2003 AMC 8 Problems/Problem 6: Difference between revisions
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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> | ||
{{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? Error creating thumbnail: Unable to save thumbnail to destination
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 | |
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