Chapter 16.4, Problem 3PS

To determine

To find: the perimeter of the quadrilateral $EFGH$ which is inscribe in the circle.

Answer to Problem 3PS

The perimeter is $60.8$ .

Explanation of Solution

Given:

  

Quadrilateral $EFGH$ having $EF=7,GH=20\text{ and EH=25}\text{.}$

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:

Quadrilateral $EFGH$ having $EF=7,GH=20\text{ and EH=25}\text{.}$

Using Ptolemy theorem:

  $\begin{array}{l}15\times 24=7\times 20+25\times x.\\ x=\mathrm{8.8.}\end{array}$

So, the perimeter of the quadrilateral is sum of all side of the quadrilateral which means:

  $EF+FG+GH+EH=7+8.8+20+25=60.8$ .

Hence, the perimeter is $60.8$ .