Chapter 9.4, Problem 20WE

To determine

To Find : The distance of plane P from center point O of the sphere .

Answer to Problem 20WE

The distance of plane p from center point O of the sphere is $\text{5}\sqrt{5}$ cm.

Explanation of Solution

Given information:

  

In above figure a plane p cuts sphere from the centre point O at the sphere in a circle which has diameter AB=20

Diameter of sphere CD= 30

In sphere diameter CD bisects plane P diameter AB=24 into two equal parts from the midpoint M which means $MA=MB$ and $AB=MA+MB$ .

Or it can be define as

  $MB=\frac{AB}{2}$

If $AB=20$ then;

  $\begin{array}{l}\begin{array}{l}MB=\frac{\text{AB}}{2}\hfill \\ MB=\frac{20}{2}\hfill \end{array}\\ MB=10\end{array}$

And if diameter of sphere CD=30 then radius

  $\begin{array}{l}OC=OD\hfill \\ =OB\hfill \\ =\frac{30}{2}\hfill \\ =15\hfill \end{array}$

To find the measure of let $OM=x$ distance of plane P from center point O of the sphere.

Use Pythagoras theorem:

  $\left(Hypotenuse\right)²=\left(Perpendicular\right)²+\left(Base\right)²$

In triangle ΔOMB.

Here $HypotenuseOB=15,PerpendicularOM=x$ and $BaseMB=10$

  $\begin{array}{l}\left(15\right)²=\left(x\right)²+\left(10\right)²\hfill \\ 225=\left(x\right)²+100\hfill \end{array}$

Subtract 100 to the both sides of the equation

  $225=\left(x\right)²+100$

  $\begin{array}{l}225–100=\left(x\right)²+\cancel{100}-\cancel{100}\hfill \\ 125=\left(x\right)²\hfill \\ \left(x\right)²=125\hfill \end{array}$

Take the square root and simplify as below:

  $\begin{array}{l}x=\sqrt{125}\hfill \\ x=\sqrt{5\times 5\times 5}\hfill \\ x=\text{ 5}\sqrt{5}\hfill \end{array}$

Therefore, measure of distance of plane P from center point O of the sphere is $\text{5}\sqrt{5}$ cm .