1993 UNCO Math Contest II Problems/Problem 10
Problem
The scalene triangle 
has side lengths 
is perpendicular to 
![[asy] draw((0,0)--(52,0)--(24,sqrt(3)*26)--cycle); draw((24,0)--(24,sqrt(3)*26)); draw((0,-8)--(52,-8),arrow=Arrow()); draw((52,-8)--(0,-8),arrow=Arrow()); draw((24,3)--(21,3)--(21,0),black); MP("B",(0,0),SW);MP("A",(24,sqrt(3)*26),N);MP("C",(52,0),SE);MP("D",(24,0),S); MP("52",(26,-8),S);MP("53",(38,sqrt(3)*13),NE);MP("51",(12,sqrt(3)*13),NW); [/asy]](http://latex.artofproblemsolving.com/1/f/5/1f591d390f2c7453f1a1db5777dcbe9eed30220a.png)
(a) Determine the length of 
(b) Determine the area of triangle 
Solution
We will solve both parts at once since it is easy to get the two answers from this all-inclusive solution.
Heron's Formula states that in a triangle with sides 
and 
the area is given by ![\[\sqrt{s(s - a)(s - b)(s - c)}.\]](http://latex.artofproblemsolving.com/3/7/7/3777ec93b0beb3be1569c6b6bec9537f88bf6d47.png)
We plug in 
![\begin{align*} s &= \dfrac{51 + 52 + 53}{2} = \dfrac{3 \cdot 52}{2} = 3 \cdot 26 = 78 \\ [ABC] &= \sqrt{78(78 - 51)(78 - 52)(78 - 53)} = \sqrt{78(27)(26)(25)} = 5 \sqrt{(3 \cdot 26)\left(3^3\right)(2 \cdot 13)} \\ &= 5 \cdot 3 \sqrt{3 \cdot 2 \cdot 13 \cdot 3 \cdot 2 \cdot 13} = 15 \sqrt{2^2 \cdot 3^2 \cdot 13^2} = 15 \sqrt{(2 \cdot 3 \cdot 13)}^2 \\ &= \boxed{1170} \end{align*}](http://latex.artofproblemsolving.com/9/5/7/95702f4f1c063eeacf355fd0c8f52baf1570fa13.png)
Since ![$[ABC] = \frac{bh}{2},$](http://latex.artofproblemsolving.com/f/7/7/f7709d83f96dddbe5ad579cd33d307d8a533cda2.png)
we know that 
Solving, we get 
Remembering our 8-15-17 Pythagorean triple, we see that 
See also
| 1993 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
