Chapter 16.4, Problem 4PS

To determine

To find: the distance $AB$ and the ratio of $AB$ to $CD$ from the given quadrilateral $ABCD$ .

Answer to Problem 4PS

  $AB=2.12$ and $\frac{AB}{CD}=\frac{1}{5}$ .

Explanation of Solution

Given:

  

  $AE=2,BE=5,CE=10,DE=4\text{ and BC=7}\text{.5}\text{.}$ .

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:

According to the given:

  $AE=2,BE=5,CE=10,DE=4\text{ and BC=7}\text{.5}\text{.}$

The distance between $AB$ can be calculated as:

  $\begin{array}{l}AC=AE+EC\\ AC=2+10.\\ AC=12.\end{array}$

Similarly,

  $\begin{array}{l}BD=BE+ED.\\ BD=5+4.\\ BD=9.\end{array}$

Using similar triangle laws:

  $\begin{array}{l}\frac{AB}{CD}=\frac{AE}{CE}.\\ \frac{AD}{BC}=\frac{AE}{BE}.\end{array}$

According to the Ptolemy’s theorem:

  $\begin{array}{l}AC\times BD=AD\times BC+AB\times CD.\\ 12\times 9=AD\times 7.5+AB\times CD.\\ 108=\left(BC\times \frac{AE}{BE}\right)\times 7.5+AB\times \left(AB\times \frac{CE}{AE}\right).\end{array}$

By substituting the values of the given value of the side.

  $\begin{array}{l}108=7.5\times \frac{2}{5}\times 7.5+{\left(AB\right)}^2\times \frac{10}{2}.\\ 108-112.5=A{B}^2.\\ AB=\sqrt{4.5}=\mathrm{2.12132.}\end{array}$

The ratio of

  $\begin{array}{l}\frac{AB}{CD}=\frac{AE}{CE}.\\ \Rightarrow \frac{2}{10}=\frac{1}{5}.\end{array}$

Hence, $AB=2.12$ and $\frac{AB}{CD}=\frac{1}{5}$ .