Chapter 16.4, Problem 5PS

To determine

To find: the possible measurement of the third side of a triangle.

Answer to Problem 5PS

  $x=4.51$ .

Explanation of Solution

Given:

Triangle is inscribed in a circle of radius $5$ .

One side of the triangle is $5\text{cm}$ and that of second triangle’s side is $6\text{cm}$ .

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:

Triangle is inscribed in a circle of radius $5$ .

One side of the triangle is $5\text{cm}$ and that of second triangle’s side is $6\text{cm}$ .

According to the Ptolemy’s theorem:

If the radius is $5\text{cm}$ diameter will be $\text{10cm}$ .

Let the other third side of the triangle be $x$ .

  $\begin{array}{l}10\times 10=\left(5+x\right)\left(6+x\right).\\ 100=30+5x+6x+x^2.\\ x^2+11x-70=0.\end{array}$

Using quadratic formula:

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.\\ x=\frac{-11\pm \sqrt{{\left(11\right)}^2-4\left(1\right)\left(-70\right)}}{2\left(1\right)}.\\ x=\frac{-11\pm \sqrt{401}}{2}.\\ x_1=4.51\text{ and }{x}_2=-\mathrm{15.51.}\end{array}$

Since. The measure of a side cannot be negative or distance cannot be negative take positive value of $x$ .

  $x=4.51$ .