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 "2003 AMC 8 Problems/Problem 21"
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
The area of trapezoid <math>ABCD</math> is <math>164</math> <math>cm^2</math>. The altitude is 8 cm, <math>AB</math> is 10 cm, <math>CD</math> is 17 cm. What is <math>BC</math>, in centimeters?  
+
The area of trapezoid <math> ABCD</math> is <math>164\text{ cm}^2</math>. The altitude is 8 cm, <math>AB</math> is 10 cm, and <math>CD</math> is 17 cm. What is <math>BC</math>, in centimeters?
 +
 
 +
<asy>/* AMC8 2003 #21 Problem */
 +
size(4inch,2inch);
 +
draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);
 +
draw((11,8)--(11,0), linetype("8 4"));
 +
draw((11,1)--(12,1)--(12,0));
 +
label("$A$", (0,0), SW);
 +
label("$D$", (31,0), SE);
 +
label("$B$", (6,8), NW);
 +
label("$C$", (16,8), NE);
 +
label("10", (3,5), W);
 +
label("8", (11,4), E);
 +
label("17", (22.5,5), E);</asy>
  
  
 
== Solution ==
 
== Solution ==
 
Using the formula for the area of a trapezoid, we have <math>164=8(\frac{BC+AD}{2})</math>. Thus <math>BC+AD=41</math>. Drop perpendiculars from <math>B</math> to <math>AD</math> and from <math>C</math> to <math>AD</math> and let them hit <math>AD</math> at <math>E</math> and <math>F</math> respectively. Note that each of these perpendiculars has length <math>8</math>. From the Pythagorean Theorem, <math>AE=6</math> and <math>DF=15</math> thus <math>AD=BC+21</math>. Substituting back into our original equation we have <math>BC+BC+21=41</math> thus <math>BC=10\Rightarrow \boxed{B}</math>
 
Using the formula for the area of a trapezoid, we have <math>164=8(\frac{BC+AD}{2})</math>. Thus <math>BC+AD=41</math>. Drop perpendiculars from <math>B</math> to <math>AD</math> and from <math>C</math> to <math>AD</math> and let them hit <math>AD</math> at <math>E</math> and <math>F</math> respectively. Note that each of these perpendiculars has length <math>8</math>. From the Pythagorean Theorem, <math>AE=6</math> and <math>DF=15</math> thus <math>AD=BC+21</math>. Substituting back into our original equation we have <math>BC+BC+21=41</math> thus <math>BC=10\Rightarrow \boxed{B}</math>

Revision as of 12:22, 11 March 2012

Problem

The area of trapezoid $ABCD$ is $164\text{ cm}^2$. The altitude is 8 cm, $AB$ is 10 cm, and $CD$ is 17 cm. What is $BC$, in centimeters?

[asy]/* AMC8 2003 #21 Problem */ size(4inch,2inch); draw((0,0)--(31,0)--(16,8)--(6,8)--cycle); draw((11,8)--(11,0), linetype("8 4")); draw((11,1)--(12,1)--(12,0)); label("$A$", (0,0), SW); label("$D$", (31,0), SE); label("$B$", (6,8), NW); label("$C$", (16,8), NE); label("10", (3,5), W); label("8", (11,4), E); label("17", (22.5,5), E);[/asy]


Solution

Using the formula for the area of a trapezoid, we have $164=8(\frac{BC+AD}{2})$. Thus $BC+AD=41$. Drop perpendiculars from $B$ to $AD$ and from $C$ to $AD$ and let them hit $AD$ at $E$ and $F$ respectively. Note that each of these perpendiculars has length $8$. From the Pythagorean Theorem, $AE=6$ and $DF=15$ thus $AD=BC+21$. Substituting back into our original equation we have $BC+BC+21=41$ thus $BC=10\Rightarrow \boxed{B}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_21&oldid=45360"