Chapter 15, Problem 21RP

To determine

To Justify: The perimeter is greater than the sum of the diagonals of quadrilaterals.

Explanation of Solution

Given:

Any Quadrilaterals

Formula Used:

Triangle-Inequality Theorem: It states that sum of two sides of a triangle is always greater than the third.

  $perimeterofquadrilateral=\text{Sum of all sides}$

Calculation:

  

  $\begin{array}{l}In\Delta AOB\\ AB<OA+OB\left(1\right)\\ In\Delta BOC\\ BC<OB+OC\left(2\right)\\ In\Delta COD\\ CD<OC+OD\left(3\right)\\ In\Delta AOD\\ AD<OA+OD\left(4\right)\\ \text{On adding all inequalities one will get}\\ AB+BC+CD+AD<OA+OB+OB+OC+OC+OD+OA+OD\\ \Rightarrow AB+BC+CD+AD<2\left(OA+OB+OC+OD\right)\\ \Rightarrow AB+BC+CD+AD<2\left(AC+BD\right)\\ \Rightarrow AB+BC+CD+AD>AC+BD\end{array}$

Conclusion:

Hence, Proved.