Chapter 8.2, Problem 39WE

To determine

To find: the $OE$ , the distance from $O$ to the plane of $\Delta VBC$ .

Answer to Problem 39WE

The length of $OE=h=12$

Explanation of Solution

Given Information:

  $O$ is the center of square $ABCD$ and $\overline{VO}$ is perpendicular to the plane of the square.

Given figure,

  

Calculation:

Consider the figure as shown below:

  

Consider the triangle $VOZ$ .

Therefore, by Pythagorean Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.

So here,

  $\begin{array}{c}V{O}^2+O{Z}^2=V{Z}^2\\ {20}^2+{15}^2=V{Z}^2\\ 625=V{Z}^2\\ VZ=25\end{array}$

Suppose, $OE=h,EZ=x$ and $VE=25-x$ .

Consider triangles $VOE$ and $OEZ$ ,

  $\begin{array}{c}O{Z}^2=O{E}^2+E{Z}^2\\ 225=h^2+x^2\mathrm{..........}(1)\end{array}$

Similarly,

  $\begin{array}{c}V{O}^2=O{E}^2+V{E}^2\\ 175=h^2+{\left(25-x\right)}^2\mathrm{.......}(2)\end{array}$

Solve the above equations to get,

  $x=9$ and $h=12$

Hence, the length of $OE=h=12$