2
$\begingroup$

The vertex $B$ is surrounded by $A, C, B_1$

I'm stuck here. I feel like the problem can be solved with analytical geometry by searching from the distance between a point and a plane, but that hasn't been part of our program so I assume it can be done much more simply.

I can calculate the sides of the cuboid from the diagonals, but all my attempts after that have resulted in the wrong solution. I feel like the projection of $B$ on the $B1AC$ plane is an important point on a triangle, but I'm not sure if it has to be and if it is, which one.

  • 0
    So, you have applied Pythagoras' theorem ? Could you give us the values of the sides you have found and where you are blocked. – 2017-01-19
  • 0
    Michael calculated them correctly in his answer. I got the same thing, but couldn't progress further. – 2017-01-19
  • 0
    @John Doe I am ready to help. What is your question? – 2017-01-19
  • 0
    @MichaelRozenberg Took me a bit to get it, thank you :) – 2017-01-19

1 Answers 1

2

Let $h$ be our height, $BA=x$, $BB_1=y$ and $BC=z$.

Hence, $S_{\Delta ACB_1}=\sqrt{12\cdot5\cdot4\cdot3}=12\sqrt5$.

In another hand, $x^2+y^2=49$, $x^2+z^2=64$ and $y^2+z^2=81$, which gives

$x^2+y^2+z^2=97$ and $x=4$, $y=\sqrt{33}$ and $z=4\sqrt3$.

Thus, $\frac{1}{3}h\cdot12\sqrt5=\frac{1}{6}\cdot4\cdot\sqrt{33}\cdot4\sqrt3$, which gives the answer: $$h=2\sqrt{\frac{11}{5}}$$

  • 0
    What is the right hand side in the last equation before the solution? Edit: Rather, why is the volume of the pyramid necessarily 1/6th of the volume of the cuboid? – 2017-01-19
  • 0
    @John Doe it's a volume of the pyramid. – 2017-01-19
  • 0
    Why is the volume of the pyramid necessarily 1/6th of the volume of the cuboid? – 2017-01-19
  • 0
    @John Doe It's just $\frac{AB\cdot S_{\Delta BB_1C}}{3}$ – 2017-01-19