Difference between revisions of "2003 AMC 8 Problems/Problem 6"
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The sides of the squares are <math> 5, 12 </math> and <math> 13 </math> for the square with area <math> 25, 144 </math> and <math> 169 </math>, respectively. The legs of the interior triangle are <math> 5 </math> and <math> 12 </math>, so the area is <math> \frac{5 \times 12}{2}=\boxed{\mathrm{(B)}\ 30} </math> | The sides of the squares are <math> 5, 12 </math> and <math> 13 </math> for the square with area <math> 25, 144 </math> and <math> 169 </math>, respectively. The legs of the interior triangle are <math> 5 </math> and <math> 12 </math>, so the area is <math> \frac{5 \times 12}{2}=\boxed{\mathrm{(B)}\ 30} </math> | ||
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| + | {{AMC8 box|year=2003|num-b=5|num-a=7}} | ||
Revision as of 08:50, 25 November 2011
Problem
Given the areas of the three squares in the figure, what is the area of the interior triangle? 
Solution
The sides of the squares are 
and 
for the square with area 
and 
, respectively. The legs of the interior triangle are 
and 
, so the area is 
| 2003 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
