Chapter 6.4, Problem 20WE

To determine

To discover, state, and prove a theorem that compares the perimeter of a quadrilateral with the sum of the lengths of the diagonals.

Explanation of Solution

Consider the theorem,

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

  $\begin{array}{l}AB+BC>AC\\ AB+AC>BC\\ AC+BC>AB\end{array}$

  

Consider the following figure,

  

Apply above theorem,

In $\Delta ABC$ ,

  $AB+BC>AC\mathrm{........}(1)$

Similarly, in $\Delta ABD$ ,

  $BA+AD>BD\mathrm{........}(2)$

Perimeter of a quadrilateral is $AB+BC+CD+DA$

Sum of the lengths of the diagonals $=AC+BD$

Add equation (1) and (2) to get,

  $AB+BC+BA+AD>\left(AC+BD\right)$

Therefore, the perimeter of a quadrilateral is always greater than sum of the lengths of the diagonal.