Difference between revisions of "2003 AMC 8 Problems/Problem 21"
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label("10", (3,5), W); | label("10", (3,5), W); | ||
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 == | ||
− | 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= | + | 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 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.