Chapter 16.4, Problem 2PS

To determine

To find: the distance between $EF$ if point $F$ is located in the exterior of the square.

Answer to Problem 2PS

The distance, between $EF$ is $7\sqrt{2}$ .

Explanation of Solution

Given:

  

  $AF=6$ and $BF=8$ .

Concept used:

Ptolemy’s theorem:

For a cyclic quadrilateral or quadrilateral which inscribe in a circle then the sum of the products of the two pairs opposite side equals the product of the diagonal.

Mathematically,

  $AC.BD=AB.CD+AD.BC$ .

Pythagoras theorem:

  $H^2=P^2+B^2$ .

Calculation:

Using Pythagoras theorem:

  $\begin{array}{l}{\left(AB\right)}^2={\left(AF\right)}^2+{\left(BF\right)}^2.\\ {\left(AB\right)}^2={\left(6\right)}^2+{\left(8\right)}^2.\\ {\left(AB\right)}^2=36+64.\\ AB=10.\end{array}$

Since, square has equal side $BC=AB=BD=AD=10$ .

Again, using Pythagoras theorem $\Delta ABD$ :

  $\begin{array}{l}AC=\sqrt{{10}^2+{10}^2}=10\sqrt{2.}\\ AE=\frac{AC}{2}=\frac{10\sqrt{2}}{2}=5\sqrt{2}.\end{array}$

Similarly, $BE=5\sqrt{2}.$

Using Ptolemy’s theorem.

  $\begin{array}{l}AB\times EF=AF\times BE+BF\times AE.\\ 10\times EF=6\left(5\sqrt{2}\right)+8\left(5\sqrt{2}\right).\\ EF=7\sqrt{2}.\end{array}$

Hence, the distance, between $EF$ is $7\sqrt{2}$ .