Chapter 9.4, Problem 2PSA

To determine

To find: The length of the diagonal of a square.

Answer to Problem 2PSA

The length of diagonal of a square = 3 $\sqrt{2}$ cm

Explanation of Solution

Given information :

The perimeter of a square = 12 cm

Formula used :

Use Pythagoras theorem.

Diagonals of a square bisect each other at right angles.

Calculation :

There are a square $ABCD$ , whose diagonals bisects each other at O .

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

Perimeter of the square = 12 cm

  $AB=BC=CD=DA=3\text{ cm}$

  

Let $OA=OB=OC=OD=a\text{ units}$

  $\therefore$ In $\Delta AOB$ , by Pythagoras theorem,

  $\begin{array}{l}O{A}^2+O{A}^2=A{B}^2\\ \Rightarrow a^2+a^2=3^2\\ \Rightarrow 2a^2=9\\ \Rightarrow a^2=\frac{9}{2}\\ \Rightarrow a=\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}}=\frac{1}{2}\left(3\sqrt{2}\right)\\ \Rightarrow 2a=3\sqrt{2}\end{array}$

Hence, the length of each diagonal is $3\sqrt{2}$ cm respectively.