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> | ||
| − | (note: the ab/2 area method only works because the converse Pythagorean theorem holds for the triple [5, 12, 13]. Therefore, we can find the solution this way because we know the triangle is right ~megaboy6679) | + | (note: the ab/2 area method only works because the converse Pythagorean theorem holds for the triple [<math>5</math>, <math>12</math>, <math>13</math>]. Therefore, we can find the solution this way because we know the triangle is right ~megaboy6679) |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2003|num-b=5|num-a=7}} | {{AMC8 box|year=2003|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 17:06, 24 May 2025
Problem
Given the areas of the three squares in the figure, what is the area of the interior triangle?
![[asy] draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1)); label("$25$",(14.5,1),N); label("$144$",(6,-7.5),N); label("$169$",(3.5,7),N); [/asy]](http://latex.artofproblemsolving.com/3/7/2/372ee1f13836b7ad774c0aa23e598d4f8b07cbb6.png)
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 
(note: the ab/2 area method only works because the converse Pythagorean theorem holds for the triple [, , ]. Therefore, we can find the solution this way because we know the triangle is right ~megaboy6679)
See Also
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
