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
Let BD=x. First, we notice that line segment AD is common to both right triangles ADB and ADC. We can therefore use the Pythagorean Theorem to say that (AD)^2 = 51^2 - x^2 = 53^2 - (52-x)^2. Solving for x, we get x=24.
To find the area, we simply use Heron's Formula to get 1170. (semi-perimeter is 78).
See also
| 1993 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
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| All UNCO Math Contest Problems and Solutions | ||
