Chapter 9.4, Problem 4PSA

To determine

To find: The perimeter of given rectangle $ABCD$ .

Answer to Problem 4PSA

The perimeter of given rectangle is 46 mm.

Explanation of Solution

Given information :

The diagonals and base of the rectangle are 17 mm and 15 mm

Formula used :

Use Pythagoras theorem.

Calculation :

Let $ABCD$ is a rectangle.

Given,

  $\text{Base (}AB)=15\text{ mm},AC=BD=17\text{ mm}$

  

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

  $\begin{array}{l}B{D}^2=A{B}^2+A{D}^2\\ \Rightarrow {17}^2={15}^2+A{D}^2\\ \Rightarrow A{D}^2=289-225=64\\ \Rightarrow AD=8\text{ mm}\end{array}$

  $\begin{array}{l}\because \text{Perimeter of rectangle }ABCD=2(\text{Length+Width})\\ =2(AB+AD)\text{ (Since, }AB=CD\text{ and }AD=BC)\\ =2(15+8)=2\times 23=46\text{ mm}\end{array}$