Dividing a Line Segment into Seven Equal PartsDate: 07/04/2003 at 01:36:51 From: Erik Subject: Dividing a line segment into seven equal parts I need to divide a segment with the length of X into seven equal parts using only a compass and straightedge. I also have to construct a line segment the length squareroot of X. Date: 07/04/2003 at 12:12:45 From: Doctor Jaffee Subject: Re: Dividing a line segment into seven equal parts Hi Erik, Suppose you have a segment AB whose length is X, and you want to divide it into seven equal parts. Draw AB along the bottom of a sheet of paper. Now draw the ray AC, where C is close to the top right corner of the paper. Locate point A1 on the ray about 1 inch from A. It doesn't have to be exact. Then mark off points A2,A3,A4,A5,A6, and A7 on ray AC so that the distance between each pair of points is exactly the same as the distance from A to A1. Connect A7 to B. Then construct a line that goes through A6 and is parallel to BA7. Where this line intersects AB, call the point P. The distance from P to B should be exactly 1/7 of the segment AB. You might also see Paul Kunkel's Geometry Construction Reference: Divide a line segment into n congruent line segments. http://whistleralley.com/construction/c8.htm Now for the second problem. If you have two legs of a right triangle whose lengths are sqrt(x - 1) and 1, the length of the hypotenuse will be sqrt(x). I assume that you can construct a segment whose length is 1. Then you can construct any segment whose length is a whole number. So, suppose you want to construct a segment whose length is sqrt(11). According to my instructions, you would need a right triangle whose legs measure 1 and sqrt(10). You can construct the leg of length 1, but not the leg of length sqrt(10). That is, not unless you construct another right triangle whose legs have lengths 1 and sqrt(9). This you can do since sqrt(9) = 3. Give it a try and if you want to check your solutions with me or if you have difficulties or other questions write back to me and I'll try to help you some more. Good luck, - Doctors Jaffee and Sarah, The Math Forum http://mathforum.org/dr.math/ |
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