Question

If AB = 6 m, then the perimeter of the $△$ ADB is .

• A. $8+\sqrt{37}$ m
• B. $7+\sqrt{37}$ m
• C. $7+\sqrt{39}$ m

Answer 1

The correct option is B $7+\sqrt{37}$ m

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

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

$BD=\sqrt{1^2+6^2}$

= $\sqrt{1+36}$ = $\sqrt{37}$ m

$\therefore$ Perimeter of the $△ABD$
$=AD+AB+BD=1+6+\sqrt{37}$

$=7+\sqrt{37}$ m