Difference between revisions of "2003 AMC 8 Problems/Problem 21"
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label("8", (11,4), E); | label("8", (11,4), E); | ||
label("17", (22.5,5), E);</asy> | label("17", (22.5,5), E);</asy> | ||
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| + | <math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20</math> | ||
== Solution == | == Solution == | ||
Revision as of 01:07, 28 October 2019
Problem
The area of trapezoid 
is 
. The altitude is 8 cm, 
is 10 cm, and 
is 17 cm. What is 
, 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]](http://latex.artofproblemsolving.com/c/6/6/c66899b4bcfdeca2247f2e6ca1ab430486b95306.png)
Solution
Using the formula for the area of a trapezoid, we have 
. Thus 
. Drop perpendiculars from 
to 
and from 
to and let them hit at 
and 
respectively. Note that each of these perpendiculars has length 
. From the Pythagorean Theorem, 
and 
thus 
. Substituting back into our original equation we have 
thus 
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. 
