Question

In a rhombus PQRS, the diagonals intersect at O. If $\angle P={120}^0$ and OP = 4 cm, then the side of the rhombus is equal to

• A. 2 cm
• B. 8 cm
• C. 5 cm
• D. 6 cm

Answer 1

The correct option is B 8 cm


The diagonals of a rhombus bisect each other at $90º$.
Also, the diagonals of a rhombus bisect the vertex angles.
$\therefore \angle OPQ=\angle SPO=\frac{{120}^0}{2}={60}^0$
Now, the opposite angles in a rhombus are equal, therefore, $\angle SRP=60°$.
Now in $\Delta SPR$, by Angle sum property of triangle,
$\angle SPR+\angle SRP+\angle PSR=180°$
$\Rightarrow 60°+60°+\angle PSR=180°$
$\Rightarrow \angle PSR=60°$
Thus, $\Delta PSR$ is an equilateral triangle.
$\therefore PS=SR=PR$
$PR=2OP=2\times 4=8$cm ($\because OP=4$cm)
$\therefore PS=PR=8$cm
Thus, the side of rhombus PQRS is 8 cm.
Hence, the correct answer is option (b).