Chapter 16.4, Problem 1PS

To determine

To find: the distance $AC$ from the cyclic quadrilateral $ABCD$ .

Answer to Problem 1PS

  $AC=\frac{2\sqrt{51}}{3}$ .

Explanation of Solution

Given:

  

  $AB=2,BC=3,CD=6,AD=8\text{ and BD=}\frac{18\sqrt{51}}{17}.$

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,

  $AC.BD=AB.CD+AD.BC$ .

Calculation:

Applying the Ptolemy’s theorem:

  $AC.BD=AB.CD+AD.BC\mathrm{...........}\left(i\right)$ .

From the given:

  $AB=2,BC=3,CD=6,AD=8\text{ and BD=}\frac{18\sqrt{51}}{17}.$

By putting the given data in equation $\left(i\right)$ :

  $\begin{array}{l}AC.\left(\frac{18\sqrt{51}}{17}\right)=2\times 6+8\times 3.\\ \frac{18\sqrt{51}}{17}AC=36.\\ AC=\frac{36}{\frac{18\sqrt{51}}{17}}=\frac{34}{\sqrt{51}}.\\ AC=\frac{2\sqrt{51}}{3}.\end{array}$

Hence, $AC=\frac{2\sqrt{51}}{3}$ .