Difference between revisions of "2011 AMC 8 Problems/Problem 16"
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We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the [[Pythagorean Theorem]], we have | We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the [[Pythagorean Theorem]], we have | ||
| − | + | <cmath> 15^2 + x^2 =25^2 </cmath> | |
| − | + | <cmath> \x^2 = 25^2 - 15^2</cmath> | |
| − | + | <cmath>x^2 = (25 + 15)(25-15)</cmath> | |
| − | + | <cmath>x^2= 40\cdot 10</cmath> | |
| − | + | <cmath>x^2= 400</cmath> | |
| − | + | <cmath>x = \sqrt{400}</cmath> | |
| − | + | <cmath>x= 20</cmath> | |
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Let the area of a 15-20-25 right triangle be <cmath>x</cmath>. | Let the area of a 15-20-25 right triangle be <cmath>x</cmath>. | ||
| − | + | <cmath>a = 2x</cmath> | |
| − | + | <cmath> b = 2x</cmath> | |
| − | + | <cmath>\textbf{C) }\boxed{a = b} </cmath> | |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=15|num-a=17}} | {{AMC8 box|year=2011|num-b=15|num-a=17}} | ||
Revision as of 09:56, 27 November 2011
Let 
be the area of the triangle with sides of length 
, and 
. Let 
be the area of the triangle with sides of length 
and 
. What is the relationship between and ?
Solution
25-25-30
We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the Pythagorean Theorem, we have
![\[15^2 + x^2 =25^2\]](http://latex.artofproblemsolving.com/0/0/b/00b2e5a4217d732c1e2542e1594940dac0c2190a.png)
\[\x^2 = 25^2 - 15^2\] (Error compiling LaTeX. Unknown error_msg)
![\[x^2 = (25 + 15)(25-15)\]](http://latex.artofproblemsolving.com/c/9/6/c96d946e9ed6cbba439bb205db2d285e76de80b9.png)
![\[x^2= 40\cdot 10\]](http://latex.artofproblemsolving.com/d/a/1/da1bba294ea8f5a3b77b090dc883eb2d26625710.png)
![\[x^2= 400\]](http://latex.artofproblemsolving.com/e/6/e/e6ebbd7ba07f85052c5fa3ffa25547aae332a1a8.png)
![\[x = \sqrt{400}\]](http://latex.artofproblemsolving.com/c/6/5/c6515fac376dc34f6af7fda7938e3c39c3bb87af.png)
![\[x= 20\]](http://latex.artofproblemsolving.com/d/a/6/da648e1ee0029c7c1db961c89f1c6d031fe5d2dc.png)
Thus we have two 15-20-25 right triangles.
25-25-40
We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles.
Let the area of a 15-20-25 right triangle be ![\[x\]](http://latex.artofproblemsolving.com/8/e/7/8e7ba7578864781f5052ab8e190bd855a9a77c82.png)
.
![\[a = 2x\]](http://latex.artofproblemsolving.com/d/2/5/d2522bf4fd5c76a00d3a0b8f6f262c7dd62f86f8.png)
![\[b = 2x\]](http://latex.artofproblemsolving.com/7/8/a/78a5de5761fb75d82891f5c478373023b7accd5d.png)
![\[\textbf{C) }\boxed{a = b}\]](http://latex.artofproblemsolving.com/8/9/3/893b9ed100f4b5d7526e19af1ff134d0b8092647.png)
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
