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
To find BD, we need to first find the altitude. If we have the angles of the triangle, we can use law of sines to find the length of the altitude. We know that the total degrees in a triangle is 180, so we can use law of sines to set up a proportion. The sum of the side lengths is 156. Diving 180 by 156 gives us each 15/13 for each side unit. Multiplying by 53, we get angle B is 795/13 degrees. Using law of sines gives us AD is about 44.67. Using pythagorean theorem, we find that BD is about 24.6.
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 | ||
