Perimeter Equals Area in a TriangleDate: 4/2/96 at 11:19:9 From: Anonymous Subject: Right triangles where area = perimeter When will the area and perimeter of a right triangle be numerically equal? Date: 4/3/96 at 12:54:16 From: Doctor Ken Subject: Re: Right triangles where area = perimeter Hello! Well, let's see. Let's say we know the lengths (x and y) of the two legs of the triangle. Then the area of the triangle will be xy/2, and the perimeter will be x + y + Sqrt{x^2 + y^2}. So we're looking for solutions to the equation x * y ----- = x + y + Sqrt{x^2 + y^2}. 2 Now how are we going to solve that thing? Well, move the x and y over to the left side to get this: x * y ----- - x - y = Sqrt{x^2 + y^2}. 2 Now square both sides: x^2 + 2xy - x^2 y + y^2 - xy^2 + (x^2 y^2)/4 = x^2 + y^2 Cancel: 2xy - x^2 y - xy^2 + (x^2 y^2)/4 = 0 So either x = 0 or y = 0 or we have 2 - x - y + xy/4 = 0 x(y/4 - 1) = y - 2 y - 2 4y - 8 x = ------- = ------ y/4 - 1 y - 4 So I guess that's our answer. Ether x or y is 0 (so we don't have a triangle at all), or x = (4y - 8)/(y - 4). I encourage you to check out The Geometer's Sketchpad to see what intuition it can give you about this problem. You can get a demo version from the page http://mathforum.org/dynamic.html Have fun! -Doctor Ken, The Math Forum |
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