SOLUTION: A triangular sheet of metal has sides 16 cm, 18 cm, and 20 cm long respectively. How far from the vertex must it be cut to the 20 cm side so that the two points will have the same

Algebra ->  Surface-area -> SOLUTION: A triangular sheet of metal has sides 16 cm, 18 cm, and 20 cm long respectively. How far from the vertex must it be cut to the 20 cm side so that the two points will have the same      Log On


   

Question 980471: A triangular sheet of metal has sides 16 cm, 18 cm, and 20 cm long respectively. How far from the vertex must it be cut to the 20 cm side so that the two points will have the same area?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Your wording is not right. You are suggesting to draw a perpendicular segment from the 20 cm side to meet the 18 cm side, and this perpendicular segment must cut the TRIANGLE into two equal areas. One of these pieces will be a triangle and the other piece will be a trapezoid.

An objective is to find the actual altitude of the original triangle, using 20 cm as the base, and you may pick alpha as the angle between the 16 cm and the 20 cm sides. I skip some of the steps here, but cos%28alpha%29=83%2F144, and Law Of Cosines was used.

The Pythagorean trigonometry identify for unit circle lets us do this:
cos%5E2%28alpha%29%2Bsin%5E2%28alpha%29=1
sin%28alpha%29=sqrt%281-cos%5E2%28alpha%29%29
sin%28alpha%29=sqrt%281-%2883%2F144%29%5E2%29
highlight_green%28sin%28alpha%29=sqrt%2813847%2F20736%29%29 not simplifiable; you might want to use scientific or graphing calculator or table of trig values.

If let a=altitude from "vertex" to the base,
then
a=16sin%28alpha%29
highlight_green%28a=16%2Asqrt%2813847%2F20736%29%29, and you use this for finding the area of the original triangle.

If let A be area for the original uncut triangle, then A=%281%2F2%2920%2A%2816%2Asqrt%2813847%2F20736%29%29

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