Difference between revisions of "2016 UNCO Math Contest II Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | <math>\fbox{84}</ | + | Call the lengths of the legs of the triangle a and b. Call the length of the hypotenuse c. Then, by the Pythagorean Theorem, |
| + | <cmath>a^2+b^2=c^2</cmath> | ||
| + | From the problem statement, we also know that | ||
| + | <cmath>a^2+b^2+c^2=1250</cmath> | ||
| + | By substitution, we have | ||
| + | <cmath>2c^2=1250</cmath> | ||
| + | <cmath>c^2=625</cmath> | ||
| + | Since c is positive, <math>c=25</math>. We are also given that | ||
| + | <cmath>a+b+c=56</cmath> | ||
| + | Substituting, we get | ||
| + | <cmath>a+b=31</cmath> | ||
| + | Therefore, | ||
| + | <cmath>(a+b)^2=a^2+2ab+b^2=31^2=961</cmath> | ||
| + | We also know that | ||
| + | <cmath>a^2+b^2+625=1250</cmath> | ||
| + | <cmath>a^2+b^2=625</cmath> | ||
| + | Combining the two equations, we get that | ||
| + | <cmath>(a^2+2ab+b^2)-(a^2+b^2)=961-625</cmath> | ||
| + | <cmath>2ab=336</cmath> | ||
| + | Since the legs of a right triangle are perpendicular, the area of the triangle is <math>\frac{1}{2}ab</math>. | ||
| + | We can divide both sides of our equation by 4 to get this, which yields | ||
| + | <cmath>\frac{1}{2}ab=\fbox{84}</cmath> | ||
== See also == | == See also == | ||
{{UNCO Math Contest box|year=2016|n=II|before=First Question|num-a=2}} | {{UNCO Math Contest box|year=2016|n=II|before=First Question|num-a=2}} | ||
| − | [[Category:Intermediate Geometry]] | + | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 20:26, 16 April 2017
Problem
The sum of the lengths of the three sides of a right triangle is 56. The sum of the squares of the lengths of the three sides of the same right triangle is 1250. What is the area of the triangle?
Solution
Call the lengths of the legs of the triangle a and b. Call the length of the hypotenuse c. Then, by the Pythagorean Theorem,
![\[a^2+b^2=c^2\]](http://latex.artofproblemsolving.com/f/f/4/ff4df74c385f3411d529f28ebfc7f383e34660bc.png)
From the problem statement, we also know that
![\[a^2+b^2+c^2=1250\]](http://latex.artofproblemsolving.com/c/8/9/c895bdf2327775425c22d7b3c37e9613735eb5a0.png)
By substitution, we have
![\[2c^2=1250\]](http://latex.artofproblemsolving.com/d/4/6/d46fdf8b19b2f5e00431ed70c3eea47bdf8be7e1.png)
![\[c^2=625\]](http://latex.artofproblemsolving.com/6/f/b/6fb80762c14c33cfea13781e75455c840ae18f39.png)
Since c is positive, 
. We are also given that
![\[a+b+c=56\]](http://latex.artofproblemsolving.com/b/b/0/bb0cc3b20c0a2152e291cc263757a04bec4ed6c7.png)
Substituting, we get
![\[a+b=31\]](http://latex.artofproblemsolving.com/0/3/0/0309e52d7ace18700fabddc35238f8b259422715.png)
Therefore,
![\[(a+b)^2=a^2+2ab+b^2=31^2=961\]](http://latex.artofproblemsolving.com/c/d/d/cddbe6c5cd698783ce5350d92ecf24d5f7e4f649.png)
We also know that
![\[a^2+b^2+625=1250\]](http://latex.artofproblemsolving.com/a/e/0/ae0ab9e13a8c6b33998e36076894b9385ba109b8.png)
![\[a^2+b^2=625\]](http://latex.artofproblemsolving.com/e/5/5/e55c946061b001156e5bcb772da3058f0489f5fe.png)
Combining the two equations, we get that
![\[(a^2+2ab+b^2)-(a^2+b^2)=961-625\]](http://latex.artofproblemsolving.com/4/5/a/45a79140b3edb7e64a0a2e27735992c68868e5d7.png)
![\[2ab=336\]](http://latex.artofproblemsolving.com/9/b/7/9b7de54be95a8357ef78793d22753d65505f06d2.png)
Since the legs of a right triangle are perpendicular, the area of the triangle is 
.
We can divide both sides of our equation by 4 to get this, which yields
![\[\frac{1}{2}ab=\fbox{84}\]](http://latex.artofproblemsolving.com/d/9/d/d9debe2495dcec7300217bb9d3037a0a09beed50.png)
See also
| 2016 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
