{1}{2}\cdot 6 \cdot 2 + 6\cdot 2</math>
 +
 
 +
<math>A_{trap}  = 6 + 12 = 18 \rightarrow \boxed{C}</math>
 +
 
 +
==Video Solution==
 +
https://youtu.be/5gldUJaZZCg  Soo, DRMS, NM
 +
 
  
 +
== See Also ==
 
{{AMC8 box|year=2001|num-b=10|num-a=12}}
 
{{AMC8 box|year=2001|num-b=10|num-a=12}}
 +
{{MAA Notice}}

Latest revision as of 18:36, 24 December 2023

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.

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

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

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_AMC_8_Problems/Problem_11&oldid=208316"