Area, Perimeter, and Congruency of Quadrilaterals
Date: 06/23/2003 at 21:32:14 From: Martha Subject: Area, Perimeter, and congruency of quadrilaterals I'm trying to find out how to prove or disprove if it is always true that if two quadrilaterals have the same area and the same perimeter, they will always be congruent. I know how to figure it out for rectangles. If I have a fixed perimeter, I can make a table to show that as one side increases, the area increases until the side becomes 1/4 of the perimeter (and your rectangle becomes a square) - so if a rectangle has the same perimeter AND area, it has to be congruent. I don't know how to go about it for other quadrilaterals, though. In order to find the perimeters of parallelograms and trapezoids with slanty sides, I'm going to be using the Pythagorean theorem a lot and therefore dealing with irrational numbers. It's a lot more complicated than comparing the perimeter and areas of rectangles to see if they are always congruent. I'm not sure where to start on this part. (I'm an 8th grade teacher, by the way.)
Date: 06/23/2003 at 23:23:13 From: Doctor Peterson Subject: Re: Area, Perimeter, and congruency of quadrilaterals Hi, Martha. Interesting question! I'm inclined to doubt that this is true for quadrilaterals in general; it sounds as if it will be barely true for rectangles, and since quadrilaterals have more wiggle, they will probably slip out of our grasp. So let's try making a proof for rectangles, and then a counterexample for quadrilaterals in general. For rectangles with sides a by b and c by d, if the areas and perimeters are both equal, we have ab = cd a+b = c+d Take a and b as given, and we have two equations in two unknowns, from which we should be able to solve for c and d. (That's the basic observation that leads me to expect this to be true.) In fact, ab = c(a+b-c) c^2 - (a+b)c + ab = 0 (c - a)(c - b) = 0 and c is equal to either a or b. (That's a neat bit of work: it's not often that you can factor an equation so neatly, but in this case I expected this answer so it was easy to see. I almost made a fool of myself by using the quadratic formula!) This is what we wanted: c and d are either a and b, or b and a, respectively. Now how about a counterexample for two arbitrary quadrilaterals? I'll try making one a rectangle and the other an isosceles trapezoid, to keep it simple: d +-------------+ +-------+ | | /| \ | |b / |h \ | | / | \ +-------------+ +---+-----------+ a c Our equations now are ab = (c+d)h/2 2(a+b) = c+d + 2 sqrt(h^2+(c-d)^2/4) The mere fact that we now have two equations with three unknowns (c, d, and h), is the "wiggle" I mentioned: there are more variables even under the constraint I've given myself in specifying what kind of quads we have! This will take a lot more work to solve. My claim is that I should be able to specify the value of one variable (say, h) and find at least one pair of c and d that will make both equations true. If it turned out that this would only work if h=b and c=d=a, then I would be proven wrong. I'd better make sure my guess is right. We can first plug the value for c+d from the first equation into the second: 2(a+b) = 2ab/h + 2 sqrt(h^2+(c-d)^2/4) Let's multiply this by h: 2(a+b)h - 2ab = h sqrt(4h^2 + (c-d)^2) Square both sides and do a little more work and 4[(a+b)h - ab]^2 = h^2(4h^2 + (c-d)^2) 4[(a+b)h - ab]^2/h^2 = 4h^2 + (c-d)^2 c-d = 2 sqrt([(a+b)h - ab]^2/h^2 - h^2) Combining this with our equation for c+d, we can solve for each variable; for example, c = [(c+d)+(c-d)]/2 = ab/h + sqrt([(a+b)h - ab]^2/h^2 - h^2) As a check, if h=b(making the trapezoid as high as the rectangle), we get c = a + sqrt(b^2 - b^2) = a as I would expect. --------------------- As a follow-up, I wanted to find a nice specific counterexample, which would have whole numbers for all the sides. I realized that if I took c, d, and h as the givens, I could choose them so they were all whole and then solve for a and b; and if I took h and c-d instead as given, I could then adjust d so that a and b would be whole. I took h=4 and c-d=6 so that I had a 3-4-5 right triangle at each end of the trapezoid; then the area of the trapezoid is 4d+12 and the perimeter is 2d+16. Solving for a and b, I found that d=3 made a good choice. That gives these two quadrilaterals that have the same area (24) and perimeter (22): 3 +---+ +---------+ /| \ | |3 5/ |4 \ | | / | \ +---------+ +---+-------+ 8 9 - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 06/25/2003 at 18:50:11 From: Martha Subject: Area, Perimeter, and congruency of quadrilaterals Dear Dr. Peterson, Thanks so much for your reply - it was very helpful and thanks also for the counterexample - I had found the same one by doing some guessing and testing! One more thing I discovered to help me out by reading another Dr. Math archive is that if you draw and cut out a parallelogram, cut it along the diagonal, reflect one half of it, and put it back together along the diagonal, you have another quadrilateral of equal area and perimeter. Depending upon which diagonal you use, your new quadrilateral may be a kite or a chevron. Another pretty quick counterexample. Thanks again!
Date: 06/25/2003 at 22:41:39 From: Doctor Peterson Subject: Re: Area, Perimeter, and congruency of quadrilaterals Hi, Martha. Yes, that's a nice simple counterexample. And I misread it the first time, giving me another: +--+-----+ +--+--+ / | / / | \ / | / / | \ +-----+--+ +-----+-----+ There you have a parallelogram and a trapezoid with the same area and perimeter. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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