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# 3 Dimensional Problem

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A "rectangular" parellelpiped has rectangles for all its faces. The sum of the lengths of the four diagonals of a "rectangular" parrallelpeped is 28. The diagonal of one face has length $$2\sqrt{10}$$, while hte diagonal of another face has length $$3\sqrt{5}$$. Find the number of cubic units in the volume of the rectangular parellelpiped. (Mathcounts Competitions)

So this is how I think I set up equations

x+ y= 40

y+ z= 45

$$(40+z^2)=49$$

Lol third equation took me intense thinking to come up with

The text highlighted in yellow I am confused if it is talking about the diagonals of the entire three dimensional figure, because thats how I set up my equation for. Can someone check if the equation I set up is correct or if I misinterpreted the question?

Sep 17, 2019
edited by CalculatorUser  Sep 17, 2019
edited by CalculatorUser  Sep 17, 2019
edited by CalculatorUser  Sep 17, 2019
edited by CalculatorUser  Sep 17, 2019
edited by CalculatorUser  Sep 17, 2019

#1
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$$4\sqrt{x^2+y^2+z^2} = 28\\ x^2 + y^2 + z^2 = 49\\ \text{The other two equations are correct}$$

$$z^2 = 49-40 = 9\\ z=3\\ y^2 = 45-9=36\\ y=6\\ x^2 = 49-36-9 = 4\\ x=2\\ xyz = 36$$

.
Sep 17, 2019

#1
+6192
+1

$$4\sqrt{x^2+y^2+z^2} = 28\\ x^2 + y^2 + z^2 = 49\\ \text{The other two equations are correct}$$

$$z^2 = 49-40 = 9\\ z=3\\ y^2 = 45-9=36\\ y=6\\ x^2 = 49-36-9 = 4\\ x=2\\ xyz = 36$$

Rom Sep 17, 2019
#2
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My last equation is correct too I just substituted in for x^2 + y^2, but thank you though!

CalculatorUser  Sep 19, 2019
edited by CalculatorUser  Sep 19, 2019