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- #1

#### ghostfirefox

##### New member

- Nov 13, 2019

- 14

- Thread starter ghostfirefox
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- Thread starter
- #1

- Nov 13, 2019

- 14

- Mar 1, 2012

- 977

let the four corners of the square base with side length $s$ lie in the x y plane with positions

$(0,0,0)$, $(s,0,0)$, $(s,s,0)$, and $(0,s,0)$

let the apex of the pyramid be at position $(a,b,c)$

using the distance formula between two points in space yields the following equations... assume that edges 6016 and 2370 extend from the opposite tops of the base

$a^2+b^2+c^2 = 6016^2$

$(a-s)^2+(b-s)^2+c^2 = 2370^2$

$(a-s)^2+b^2+c^2=4350^2$

$a^2+(b-s)^2+c^2 = d^2$, where $d$ is the length of the fourth edge

use the system of equations to solve for $d$ ... I get a unique integral value for $d$ such that $4350 < d < 6016$

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- #3

- Nov 13, 2019

- 14

I tried to calculate this, but there are too many variables in this system of equations.

- Mar 1, 2012

- 977

Subtract the 2nd equation from the 4th.

Work the solution for $d$ from the resulting two equations.