1993 UNCO Math Contest II Problems/Problem 10

Revision as of 22:44, 23 December 2014 by Pi3point14 (talk | contribs) (Solution)

Problem

The scalene triangle $ABC$ has side lengths $51, 52, 53.$ $AD$ is perpendicular to $BC.$ [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]

(a) Determine the length of $BD.$

(b) Determine the area of triangle $ABC.$


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 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Last Question
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All UNCO Math Contest Problems and Solutions

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