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The volume of a rectangular parallelepiped is given as 144 cc. Its surface area is given as 192 sq cm. And its corner to corner diagonal is given as 13 cm. How do I find out the three sides.

I have assumed the sides to be $a,b,c$. Now, $a^2+b^2+c^2=169, 2(ab+bc+ca)=192$, and $abc=144$. How do I solve the three equations without forming a cubic equation? (This is a class 9 problem so i am not suppoesd to use solution to cubic equations.)

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Look up Vieta's formulae and the Newton-Girard formulae. To that effect, here's one of the Newton-Girard formulae: $$a^2+b^2+c^2=(a+b+c)^2-2(a b+a c+b c)$$ –  J. M. Aug 25 '12 at 1:17
    
I have used this formula but I am getting a cubic equation. –  SN77 Aug 25 '12 at 1:21
    
You could try using the rational root theorem on $x^3-19x^2+96x-144$, no? –  J. M. Aug 25 '12 at 1:22
    
I am not meant to solve this by cubic equation.I know this can be solved but I wanted to know if I can solve this without using any solution for cubic equation. –  SN77 Aug 25 '12 at 1:26
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Whenever I see a right angle and $13$ together, I think about $5$ and $12$. And when I see a right angle and $5$, I think about $3$ and $4$. Bingo. –  Ross Millikan Aug 25 '12 at 1:43

2 Answers 2

up vote 3 down vote accepted

The quickest way is just to reacall pythagorean triples $(3,4,5)$ and $(5,12,13)$ due to 5 being common to both triples and 13 being the diagonal length thus establishing that 3, 4,and 12 satisfy $3^2+4^2+12^2=13^2$ as desired for the correct diagonal length. Notice that the product of these three numbers in the triples is 144 and they give the correct surface area.

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It seems likely that you are expected to use the "guess and check" procedure.

The product is $144$, not too many integer possibilities for $a$, $b$, $c$. Without loss of generality you may assume $a \ge b\ge c$. Also, the sum of squares condition tells you that $a \le 12$.

Why integers? Because the problem is meant to be solved easily by guess and check.

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So,can there be no method apart from hit and trial. In fact we can do it like that but i was expecting some better procedure to do this. –  SN77 Aug 25 '12 at 1:48
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@SN77: The approach through the cubic is certainly more general. But in the question and comments, using the cubic was explicitly ruled out. –  André Nicolas Aug 25 '12 at 1:52

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