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Difference between revisions of "2003 AMC 8 Problems/Problem 21"
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== Problem ==
 
== Problem ==
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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?  
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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);
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draw((11,8)--(11,0), linetype("8 4"));
 +
draw((11,1)--(12,1)--(12,0));
 +
label("$A$", (0,0), SW);
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label("$D$", (31,0), SE);
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label("$B$", (6,8), NW);
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label("$C$", (16,8), NE);
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label("10", (3,5), W);
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label("8", (11,4), E);
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label("17", (22.5,5), E);</asy>
 +
 +
<math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20</math>
  
 
== 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=\boxed{\text{(B)}\ 10}</math>
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2003|num-b=20|num-a=22}}
 +
{{MAA Notice}}

Latest revision as of 09:37, 14 June 2024

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]

$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

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 $BC=\boxed{\text{(B)}\ 10}$

See Also

2003 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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

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