Question

If AB = 2 m, then find the perimeter of the $△$ ADB.

• A. 3 + $\sqrt{5}$ m
• B. 2 + $\sqrt{5}$ m
• C. 4 + $\sqrt{5}$ m
• D. $2\sqrt{5}$ m

Answer 1

The correct option is A

3 + $\sqrt{5}$ m

We have to find the perimeter of the following triangle :

AD and AB are given. We can find BD using Pythagoras theorem.

BD = $\sqrt{A{D}^2+A{B}^2}$

BD = $\sqrt{1^2+2^2}$

= $\sqrt{1+4}$ = $\sqrt{5}$ m

$\therefore$ Perimeter of the $△$ ABD
= AD + AB + BD
= 1 + 2 + $\sqrt{5}$

= 3 + $\sqrt{5}$ m