Question : If one diagonal of a rhombus is equal to its side, then the diagonals of the rhombus are in the ratio of:
Option 1: $\sqrt{3}:\sqrt{6}$
Option 2: $\sqrt{2}:1$
Option 3: $\sqrt{2}:\sqrt{3}$
Option 4: $1:\sqrt{3}$
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Correct Answer: $1:\sqrt{3}$
Solution :
Let the side of the rhombus be $a$.
In a rhombus, the diagonals bisect each other at 90°.
Here, PR = $a$, So, OP = $\frac{a}{2}$
Also OQ = $\frac{SQ}{2}$
Now, $\triangle$POQ is a right-angled triangle.
On applying Pythagoras theorem, we get,
$PQ^2=OP^2+OQ^2$
$⇒a^2=(\frac{a}{2})^2+(\frac{SQ}{2})^2$
$⇒(\frac{SQ}{2})^2=\frac{3a^2}{4}$
$⇒\frac{SQ}{2}=\frac{\sqrt3}{2}a$
$\therefore SQ=\sqrt3 a$
So, Second diagonal = $\sqrt3a$
Ratio of first and second diagonal $= \frac{a}{\sqrt{3}a} = \frac{1}{\sqrt3}=1:\sqrt3$
Hence, the correct answer is ${1}:{\sqrt3}$.
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