Question

What is the perimeter of the given figure ABCD ? ( Take $\sqrt{2}$ = 1.4 and $\sqrt{5}$ = 2.2 )

• A. 5 m
• B. 6 m
• C. 6$\sqrt{2}$ m
• D. 6.8 m

Answer 1

The correct option is B

6 m

ABC is a right - angled isosceles triangle.

The sides are in the ratio $1:1:\sqrt{2}$

Let the sides be $x,x,\sqrt{2}x$

Here, $x=\sqrt{2}$

$\therefore$ The Length of AC = $\sqrt{2}x$

= $\sqrt{2}\times \sqrt{2}$ = 2 m

Hence, using Pythagoras theorem we can say:

AD = $\sqrt{1^2+2^2}$ = $\sqrt{5}$ m

$\therefore$ Perimeter of the figure ABCD
= AB + BC + CD + AD

= $\sqrt{2}+\sqrt{2}+1+\sqrt{5}$

= 1.4 + 1.4 + 1 + 2.2

= 6 m