Difference between revisions of "2003 AMC 8 Problems/Problem 6"
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| − | ==Problem | + | ==Problem== |
| − | Given the areas of the three squares in the figure, what is the area of the interior triangle? | + | 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> | ||
<math>\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800</math> | <math>\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800</math> | ||
| Line 7: | Line 14: | ||
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) | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2003|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 22:02, 29 August 2024
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 [5, 12, 13]. 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
