Difference between revisions of "2003 AMC 8 Problems/Problem 21"
(Created page with "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>...") |
|||
| Line 1: | Line 1: | ||
| + | == 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 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 B</math> | ||
Revision as of 03:50, 11 March 2012
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 
