Chapter 9.4, Problem 3PSA

To determine

To find: The perimeter of a given rhombus.

Answer to Problem 3PSA

The perimeter of a given rhombus is 40 km.

Explanation of Solution

Given information :

The diagonals of rhombus are 12 km and 16 km respectively.

Formula used :

Use Pythagoras theorem

Diagonals of a rhombus bisect each other at right angles.

Calculation :

Let $ABCD$ be a rhombus.

Its diagonals bisect each other at point $O$ .

  $OA=OC=6\text{ km and }OB=OD=8\text{ km}$

  $\angle AOB=\angle BOC=\angle COD=\angle DOA=90°$

  

On using Pythagoras theorem, in $\Delta AOB$ , we have

  $\begin{array}{l}O{A}^2+O{B}^2=A{B}^2\\ \Rightarrow A{B}^2=6^2+8^2=36+64=100\\ \Rightarrow AB=10\text{ km}\end{array}$

  $\therefore$

  $\text{Perimeter of a rhombus}=4\times \text{Side}$

  $\begin{array}{l}=4\times 10\\ =40\text{ km}\end{array}$