Difference between revisions of "2003 AMC 8 Problems/Problem 21"
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label("$A$", (0,0), SW); | label("$A$", (0,0), SW); | ||
label("$D$", (31,0), SE); | label("$D$", (31,0), SE); | ||
− | |||
label("$B$", (6,8), NW); | label("$B$", (6,8), NW); | ||
label("$C$", (16,8), NE); | label("$C$", (16,8), NE); | ||
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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> | ||
+ | |||
+ | <math>\textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20</math> | ||
== Solution == | == Solution == |
Latest revision as of 10:37, 14 June 2024
Problem
The area of trapezoid is
. The altitude is 8 cm,
is 10 cm, and
is 17 cm. What is
, in centimeters?
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.