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estudier
Group Title
All Pythagorean triples where area = perimeter (numerically)?
 one year ago
 one year ago
estudier Group Title
All Pythagorean triples where area = perimeter (numerically)?
 one year ago
 one year ago

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estudier Group TitleBest ResponseYou've already chosen the best response.1
More algebra...:)
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
5,12,13 using trial and error method ^^
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
That's one.....
 one year ago

CheetahBooks1 Group TitleBest ResponseYou've already chosen the best response.0
what do you mean?
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
one more: 6,8,10
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
That's two....
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
ok, general form tripel pythagorean number is : a=x b=(x^21)/2 c=(x^2+1)/2 with a<b<c (a,b,c are sides of a righttriangle) perimeter = area x + (x^21)/2 + (x^2+1)/2 = x(x^21)/2, simplify it we get : x^3  4x^2  5x =0 x(x^2  4x  5) = 0 x(x5)(x+1)=0 x=5 satifisfy it (so, a=5, b=12, c=13) with same way, for a=2x b=x^21 c=x^2+1 perimeter = area 2x+x^21+x^2+1 = 2x(x^21)/2, simplify it we get : x^3  2x^2  3x = 0 x(x^2  2x  3) =0 x(x3)(x+1)=0 x=3 satifisfy it (so, a=6, b=8, c=10) i have done for a=kx b=k(x^21) c=k(x^2+1) with k=3,4,5,6... there are x satisfy it but the result just reflication of on... and its rest not satisfy for x which make for a,b,c integer number
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
so, just 2 triple only...
 one year ago

CheetahBooks1 Group TitleBest ResponseYou've already chosen the best response.0
3,4,5 is a triple, too!
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
but, its area not equal its perimeter...
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Yes, good enough:)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
you can have an easier algebra by eliminating z from x^2+y^2=z^2 and xy/2 = x + y +z leads to xy 4x 4y +8 = 0 > (x4)(y4) = 8 etc.
 one year ago

RadEn Group TitleBest ResponseYou've already chosen the best response.1
just combine of (x4)(y4) = 8, and we will get 2 triple only are : (x,y,z) = 5,12,13 and 6,8,10 but yours easier than me :)
 one year ago
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