Question

What is the perimeter of the triangle ACD?

• A. $1+\sqrt{2}+\sqrt{3}$ m
• B. $2(1+\sqrt{2}+\sqrt{3})$ m
• C. $3\sqrt{2}$
• D. $2(\sqrt{2}+\sqrt{3})$ m

Answer 1

The correct option is B

$2(1+\sqrt{2}+\sqrt{3})$ 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 = $2\sqrt{2}$ and CD = 2

Using Pythagoras theorem,

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

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

= $\sqrt{8+4}$

= $\sqrt{12}$

= $2\sqrt{3}$ m

$\therefore$ Perimeter of the triangle ACD = $2\sqrt{2}$ + 2 + $2\sqrt{3}$ = $2(1+\sqrt{2}+\sqrt{3})$ m