Question

What is the Perimeter of the given figure ABCD ?

• A. $6\sqrt{3}$ m
• B. 3 + $2\sqrt{3}$ m
• C. 6 + $2\sqrt{3}$ m
• D. $3\sqrt{6}$ m

Answer 1

The correct option is C

6 + $2\sqrt{3}$ m

In the given figure ABC is a right-angled isosceles triangle

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

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

Here x = 2 m

$\therefore$ AC = $\sqrt{2}$x = $2\sqrt{2}$ m

Now, in $△$ ACD,

AC = $2\sqrt{2}$ m and CD = 2 m

Using Pythagoras theorem,

AD = $\sqrt{A{C}^2+C{D}^2}$

= $\sqrt{(2\sqrt{2}{)}^2+2^2}$

= $\sqrt{8+4}$

= $\sqrt{12}$

= $2\sqrt{3}$ m

$\therefore$ Perimeter of the figure ABCD = 2+ 2 + 2 + $2\sqrt{3}$ = 6 + $2\sqrt{3}$ m