Constructing a Line to Divide Area of a Triangle in Half
Date: 05/13/98 at 01:53:02 From: Nathan Rabiroff Subject: Similar Triangles Here is the problem I was asked to solve: C /\ / \ / \ / \ / \ /__________\ A B Can you cut triangle ABC into two pieces of equal area by drawing a line parallel to one of the sides? How would you find it? I know there is a way to divide it, but I can't figure out how to do it. I have spent hours trying to figure it out, and desperately need help. I already stumped my math tutor with it, and need someone else to look at it. Thanks, Nate
Date: 05/13/98 at 11:58:36 From: Doctor Rob Subject: Re: Similar Triangles Drop a perpendicular from C to AB, meeting AB at D. On that perpendicular, mark a point P such that PC/PD = sqrt(2)/2. Construct a line through P parallel to AB (and perpendicular to PD), and that will do it. (Why? You supply the reason. It has to do with similar triangles having proportional sides, and with the formula for the area of a triangle as b*h/2.) Here is another construction of the same line. On side AC, mark point A' such that A'C/AC = sqrt(2)/2. On side BC, mark point B' such that B'C/BC = sqrt(2)/2. Connect A' and B', and this line is the right one. If you need to know how to construct P (or A', or B') using straightedge and compass, here is how. Draw the perpendicular bisector of the line segment CD (or AC, or BC). Using the midpoint of the segment as a center, draw a circle with the segment as a diameter. Draw a circular arc whose center is C and whose radius is the distance from C to the intersection of the circle and the perpendicular bisector. Its intersection with the segment will be P (or A', or B'). -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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