Question

In the given figure, if ABCD is a square with BC$=x\text{units}$, then $OD:AB$ will be equal to ___.

• A. $1:2$
• B. $1:\sqrt{2}$
• C. $1:1$
• D. $\sqrt{2}:1$

Answer 1

The correct option is B $1:\sqrt{2}$

We have, AB = BC = CD = DA = $xunits$
We know that the diagonals of a square are equal and bisect each other at right angle.
$\therefore DO=AO=BO=CO$

In $△AOD$, right angled at $O$,
$D{A}^2=D{O}^{+}A{O}^2$
[By Pythagorus theorem]
$\Rightarrow x^2=D{O}^{+}D{O}^2$
$\Rightarrow x^2=2D{O}^2$
$\Rightarrow D{O}^2=\frac{x^2}{2}$
$\Rightarrow DO=\sqrt{\frac{x^2}{2}}=\frac{x}{\sqrt{2}}$

Now we can write,
$\frac{OD}{AB}=\frac{\frac{x}{\sqrt{2}}}{x}$
$\Rightarrow \frac{OD}{AB}=\frac{1}{\sqrt{2}}$
$\Rightarrow OD:AB=1:\sqrt{2}$
Hence, $OD:AB$ will be equal to $1:\sqrt{2}$