Question

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

• A. $\sqrt{2}$ m
• B. $\sqrt{3}$ m
• C. $\sqrt{5}$ m
• D. $\sqrt{7}$ m

Answer 1

The correct option is C

$\sqrt{5}$ 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